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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →
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1IBMI-033-03-r2-02Normal distribution and chi-squared testsPaper 1Easy5 marks
A machine fills bottles of sparkling water. The volume, V ml, in a bottle is normally distributed with mean 500 ml and unknown standard deviation σ ml. Quality checks show that 3% of bottles contain less than 490 ml.
(a)
Find the value of σ.
(3)
(b)
The machine is adjusted so that its mean changes but σ does not. After the adjustment only 1% of bottles contain less than 490 ml. Find the new mean volume.
(2)
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Part (a)
Inverse normal of 0.03 gives z = −1.8808 (A1)
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Setting up (490 − 500)/σ = −1.8808 (M1)
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σ = 5.32 ml (5.3169…) (A1)
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Part (b)
Using z = −2.3263 with (490 − μ)/5.3169 = −2.3263 (M1)
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μ = 502 ml (502.37…) (A1)
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Common error: Use the unrounded value of σ from part (a) in part (b); using 5.3 instead of 5.3169 shifts the new mean noticeably.
2IBMI-033-04-r2-03T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
Each working day Priya travels to work either by bicycle (C) or by bus (B). If she cycles one day, the probability that she cycles the next day is 0.7. If she takes the bus one day, the probability that she cycles the next day is 0.4.
(a)
Write down the transition matrix T for this Markov chain, with columns representing today's state (C, B) and rows representing tomorrow's state (C, B).
(2)
(b)
Priya cycles on Monday. Find the probability that she cycles on Thursday of the same week.
(2)
(c)
Find the long-term proportion of working days on which Priya cycles.
Use of \(\mathbf{T}^3\begin{pmatrix}1\\0\end{pmatrix}\)
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0.583
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Part (c)
Steady state: solve \(\mathbf{T}\mathbf{s}=\mathbf{s}\), e.g. \(p=0.7p+0.4(1-p)\)
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\(p=\frac{4}{7}\) = 0.571
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Exam tip: From Monday to Thursday is three transitions, so use T³. Check that each column of your matrix sums to 1.
3IBMI-033-04-r1-02T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A machine fills bags of rice. The label states that each bag contains 500 g. The masses of the bags filled are normally distributed with a known standard deviation of 4.8 g. A consumer group weighs a random sample of 40 bags and finds a sample mean of 502.3 g.
(a)
Find a 95% confidence interval for the mean mass of rice in a bag. Give the bounds to 1 decimal place.
(3)
(b)
The consumer group suspects that the machine is putting more than 500 g of rice in the bags on average. Comment on this suspicion, referring to your interval.
(2)
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Part (a)
M1: uses \(502.3 \pm 1.960 \times \frac{4.8}{\sqrt{40}}\) or a GDC z-interval
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A1: lower bound 500.8 g
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A1: upper bound 503.8 g
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Part (b)
R1: 500 g lies below the whole interval / is not contained in the interval
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A1: so the data support the suspicion that the mean mass is greater than 500 g
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Common error: Here the population standard deviation is known, so a z-interval is used, not a t-interval. Using 4.8 without dividing by \(\sqrt{40}\) gives an interval far too wide.
4IBMI3H-4-08-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A helpline receives calls at random at an average rate of 3.2 calls per 10-minute period. Let X be the number of calls in a 10-minute period.
(a)
Find the probability that exactly 4 calls are received in a 10-minute period.
(2)
(b)
Find the probability that at least 2 calls are received in a 10-minute period.
(2)
(c)
Find the probability that at most 6 calls are received in a 25-minute period.
(2)
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Part (a)
M1 \(X\sim\text{Po}(3.2)\), \(P(X=4)\)
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A1 0.178
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Part (b)
M1 \(1-P(X\le1)\)
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A1 0.829
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Part (c)
M1 new mean \(3.2\times2.5=8\)
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A1 0.313
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Common error: When the time interval changes, the Poisson mean must be scaled to match it before any probability is calculated.
5IBMI3H-4-08-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy8 marks
A bus journey has two stages. The time for stage 1, A minutes, is normally distributed with mean 12 and standard deviation 1.5. The time for stage 2, B minutes, is normally distributed with mean 8 and standard deviation 2. The two times are independent. Let T = A + B.
(a)
Find E(T) and Var(T).
(3)
(b)
Find the probability that the total journey time exceeds 24 minutes.
(2)
(c)
State why the variances can be added in part (a).
(1)
(d)
Find the journey time that is not exceeded on 95% of journeys.
(2)
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Part (a)
A1 E(T) = 20
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M1 \(\text{Var}(T)=1.5^2+2^2\)
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A1 Var(T) = 6.25
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Part (b)
M1 \(T\sim N(20,6.25)\) and \(P(T>24)\)
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A1 0.0548
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Part (c)
R1 the two stage times are independent
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Part (d)
M1 \(z=1.645\) with standard deviation 2.5
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A1 24.1 minutes
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Common error: Variances add for a sum of independent variables; standard deviations do not. Adding 1.5 and 2 to get 3.5 is a frequent slip.
6IBMI3H-4-08b-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
During a meteor shower, meteors are seen by an observer at random at an average rate of 6 per hour. Assume the number of meteors seen in any period follows a Poisson distribution.
(a)
Let X be the number of meteors seen in 30 minutes. State the distribution of X and find P(X = 2).
(3)
(b)
Find the probability that at least one meteor is seen in a 20-minute period.
(3)
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Part (a)
A1 X ~ Po(3)
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M1 attempt to use Poisson probability e−3 32 / 2!
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A1 0.224
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Part (b)
A1 mean for 20 minutes is 2
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M1 1 − P(0) = 1 − e−2
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A1 0.865
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Exam tip: Scale the Poisson mean to match the length of the time period before calculating anything.
7IBMI3H-4-08b-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The peak running speeds of 16 randomly chosen players in a youth football league are measured. The sample mean is 31.4 km h−1 and the sample standard deviation is 1.8 km h−1. The speeds may be assumed to be normally distributed.
(a)
Explain why a t-interval rather than a z-interval is used.
(1)
(b)
Find a 95% confidence interval for the population mean peak speed.
(3)
(c)
A coach claims that the population mean is 33 km h−1. State, with a reason, whether the interval supports the claim.
(2)
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Part (a)
R1 the population standard deviation is unknown and estimated from a small sample
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Part (b)
M1 t critical value with 15 degrees of freedom = 2.131
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A1 half-width 0.959
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A1 (30.4, 32.4) km h−1
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Part (c)
M1 compares 33 with the interval (30.4, 32.4)
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R1 33 lies outside the interval, so the claim is not supported at the 95% level
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Exam tip: The degrees of freedom for a one-sample t-interval are n − 1, and the half-width is the critical value × s/√n.
8IBMI3H-4-08b-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy8 marks
Each day Noor travels to the sports centre either by train or by bus. If she takes the train one day, the probability she takes the train the next day is 0.7. If she takes the bus one day, the probability she takes the train the next day is 0.4. She takes the train on day 1.
(a)
Write down the transition matrix T, with rows for today's mode and columns for tomorrow's, in the order train, bus.
(2)
(b)
Find the probability that Noor takes the train on day 3.
(3)
(c)
Find the long-term proportion of days on which Noor takes the train.
(3)
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Part (a)
A1 first row 0.7, 0.3
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A1 second row 0.4, 0.6
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Part (b)
M1 two-step transitions from the train state
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A1 0.7 × 0.7 + 0.3 × 0.4
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A1 0.61
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Part (c)
M1 steady-state equations πT = π with π1 + π2 = 1
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M1 0.3π1 = 0.4π2
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A1 π1 = 4/7 = 0.571
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Exam tip: Every row of a transition matrix (rows = from) must sum to 1. Use this to check your matrix.
9IBMI3H-4-08b-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Easy8 marks
The area N cm2 covered by algae in a pond sample, t days after it was first measured, is modelled by N = 120 × 1.18t. For this model R2 = 0.994.
(a)
Interpret the value 1.18 in the context of the model.
(2)
(b)
Estimate the area covered after 10 days.
(2)
(c)
Interpret the value R2 = 0.994.
(1)
(d)
Find the time for the area to reach 1000 cm2.
(3)
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Part (a)
A1 each day the area is multiplied by 1.18
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R1 an increase of 18% per day
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Part (b)
M1 120 × 1.1810
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A1 628 cm2
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Part (c)
R1 about 99.4% of the variation in the data is explained by the model, so the fit is very good
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Part (d)
M1 120 × 1.18t = 1000
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M1 t = ln(1000/120)/ln 1.18
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A1 12.8 days
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Exam tip: To solve for an exponent, isolate the power first, then take logarithms of both sides.
10IBMI3H-4-04r7-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
Swarms leave an apiary at random times during May. The number of swarms per week is modelled by a Poisson distribution with mean 2.4.
(a)
State one condition that must hold for a Poisson model to be suitable for the number of swarms.
(1)
(b)
Find the probability that exactly 3 swarms leave in a given week.
(2)
(c)
Find the probability that at least 2 swarms leave in a given week.
(2)
(d)
Find the probability that no swarm leaves during a given two-week period.
(2)
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Part (a)
R1 swarms occur independently of one another (or at a constant average rate, or singly)
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Part (b)
M1 \(X\sim\mathrm{Po}(2.4)\) and \(P(X=3)\)
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A1 0.209
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Part (c)
M1 \(1-P(X\le1)\)
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A1 0.692
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Part (d)
M1 new mean \(2\times2.4=4.8\) for two weeks
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A1 0.00823
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Common error: When the time interval changes, scale the mean to match it; do not reuse the mean for one week over a two-week period.
11IBMI3H-4-04r7-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A beekeeper fills jars with honey. The mass of honey in a jar is normally distributed with mean 454 g and standard deviation 5 g. The mass of an empty jar is normally distributed with mean 310 g and standard deviation 4 g. The two masses are independent. Let \(M\) be the mass of a full jar.
(a)
Find \(E(M)\) and \(\mathrm{Var}(M)\).
(2)
(b)
Find the probability that a full jar has mass greater than 775 g.
(2)
(c)
Six full jars, chosen at random, are packed in a box. Find the mean and standard deviation of the total mass of the six jars.
(3)
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Part (a)
A1 \(E(M)=764\) g
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A1 \(\mathrm{Var}(M)=25+16=41\) g2
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Part (b)
M1 \(M\sim N(764,41)\) with standard deviation \(\sqrt{41}=6.40\)
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A1 0.0429
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Part (c)
A1 mean \(=6\times764=4584\) g
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M1 variance \(=6\times41=246\)
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A1 standard deviation \(=15.7\) g
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Exam tip: The total of six independent jars has variance \(6\sigma^2\), not \(36\sigma^2\); the factor squares only for a single quantity multiplied by 6.
12IBMI3H-4-04r7-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Each Grand Prix weekend race day is classed as Dry (D) or Wet (W) at a certain circuit. Weather on successive race days is modelled by a Markov chain with \(\mathbf{s}_{n+1}=\mathbf{T}\mathbf{s}_n\), in which each column of \(\mathbf{T}\) gives the probabilities of tomorrow's weather. In the order D, W: \(\mathbf{T}=\begin{pmatrix}0.7&a\\0.3&0.6\end{pmatrix}\).
(a)
Find the value of \(a\).
(1)
(b)
Race 1 is Dry. Find the probability that race 3 is Dry.
(2)
(c)
Find the long-term proportion of race days that are Dry, correct to 3 significant figures.
(3)
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Part (a)
A1 \(a=0.4\) (columns sum to 1)
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Part (b)
M1 attempt to find \(\mathbf{T}^2\) or add the two paths, \(0.7\times0.7+0.4\times0.3\)
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A1 0.61
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Part (c)
M1 solve \(\mathbf{T}\mathbf{s}=\mathbf{s}\) or \(0.3d=0.4(1-d)\)
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A1 \(d=\frac47\)
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A1 0.571
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Exam tip: Each column of a transition matrix must sum to 1, which is a quick check when one entry is unknown.
13IBMI3H-4-04r7-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The waiting times, in minutes, of a random sample of 20 patients at a clinic have mean 41.6 and a sample standard deviation (dividing by \(n\)) of 7.8. The waiting times are assumed to be normally distributed.
(a)
Find an unbiased estimate of the population variance.
(2)
(b)
Find a 90% confidence interval for the mean waiting time. Give the upper limit to 3 significant figures.
(3)
(c)
State, with a reason, whether a 99% confidence interval would be wider or narrower than your interval.
(1)
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Part (a)
M1 \(\frac{20}{19}\times7.8^2\)
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A1 64.0
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Part (b)
M1 use of \(t\)-distribution with 19 degrees of freedom, \(t=1.729\)
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A1 lower limit 38.5
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A1 upper limit 44.7
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Part (c)
R1 wider, since a higher confidence level needs a larger critical value
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Common error: The standard deviation on many calculators can be the sample value (\(\div n\)) rather than the unbiased one (\(\div (n-1)\)); check which you are using before building an interval.
14IBMI3H-4-03r9-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Explosive eruptions at Mount Thessa occur independently and at random, at a constant mean rate of 2.4 per year. The number of eruptions in a given period is modelled by a Poisson distribution. Give all probabilities to 3 significant figures.
(a)
Find the probability that exactly 3 eruptions occur in a given year.
(2)
(b)
Find the probability that at least 2 eruptions occur in a given year.
(2)
(c)
Find the probability that no eruption occurs during a given six-month period.
(2)
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Part (a)
M1 Let X ~ Po(2.4) for one year; attempt to use the Poisson probability
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A1 P(X = 3) = 0.209
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Part (b)
M1 P(X ≥ 2) = 1 − P(X ≤ 1)
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A1 = 0.692
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Part (c)
A1 Mean for six months = 1.2
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A1 P(Y = 0) = e−1.2 = 0.301
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Exam tip: When the time interval changes, scale the Poisson mean in proportion before calculating anything.
15IBMI3H-4-03r9-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A volcanologist records the mass of sulfur dioxide released by the Orlo volcano, in tonnes, on 12 randomly chosen days: 412, 389, 455, 431, 398, 467, 420, 441, 405, 436, 473, 394. The daily mass may be assumed to be normally distributed.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation.
(2)
(b)
Find a 95% confidence interval for the mean daily mass of sulfur dioxide, giving the limits to 1 decimal place.
(3)
(c)
A colleague claims that the true mean is 450 tonnes. State, with a reason, whether the interval supports this claim.
(1)
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Part (a)
A1 Sample mean = 427 tonnes (exactly 426.75)
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A1 sn−1 = 28.5 tonnes
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Part (b)
M1 Use of the t-distribution with 11 degrees of freedom, critical value 2.201
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M1 Attempt to use x̄ ± t × s/√12
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A1 (408.7, 444.8)
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Part (c)
R1 450 lies outside the interval, so the data do not support the claim
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Exam tip: With an unknown population variance and a small sample, use the t-distribution, not the normal distribution, for the interval.
16IBMI3H-4-03r9-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A theatre sells season subscriptions. At the end of each season, a fraction a of subscribers do not renew, and 15% of non-subscribers take out a subscription. Each person is either a subscriber or a non-subscriber, and the changes are modelled by a Markov chain with transition matrix T, with the first row and column representing subscribers and the second non-subscribers; columns give the probabilities of moving from each state.
(a)
Write down T in terms of a.
(2)
(b)
In the long term, 75% of people are subscribers. Find the value of a.
(3)
(c)
With a = 0.05, 40% of people are subscribers this season. Find the proportion of subscribers next season.
(2)
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Part (a)
A1 First column (1 − a, a)
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A1 Second column (0.15, 0.85)
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Part (b)
M1 Invariant state (0.75, 0.25) satisfies Ts = s
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M1 0.75(1 − a) + 0.15 × 0.25 = 0.75
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A1 a = 0.05
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Part (c)
M1 Proportion = 0.95 × 0.4 + 0.15 × 0.6
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A1 = 0.47
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Exam tip: Write the matrix columns first, as each column describes where one starting state goes and must sum to 1.
17IBMI3H-4-03r12-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
At an archaeological dig, the number of bronze coins found by a team in one day is modelled by a Poisson distribution with mean 3.2. Daily counts are independent. Give answers to 3 significant figures.
(a)
Find the probability that exactly 4 coins are found on a given day.
(2)
(b)
Find the probability that at least 2 coins are found on a given day.
(3)
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Part (a)
M1 use of Poisson probability, \(e^{-3.2}\times\frac{3.2^4}{4!}\)
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A1 0.178
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Part (b)
M1 recognises \(P(X\geq 2)=1-P(X\leq 1)\)
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M1 \(P(X=0)+P(X=1)=e^{-3.2}(1+3.2)\) (or use of cumulative Poisson function)
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A1 0.829
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Exam tip: "At least 2" means 2 or more, so subtract P(0) and P(1) from 1; subtracting only P(0) is a common slip.
18IBMI3H-4-03r12-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
In a cycling time trial, a rider's time for stage A, A minutes, has mean 42 and standard deviation 3.5. Her time for stage B, B minutes, has mean 55 and standard deviation 4.2. The two stage times are independent. Let \(T=A+B\) be her total time.
(a)
Find the mean of T.
(1)
(b)
Find the standard deviation of T, giving your answer to 3 significant figures.
(3)
(c)
Let \(D=A-B\). Find \(\text{Var}(D)\).
(2)
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Part (a)
A1 \(E(T)=42+55=97\) minutes
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Part (b)
M1 \(\text{Var}(T)=3.5^2+4.2^2\) (variances of independent variables add)
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A1 \(\text{Var}(T)=29.89\)
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A1 \(\sigma=5.47\) minutes
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Part (c)
M1 \(\text{Var}(A-B)=\text{Var}(A)+\text{Var}(B)\) (no subtraction of variances)
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A1 \(12.25+17.64=29.89\) minutes2
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Common error: Variances add for both \(X+Y\) and \(X-Y\) when the variables are independent; it is the standard deviations that never add directly.
19IBMI3H-4-03r12-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
Each night the stage lighting rig in a theatre is either working (W) or faulty (F). If it is working one night, the probability that it is working the next night is 0.9. If it is faulty one night, the probability that it is repaired and working the next night is 0.6.
(a)
Write down the transition matrix T, using the order W, F for both rows and columns, with each column giving the probabilities of moving from a state.
(2)
(b)
The rig is working on Monday night. Find the probability that it is working on Wednesday night.
(2)
(c)
Find the long-term proportion of nights on which the rig is working. Give your answer to 3 significant figures.
(3)
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Part (a)
A1 first column \(0.9,\,0.1\)
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A1 second column \(0.6,\,0.4\), i.e. \(T=\begin{pmatrix}0.9&0.6\\0.1&0.4\end{pmatrix}\)
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Part (b)
M1 two steps: \(0.9\times0.9+0.1\times0.6\) (or entry of \(T^2\))
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A1 0.87
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Part (c)
M1 steady state satisfies \(T\mathbf{s}=\mathbf{s}\) with \(w+f=1\)
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M1 \(0.1w=0.6f\) so \(w=6f\) (or equivalent)
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A1 \(w=\frac{6}{7}=0.857\)
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Exam tip: In a steady state the flow out of a state equals the flow in; use this together with the fact that the proportions sum to 1.
20IBMI3H-4-03r18-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A mountain rescue team receives call-outs at random, independently, at a constant average rate of 3.2 per day during the summer season. Let X be the number of call-outs received on a randomly chosen summer day.
(a)
Find the probability that exactly 5 call-outs are received on a given day.
(2)
(b)
Find the probability that at least 2 call-outs are received on a given day.
(3)
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Part (a)
Recognises \(X \sim \mathrm{Po}(3.2)\) and uses \(P(X=5)=\frac{e^{-3.2}3.2^5}{5!}\) — M1
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\(P(X=5)=0.114\) (3 s.f.) — A1
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Part (b)
M1 uses \(P(X\ge 2)=1-P(X\le 1)\)
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M1 \(P(X\le1)=e^{-3.2}(1+3.2)=0.1712\)
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A1 \(P(X\ge2)=0.829\) (3 s.f.)
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Exam tip: "At least 2" is the complement of "0 or 1", so only two terms need to be calculated. Using the cumulative function on a calculator is equally acceptable.
21IBMI3H-4-03r18-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A school canteen offers two lunch options each day: vegetarian (V) or standard (S). Each pupil chooses one option per day. Pupils' choices follow a Markov chain with transition matrix T, in which each column gives the probabilities of the next day's choice, as shown: \(T=\begin{pmatrix}0.92&0.15\\0.08&0.85\end{pmatrix}\) where the first row and column correspond to V and the second to S. On day 0, 40% of pupils choose V and 60% choose S.
(a)
Find the proportion of pupils who choose V on day 2.
(3)
(b)
Find the long-term proportion of pupils who choose V.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathbf{s}_0=\begin{pmatrix}0.4\\0.6\end{pmatrix}\) and \(\mathbf{s}_2=T^2\mathbf{s}_0\)
✓ 1
A1 \(\mathbf{s}_1=(0.458,\,0.542)^T\) or \(T^2\) evaluated
✓ 1
A1 proportion choosing V on day 2 is \(0.503\) (3 s.f.)
✓ 1
Part (b)
M1 solves \(T\mathbf{s}=\mathbf{s}\) with components summing to 1, e.g. \(0.08v=0.15s\), \(v+s=1\)
✓ 1
A1 \(v=\frac{15}{23}=0.652\) (3 s.f.)
✓ 1
Exam tip: State which convention your matrix follows: here columns sum to 1, so the state vector is multiplied on the right of \(T\).
22IBMI3H-4-03r18-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A food scientist compares the protein content, in g per 100 g, of loaves made from two flours. Six loaves are made with flour A and seven with flour B. The results are: Flour A: 11.8, 12.4, 12.1, 11.6, 12.9, 12.3 Flour B: 12.9, 13.4, 12.7, 13.8, 13.1, 12.6, 13.5 Assume that both populations are normally distributed with equal variances. A two-tailed pooled two-sample t-test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the p-value for the test.
(2)
(c)
State the conclusion of the test, giving a reason and referring to the context.
(2)
(d)
State why the pooled test is appropriate here, rather than any other form of the test.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_A=\mu_B\) and \(H_1:\mu_A\ne\mu_B\) (mean protein content in the population of loaves)
✓ 1
Part (b)
M1 uses pooled two-sample t-test on a calculator (degrees of freedom 11, \(t=-3.82\))
✓ 1
A1 \(p=0.00286\) (3 s.f.)
✓ 1
Part (c)
M1 compares \(p=0.00286\) with 0.05
✓ 1
A1 reject \(H_0\): there is evidence that mean protein content differs between the flours (flour B higher in the sample)
✓ 1
Part (d)
R1 the two samples are independent and both populations are assumed normal with equal variances
✓ 1
Exam tip: Always compare the p-value with the significance level, then write the conclusion in the words of the problem, not just "reject \(H_0\)".
23IBMI3H-4-03r18-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A sauce is left to cool in a kitchen. The temperature excess, y °C, of the sauce above room temperature is recorded at time t minutes: t: 0, 5, 10, 15, 20, 25 y: 92, 71, 56, 43, 33, 26 The data are modelled by \(y=ab^t\).
(a)
Use exponential regression to find a and b, to 3 significant figures.
(2)
(b)
Interpret the value of b in this context.
(1)
(c)
Use the model to estimate the temperature excess after 18 minutes. Use the unrounded values from your calculator and give your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(a=92.0\)
✓ 1
A1 \(b=0.951\)
✓ 1
Part (b)
R1 each minute the temperature excess is multiplied by about 0.951, i.e. it falls by about 4.9% per minute
✓ 1
Part (c)
M1 \(y=ab^{18}\) with unrounded a and b
✓ 1
A1 \(y=36.9\) °C
✓ 1
Exam tip: Carry unrounded regression coefficients into later calculations. Using \(a=92.0\) and \(b=0.951\) would give a noticeably different value (about 37.2), so state clearly if you use stored values.
24IBMI3H-4-05r18-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
Light aircraft land at a small grass airstrip at random times, at a constant average rate of 3.2 landings per hour. Let \(X\) be the number of landings in a one-hour period.
(a)
Find the probability that exactly 5 aircraft land in a given hour.
(2)
(b)
Find the probability that at least 2 aircraft land in a given 30-minute period.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 recognise \(X \sim \mathrm{Po}(3.2)\) and attempt \(P(X=5)\)
✓ 1
A1 \(0.114\)
✓ 1
Part (b)
M1 new mean \(\lambda = 1.6\) for 30 minutes
✓ 1
M1 \(1 - P(Y \le 1)\) with \(Y \sim \mathrm{Po}(1.6)\), i.e. \(1 - e^{-1.6}(1 + 1.6)\)
✓ 1
A1 \(0.475\)
✓ 1
Exam tip: When the time interval changes, scale the Poisson mean to match it before doing any probability calculation. 'At least 2' means \(1-P(X\le 1)\), not \(1-P(X\le 2)\).
25IBMI3H-4-05r18-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
Each week, listeners of an online radio station choose either the Classical channel (C) or the Jazz channel (J). The transition matrix is \(\mathbf{T} = \begin{pmatrix} 0.8 & p \\ 0.2 & 1-p \end{pmatrix}\), where each column gives the probabilities of moving from that channel (first column from C, second from J) and the first row is the probability of moving to C. In the long run, 60% of listeners choose Classical.
(a)
Find the value of \(p\).
(3)
(b)
Use \(p = 0.3\). In week 0 there are 500 listeners: 400 on Classical and 100 on Jazz. Find the expected number on Classical in week 2.
(2)
(c)
A listener chooses Jazz in week 0. Find the probability that the listener chooses Jazz in week 2.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 steady state: \(0.8(0.6) + p(0.4) = 0.6\)
✓ 1
A1 \(0.12 = 0.4p\) (or equivalent equation)
✓ 1
A1 \(p = 0.3\)
✓ 1
Part (b)
M1 \(\mathbf{T}^2\binom{400}{100}\), or week 1 \(= \binom{350}{150}\) then week 2
✓ 1
A1 \(325\) listeners
✓ 1
Part (c)
M1 \(0.3\times0.2 + 0.7\times0.7\) (via Classical or Jazz in week 1)
✓ 1
A1 \(0.55\)
✓ 1
Common error: Check whether the columns or the rows of the matrix sum to 1 before multiplying. A steady state satisfies \(\mathbf{T}\mathbf{s} = \mathbf{s}\) with the entries of \(\mathbf{s}\) summing to 1.
26IBMI3H-4-05r18-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A flying club records the take-off mass \(m\) (in tonnes) of seven cargo aircraft and the ground run \(x\) (in metres) each needs on the same runway. \(m\): 42, 48, 55, 61, 66, 72, 79 \(x\): 620, 710, 830, 910, 1010, 1100, 1230
(a)
Find the Pearson product-moment correlation coefficient \(r\), to 3 significant figures.
(1)
(b)
Find the equation of the regression line of \(x\) on \(m\), with coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in this context.
(1)
(d)
Use the equation from (b), with the coefficients as written to 3 significant figures, to estimate the ground run for a take-off mass of 60 tonnes. Give your answer to the nearest metre.
R1 each extra tonne of take-off mass adds about 16.4 m to the ground run
✓ 1
Part (d)
A1 \(16.4\times60 - 75.9 = 908.1\), so \(908\) m
✓ 1
Exam tip: Say which variable is on which axis: the regression line of \(x\) on \(m\) predicts the ground run from the mass. Only predict within the range of the data (here 42 to 79 tonnes).
27IBMI3H-4-05r18-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The mass of a hen's egg from a large farm is modelled by a normal distribution with mean \(\mu\) g and standard deviation \(\sigma\) g. 10.6% of eggs have a mass below 52 g and 4.01% of eggs have a mass above 64 g.
(a)
Find \(\mu\) and \(\sigma\). Give each to 3 significant figures.
(4)
(b)
Find the probability that a randomly chosen egg has a mass between 55 g and 60 g. Use \(\mu = 57.0\) and \(\sigma = 4.00\).
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 use inverse normal: \(z = -1.25\) and \(z = 1.75\)
Exam tip: Turn each percentage into a \(z\)-value with the inverse normal function, taking care with the sign for the lower tail. Two conditions give two equations, which you solve simultaneously.
28IBMI3H-4-03r25-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Small explosions at a volcanic crater occur randomly and independently, at a constant mean rate of 2.5 per day. Let X be the number of explosions in a given day, and assume X ~ Po(2.5). Give all answers to 3 significant figures.
(a)
Find the probability that exactly 4 explosions occur on a given day.
(2)
(b)
Find the probability that at least 2 explosions occur on a given day.
(2)
(c)
Over a period of 30 days, find the expected number of days on which no explosion occurs.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to use the Poisson probability for X = 4 (e.g. GDC or P(X = 4) = e−2.5 × 2.54/4!)
✓ 1
A1 P(X = 4) = 0.134
✓ 1
Part (b)
M1 use of P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
A1 P(X ≥ 2) = 0.713
✓ 1
Part (c)
M1 P(X = 0) = e−2.5 = 0.0821 (or 0.0821) and multiplied by 30
✓ 1
A1 2.46 days (accept 2.46)
✓ 1
Exam tip: 'At least 2' means P(X ≥ 2) = 1 − P(X ≤ 1); make sure the cumulative function on your GDC is set to 'less than or equal'.
29IBMI3H-4-03r25-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A new musical opens at a theatre. The average nightly attendance y in each of the first eight weeks (x = 1, 2, …, 8) is shown below.
Week x
1
2
3
4
5
6
7
8
Attendance y
310
420
495
530
520
480
410
330
(a)
Find the quadratic regression model y = ax2 + bx + c. Give a, b and c to 4 significant figures.
(2)
(b)
The quadratic model has R2 = 0.990, whereas a linear model fitted to the same data has R2 = 0.000148. Explain which model is more suitable.
(2)
(c)
Using y = −17.05x2 + 153.9x + 179.2, find the greatest nightly attendance predicted by the model. Give your answer to the nearest whole number.
(3)
Show mark scheme
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Mark
Notes
Part (a)
A1 a = −17.05 and b = 153.9
✓ 1
A1 c = 179.2
✓ 1
Part (b)
A1 the quadratic model explains about 99.0% of the variation compared with almost none for the linear model
✓ 1
R1 so the quadratic model is more suitable (the data rise then fall, which a straight line cannot represent)
✓ 1
Part (c)
M1 attempt to find the maximum (GDC, or x = −b/(2a))
✓ 1
A1 x = 4.51 (accept 4.51)
✓ 1
A1 y = 526
✓ 1
Common error: Quoting R² as a probability or as 'the model is 99% accurate'; say instead that it is the proportion of the variation in y explained by the model.
30IBMI3H-4-03r25-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
The annual container throughput T (thousands of containers) of a port in year t, where t = 0 is 2015, is modelled by T = 412 × 1.058t. The model was fitted by exponential regression to data for t = 0 to 7, and has R2 = 0.962.
(a)
Interpret the value 1.058 in the model.
(1)
(b)
Use the model to estimate the throughput in 2023. Give your answer to 3 significant figures.
(2)
(c)
Interpret R2 = 0.962 in context.
(1)
(d)
Explain why the model is unlikely to be reliable for the year 2060.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 throughput increases by 5.8% each year (multiplier of 1.058 per year)
✓ 1
Part (b)
M1 substitute t = 8
✓ 1
A1 647 thousand containers (3 s.f.)
✓ 1
Part (c)
A1 96.2% of the variation in throughput is explained by the exponential model
✓ 1
Part (d)
R1 extrapolation far beyond the data range, and the port's capacity cannot grow exponentially for ever
✓ 1
Exam tip: For an exponential model, the base is the growth factor per unit of x; subtract 1 to get the percentage change.
31IBMI3H-4-03r25-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A theatre has 45 seats for a small studio performance but accepts 48 bookings. Each booked person independently fails to attend with probability 0.07. Let X be the number of people who fail to attend.
(a)
State the distribution of X and find its mean.
(2)
(b)
Find the probability that at least 5 booked people fail to attend. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that there are enough seats for everyone who attends, given that 48 people booked. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 X ~ B(48, 0.07)
✓ 1
A1 mean = 48 × 0.07 = 3.36
✓ 1
Part (b)
M1 1 − P(X ≤ 4)
✓ 1
A1 0.244
✓ 1
Part (c)
M1 everyone fits when at least 3 do not attend: P(X ≥ 3) = 1 − P(X ≤ 2)
✓ 1
A1 0.662
✓ 1
Exam tip: Define the event in terms of the random variable first (here, 'at least 3 absentees') before reaching for the GDC.
32IBMI3H-4-07r27-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
Bird strikes on aircraft at a small regional airport are recorded. The number of strikes in one week, X, is modelled by a Poisson distribution with mean 2.4.
(a)
State one assumption that must hold for this model to be appropriate.
(1)
(b)
Find the probability that exactly 3 bird strikes occur in a given week.
(2)
(c)
Find the probability that at least 2 bird strikes occur in a given week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Strikes occur independently of each other (or at a constant average rate / one at a time) – any one valid assumption
✓ 1
Part (b)
M1 use of Po(2.4) with P(X = 3)
✓ 1
A1 P(X = 3) = 0.209
✓ 1
Part (c)
M1 use of 1 − P(X ≤ 1)
✓ 1
A1 P(X ≥ 2) = 0.692
✓ 1
Common error: For "at least" probabilities with a Poisson variable, subtract the cumulative probability up to one less than the stated value, so "at least 2" uses P(X ≤ 1), not P(X ≤ 2).
33IBMI3H-4-07r27-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A cargo aircraft is loaded with 8 freight boxes. The mass of each box, in kg, is modelled as N(24, 3.52), and box masses are independent. Let T be the total mass of the 8 boxes.
(a)
Find E(T) and Var(T).
(3)
(b)
Find the probability that T exceeds 205 kg.
(2)
(c)
Find the standard deviation of the mean mass per box, X̄ = T ÷ 8, and state how it compares with the standard deviation of a single box.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 E(T) = 192 kg
✓ 1
M1 Var(T) = 8 × 3.52 (not 82 × 3.52)
✓ 1
A1 Var(T) = 98 kg2
✓ 1
Part (b)
M1 use of T ~ N(192, 98) with P(T > 205)
✓ 1
A1 0.0946
✓ 1
Part (c)
A1 SD(X̄) = 3.5 ÷ √8 = 1.24 kg
✓ 1
A1 It is smaller than 3.5 kg (the mean is less variable than a single box)
✓ 1
Common error: The variance of a sum of independent variables is the sum of the variances, so for 8 identical boxes it is 8σ², not (8σ)². Add variances, never standard deviations.
34IBMI3H-4-07r27-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
On a farm, each year a field is planted with either wheat (W) or barley (B). If the field was planted with wheat one year, the probability that it is planted with wheat the next year is 0.3. If it was planted with barley, the probability that it is planted with wheat the next year is 0.5. In year 1 the field is planted with wheat. Use a transition matrix T whose columns give the probabilities of moving from each crop, in the order W, B.
(a)
Write down the transition matrix T.
(2)
(b)
Find the probability that the field is planted with wheat in year 3.
(3)
(c)
Find the long-term proportion of years in which the field is planted with wheat.
(2)
Show mark scheme
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Notes
Part (a)
A1 first column (0.3, 0.7) (from wheat)
✓ 1
A1 second column (0.5, 0.5) (from barley)
✓ 1
Part (b)
M1 initial state vector (1, 0)
✓ 1
M1 T2 applied to initial state (or two successive steps)
✓ 1
A1 0.440
✓ 1
Part (c)
M1 steady state: Ts = s with w + b = 1 (or 0.3w + 0.5(1 − w) = w)
✓ 1
A1 5/12 = 0.417
✓ 1
Exam tip: To find a steady state, solve Ts = s together with the condition that the entries of s add to 1; either use the matrix equation or take a high power of T as a check.
35IBMI3H-4-07r27-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A grower compares two fertilisers, A and B, on tomato plants. The yield per plant (kg) is normally distributed with the same variance for both fertilisers. Summary data are: Fertiliser A: n = 10, mean 6.42, standard deviation 0.58. Fertiliser B: n = 12, mean 5.97, standard deviation 0.71. (The standard deviations are unbiased sample estimates.) The grower tests, at the 5% significance level, whether fertiliser A gives a greater mean yield than B.
(a)
State the null and alternative hypotheses.
(2)
(b)
Perform a pooled two-sample t-test and state your conclusion in context.
(3)
(c)
Calculate the pooled estimate of the common variance used in the test.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μA = μB
✓ 1
A1 H1: μA > μB
✓ 1
Part (b)
A1 p = 0.0621
✓ 1
M1 compare with 0.05 (0.0621 > 0.05)
✓ 1
A1 Do not reject H0: insufficient evidence that fertiliser A gives a greater mean yield
✓ 1
Part (c)
M1 (9 × 0.582 + 11 × 0.712) ÷ 20
✓ 1
A1 0.429
✓ 1
Common error: When a pooled test is used, the degrees of freedom are n₁ + n₂ − 2 (here 20), not n₁ + n₂ − 1.
36IBMI3H-4-04r28-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The controller of a warehouse robot arm logs software faults. The number of faults in one week may be modelled by a Poisson distribution with mean 2.4. Faults in different weeks are independent.
(a)
Find the probability that exactly 3 faults occur in a given week.
(2)
(b)
Find the probability that more than 4 faults occur in a given week.
(2)
(c)
Find the probability that exactly 4 faults occur in a given two-week period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of Poisson with mean 2.4, P(X = 3)
✓ 1
A1 0.209
✓ 1
Part (b)
M1 1 − P(X ≤ 4)
✓ 1
A1 0.0959
✓ 1
Part (c)
M1 mean for two weeks is 4.8
✓ 1
A1 0.182
✓ 1
Common error: When the time interval changes, the Poisson mean must be scaled to match; reusing 2.4 for a two-week period gives the wrong distribution.
37IBMI3H-4-04r28-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A fashion studio cuts fabric for dresses. The length of fabric used for one dress is normally distributed with mean 2.4 m and standard deviation 0.15 m. The lengths used for different dresses are independent. Let T be the total length used for three dresses.
(a)
Find E(T) and the standard deviation of T. Give the standard deviation to 3 significant figures.
(2)
(b)
Find the probability that the three dresses use more than 7.5 m of fabric.
(2)
(c)
Find the length t such that the probability that the three dresses use more than t metres is 0.05. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 E(T) = 7.2 m
✓ 1
A1 Var(T) = 3 × 0.152, so standard deviation = 0.260 m
✓ 1
Part (b)
M1 use of T ~ N(7.2, 0.0675)
✓ 1
A1 0.124
✓ 1
Part (c)
M1 uses z = 1.645 (1.6449) with the mean and standard deviation of T
✓ 1
A1 t = 7.63 m
✓ 1
Common error: The variance of a sum of three independent dresses is 3σ², not 9σ²; the latter would describe three times the length of one dress.
38IBMI3H-4-04r28-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A fashion company tests a new fabric treatment. The number of wash cycles survived before visible fading is recorded for a random sample of fabric pieces. Both populations are assumed to be normally distributed with equal variances.
Sample size
Mean
Sample standard deviation
Untreated
10
48.2
4.1
Treated
12
52.6
4.8
The company tests, at the 5% significance level, whether the treatment increases the mean number of wash cycles survived.
(a)
State the null and alternative hypotheses.
(2)
(b)
Calculate the pooled estimate of the variance and the value of the test statistic t. Give the test statistic to 3 significant figures.
(3)
(c)
The critical value for this one-tailed test with 20 degrees of freedom is 1.725. State the conclusion in context, with a reason.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μtreated = μuntreated
✓ 1
A1 H1: μtreated > μuntreated
✓ 1
Part (b)
M1 pooled variance = (9 × 4.12 + 11 × 4.82)/20
✓ 1
A1 20.2
✓ 1
A1 t = 2.28
✓ 1
Part (c)
R1 2.28 > 1.725
✓ 1
A1 reject H0: there is sufficient evidence that the treatment increases the mean number of wash cycles survived
✓ 1
Common error: A claim that a quantity has "increased" needs a one-tailed alternative hypothesis; using a two-tailed critical value would apply the wrong cut-off.
39IBMI3H-4-04r28-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A fisheries officer classifies each day at a harbour as Good (G) or Poor (P) for catches. The day-to-day changes are modelled by a Markov chain with transition matrix
from G
from P
to G
0.7
0.4
to P
0.3
0.6
Each column gives the probabilities of the next day's state, so each column sums to 1. Today is a Good day.
(a)
Write down the probability that tomorrow is a Poor day.
(1)
(b)
Find the probability that the day after tomorrow is a Good day.
(2)
(c)
Find the long-run proportion of days that are Good. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 0.3
✓ 1
Part (b)
M1 uses T2 with initial state (1, 0) or sums the two routes (0.7 × 0.7 + 0.4 × 0.3)
✓ 1
A1 0.61
✓ 1
Part (c)
M1 solves Ts = s with the entries of s summing to 1 (for example 0.3g = 0.4p, g + p = 1)
✓ 1
A1 g = 4/7 = 0.571
✓ 1
Exam tip: The steady state is the eigenvector for eigenvalue 1 of the transition matrix, scaled so that its entries add to 1.
40IBMI3H-4-05r29-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The number of radio dropouts, \(X\), between a Formula 1 car and its pit wall during a race follows a Poisson distribution with mean \(\lambda\). Data from many races show that \(P(X=0)=0.2\).
(a)
Find the value of \(\lambda\). Store the unrounded value for use in the rest of the question.
(2)
(b)
Find the probability of exactly two dropouts in a race.
(2)
(c)
Find the probability of more than one dropout in a race.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(e^{-\lambda}=0.2\)
✓ 1
A1 \(\lambda=\ln 5=1.61\)
✓ 1
Part (b)
M1 \(\frac{0.2\lambda^2}{2}\) using unrounded \(\lambda\)
✓ 1
A1 0.259
✓ 1
Part (c)
M1 \(1-0.2-0.2\lambda\)
✓ 1
A1 0.478
✓ 1
Exam tip: When a later part depends on \(\lambda\), store the unrounded value on your calculator to avoid a rounding drift in the final answer.
41IBMI3H-4-05r29-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
In a hospital outpatient clinic, the time a patient spends at triage is \(X\) minutes, with \(E(X)=12\) and \(\mathrm{Var}(X)=16\). The time in consultation is \(Y\) minutes, with \(E(Y)=20\) and \(\mathrm{Var}(Y)=25\). \(X\) and \(Y\) are independent. The clinic estimates the cost of a visit, in euros, as \(C=2X+3Y\).
(a)
Find \(E(C)\).
(2)
(b)
Find \(\mathrm{Var}(C)\).
(2)
(c)
Find the standard deviation of \(C\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2E(X)+3E(Y)\)
✓ 1
A1 84
✓ 1
Part (b)
M1 \(2^2\mathrm{Var}(X)+3^2\mathrm{Var}(Y)\)
✓ 1
A1 289
✓ 1
Part (c)
A1 17
✓ 1
Common error: Coefficients are squared in a variance: \(\mathrm{Var}(aX)=a^2\mathrm{Var}(X)\), not \(a\,\mathrm{Var}(X)\).
42IBMI3H-4-05r29-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A beekeeper models the number of bees \(y\) (in thousands) in a colony against hive volume \(x\) litres by \(y=1.8x^{0.65}\), for \(10\le x\le 60\). For this model, \(R^2=0.97\).
(a)
Use the model to estimate the colony size when the volume is 40 litres.
(2)
(b)
Interpret the value of \(R^2\).
(1)
(c)
Find the percentage increase in the predicted colony size when the volume is doubled.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.8\times40^{0.65}\)
✓ 1
A1 19.8 thousand bees
✓ 1
Part (b)
R1 97% of the variation in colony size is explained by the model (hive volume)
✓ 1
Part (c)
M1 ratio \(\frac{1.8(2x)^{0.65}}{1.8x^{0.65}}=2^{0.65}\)
✓ 1
A1 1.569
✓ 1
A1 56.9%
✓ 1
Exam tip: In a power model \(y=ax^b\), multiplying \(x\) by \(k\) multiplies \(y\) by \(k^b\), whatever \(a\) is.
43IBMI3H-4-07r33-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
At a regional clinic, patients presenting with a rare tropical infection arrive independently at a constant mean rate of 3.2 per day. The number of such patients in one day, X, is modelled by a Poisson distribution.
(a)
Find the probability that exactly 5 such patients arrive on a given day.
(2)
(b)
Find the probability that at least 2 such patients arrive on a given day.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ Po(3.2) and use of P(X = 5) = e−3.2 × 3.25 ÷ 5!
✓ 1
A1 0.114
✓ 1
Part (b)
M1 P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
M1 1 − (e−3.2 + 3.2e−3.2) or use of GDC cumulative probability (0.1712...)
✓ 1
A1 0.829
✓ 1
Common error: For "at least" probabilities, subtract the complementary cumulative probability: P(X ≥ 2) = 1 − P(X ≤ 1), not 1 − P(X ≤ 2).
44IBMI3H-4-07r33-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A retailer packs a jar of honey and a gift box together for a parcel. The mass of a jar, J, is normally distributed with mean 450 g and standard deviation 6 g. The mass of a box, B, is normally distributed with mean 85 g and standard deviation 4 g. J and B are independent. Let T = J + B be the total mass of a parcel.
(a)
Find E(T) and Var(T).
(2)
(b)
Find the probability that a parcel has mass greater than 545 g.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(T) = 450 + 85 = 535 g
✓ 1
A1 Var(T) = 62 + 42 = 52 g2
✓ 1
Part (b)
M1 T ~ N(535, 52) used, standard deviation √52 = 7.211...
✓ 1
M1 attempt at P(T > 545) using standardisation (z = 1.387...) or GDC
✓ 1
A1 0.0828
✓ 1
Common error: Variances of independent variables add; standard deviations do not. Add 6² and 4², then take the square root only when standardising.
45IBMI3H-4-07r33-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
A supermarket tracks whether shoppers buy its own-brand pasta (O) or a competitor's brand (C) each week. The transition matrix is T = (0.85, 0.30; 0.15, 0.70), written row by row, where the first column gives the probabilities of moving from O to O and from O to C, and the second column gives those for C. Each column sums to 1. At the start, 40% of shoppers buy own-brand and 60% buy the competitor's.
(a)
Find the proportion of shoppers who buy own-brand after 2 weeks.
(2)
(b)
Find the long-term proportion of shoppers who buy own-brand.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of T2 × (0.4, 0.6)T or two successive multiplications (0.52 after week 1)
✓ 1
A1 0.586
✓ 1
Part (b)
M1 steady state satisfies Ts = s with s1 + s2 = 1
✓ 1
M1 0.15s1 = 0.30s2 (or equivalent equation)
✓ 1
A1 2/3 = 0.667
✓ 1
Exam tip: The long-term proportions do not depend on the starting proportions; use T s = s together with the condition that the proportions sum to 1.
46IBMI3H-4-07r33-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy8 marks
A café manager records the number of customers, y, in the café during the x-th hour after opening on one day.
Hour x: 1, 2, 3, 4, 5, 6, 7, 8 Customers y: 42, 63, 80, 88, 91, 85, 74, 58 The manager models the data by a quadratic function y = ax2 + bx + c.
(a)
Use your calculator to find the quadratic regression equation, giving a, b and c to 3 significant figures.
(2)
(b)
The coefficient of determination is R2 = 0.998. Interpret this value.
(2)
(c)
Use the model to estimate the number of customers in the 5th hour, to the nearest whole number. Use the unrounded regression coefficients.
(2)
(d)
Explain why the model should not be used to predict the number of customers in the 12th hour.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of quadratic regression
✓ 1
A1 y = −3.29x2 + 31.8x + 13.3
✓ 1
Part (b)
A1 about 99.8% of the variation in the number of customers is explained by the quadratic model in x
✓ 1
R1 so the model is a very good fit to these data
✓ 1
Part (c)
M1 substitute x = 5 into the model
✓ 1
A1 90 (90.18; accept 90 using rounded coefficients)
✓ 1
Part (d)
A1 x = 12 gives a negative value (about −79 customers), which is impossible
✓ 1
R1 12 is outside the range of the data, so this is extrapolation and unreliable
✓ 1
Exam tip: Use unrounded regression coefficients in later calculations; rounding a, b and c first can change the third significant figure of your estimate.
47IBMI3H-4-05r34-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Faults on a stretch of railway track occur randomly and independently, at a constant average rate of 2.4 faults per 10 km.
(a)
Find the probability that exactly 3 faults are found in a 10 km section.
(2)
(b)
State one condition that must hold for a Poisson model to be suitable here.
(1)
(c)
Find the probability that at least 2 faults are found in a 5 km section.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Let X be the number of faults in 10 km; X ~ Po(2.4)
✓ 1
P(X = 3) = 0.209 (3 s.f.)
✓ 1
Part (b)
Faults occur independently of one another / at a constant average rate / singly (any one valid condition)
✓ 1
Part (c)
Mean for 5 km is 1.2, so Y ~ Po(1.2)
✓ 1
P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − (e−1.2 + 1.2e−1.2)
✓ 1
= 0.337 (3 s.f.)
✓ 1
Common error: When the interval length changes, the Poisson mean must be scaled in proportion; do not keep using the mean for the original length.
48IBMI3H-4-05r34-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
The state of a commuter train on any morning is either on time (O) or delayed (D). The transition matrix is \(\mathbf{T}=\begin{pmatrix}0.9 & 0.35\\ 0.1 & 0.65\end{pmatrix}\), where the columns give the current state (O, D) and the rows give the next day's state (O, D), so that \(\mathbf{s}_{n+1}=\mathbf{T}\mathbf{s}_n\).
(a)
Explain what the entry 0.35 represents.
(1)
(b)
The train is on time on Monday. Find the probability that it is on time on Wednesday.
(2)
(c)
Find the probability that the train is on time on Thursday of the same week, given that it is on time on Monday. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The probability that the train is on time tomorrow given that it is delayed today
Common error: Keep track of which index is the row and which is the column; with columns as current states, each column of the matrix must sum to 1.
49IBMI3H-4-05r34-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The number of daily passengers, P thousand, on a rail line is modelled by P = 62e0.018t, where t is the number of months after 1 January 2024. The model was obtained by fitting an exponential curve to monthly data, and its coefficient of determination is R2 = 0.964.
(a)
Use the model to estimate the number of daily passengers when t = 30. Give your answer to 3 significant figures.
(2)
(b)
Interpret the value of R2.
(1)
(c)
Find how many months it takes for the number of daily passengers to double. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Substitute t = 30: P = 62e0.54
✓ 1
P = 106 thousand (3 s.f.)
✓ 1
Part (b)
96.4% of the variation in the data is explained by the model
✓ 1
Part (c)
62e0.018t = 124, i.e. e0.018t = 2
✓ 1
0.018t = ln 2
✓ 1
t = 38.5 months (3 s.f.)
✓ 1
Exam tip: The doubling time of an exponential model does not depend on the starting value: solve e^(kt) = 2.
50IBMI3H-4-05r34-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
The time, T seconds, taken by swimmers in a club to swim 100 m freestyle is normally distributed with mean 58.4 and standard deviation 1.9.
(a)
Find the probability that a randomly chosen swimmer takes less than 56.0 seconds.
(2)
(b)
Six swimmers are chosen at random. Using the unrounded probability from part (a), find the probability that at least one of them takes less than 56.0 seconds.
(2)
(c)
The fastest 5% of swimmers qualify for a regional event. Find the qualifying time, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
z = (56.0 − 58.4) ÷ 1.9 = −1.26…
✓ 1
P(T < 56.0) = 0.103 (3 s.f.)
✓ 1
Part (b)
Binomial model, n = 6 and p = 0.10327…
✓ 1
1 − (1 − 0.10327…)6 = 0.480 (3 s.f.)
✓ 1
Part (c)
z = −1.645 for a lower tail of 0.05
✓ 1
T = 58.4 − 1.645 × 1.9 = 55.3 seconds (3 s.f.)
✓ 1
Exam tip: Sketch the normal curve and shade the required region before you use the inverse normal function; it shows whether z is positive or negative.
51IBMI3H-4-05r37-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A wind farm records unplanned turbine shutdowns. They occur independently and at random, at a constant mean rate of 0.8 per week. Let X be the number of unplanned shutdowns in a period of 3 weeks.
(a)
State the distribution of X, including its parameter.
(1)
(b)
Find P(X = 3).
(2)
(c)
Find the probability that there are at least 2 shutdowns in the 3-week period.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ Po(2.4)
✓ 1
Part (b)
M1 use of Poisson probability with λ = 2.4, e.g. e−2.4×2.43/3!
✓ 1
A1 0.209
✓ 1
Part (c)
M1 P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
M1 attempt to find P(X = 0) + P(X = 1) (= 0.3084)
✓ 1
A1 0.692
✓ 1
Exam tip: "At least 2" means 2 or more, so use the complement 1 − P(X ≤ 1) rather than adding many terms.
52IBMI3H-4-05r37-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
At a leisure centre, each swimmer chooses the Fast lane (F) or the Steady lane (S) for each session. If a swimmer used F last session, the probability they choose F again is 0.8. If a swimmer used S last session, the probability they choose F is p. The transition matrix, with columns giving the current state in the order F, S, is T = [[0.8, p], [0.2, 1 − p]].
(a)
In the long run, 75% of sessions are in the Fast lane. Find the value of p.
(3)
(b)
Using p = 0.6, a swimmer uses the Fast lane today. Find the probability that they use the Steady lane in two sessions' time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses steady state (0.75, 0.25)T with Ts = s
✓ 1
M1 forms equation 0.8 × 0.75 + 0.25p = 0.75
✓ 1
A1 p = 0.6
✓ 1
Part (b)
M1 considers the two paths F→F→S and F→S→S
✓ 1
M1 0.8 × 0.2 + 0.2 × 0.4
✓ 1
A1 0.24
✓ 1
Exam tip: For two-state chains the steady-state equation gives a quick check: the long-run share of F is p ÷ (p + 0.2).
53IBMI3H-4-05r37-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
The wind speed W (m s−1) at a proposed turbine site is modelled by W ~ N(9.5, 3.22). A turbine only generates power when 3.5 ≤ W ≤ 25.
(a)
Find the probability that the turbine generates power on a given day.
(2)
(b)
Find the wind speed that is exceeded on 20% of days, to 3 significant figures.
(2)
(c)
At a second site, wind speed has the same standard deviation 3.2 m s−1, and P(W < 3.5) = 0.15. Find the mean wind speed at this site, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(3.5 < W < 25) using normal cdf (P(W > 25) is negligible)
✓ 1
A1 0.970
✓ 1
Part (b)
M1 inverse normal with area 0.8, z = 0.8416
✓ 1
A1 12.2 m s−1
✓ 1
Part (c)
M1 z = −1.0364 for area 0.15
✓ 1
M1 (3.5 − μ)/3.2 = −1.0364
✓ 1
A1 μ = 6.82 m s−1
✓ 1
Exam tip: Sketch the normal curve and shade the required area before using inverse normal; it shows whether z should be positive or negative.
54IBMI3H-4-05r37-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A library records the daily rainfall x (mm) and the number of visitors y on 8 days. Rainfall x: 0, 2, 5, 8, 11, 14, 18, 22 Visitors y: 214, 205, 188, 181, 160, 157, 131, 118
(a)
Write down the product-moment correlation coefficient r, to 3 significant figures, and describe the correlation.
(2)
(b)
Find the equation of the regression line of x on y, giving coefficients to 3 significant figures.
(2)
(c)
Use unrounded regression coefficients to estimate the rainfall on a day with 170 visitors, to 3 significant figures.
(2)
(d)
Explain why the regression line of y on x should not be used for part (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = −0.995
✓ 1
A1 very strong negative linear correlation
✓ 1
Part (b)
M1 regression of x on y using a calculator
✓ 1
A1 x = −0.226y + 48.3
✓ 1
Part (c)
M1 substitutes y = 170 into the regression line of x on y
✓ 1
A1 9.83 mm
✓ 1
Part (d)
R1 the line of y on x minimises vertical distances and is for predicting y; to predict x from y use the line of x on y
✓ 1
Common error: Rearranging the y-on-x line to estimate x gives a different (less accurate) result than the x-on-y regression line.
55IBMI3H-4-04r38-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
Dolphin sightings from a research boat are modelled by a Poisson distribution with a mean of 2.4 sightings per hour. Let \(X\) be the number of sightings in one hour.
(a)
Find \(P(X=3)\).
(2)
(b)
Find the probability of at least one sighting in a given hour.
(2)
(c)
Find the probability that there are no sightings during a 90-minute survey.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of Poisson probability with \(\lambda=2.4\): \(\mathrm{e}^{-2.4}\frac{2.4^3}{3!}\)
✓ 1
A1 \(0.209\)
✓ 1
Part (b)
M1 \(1-P(X=0)=1-\mathrm{e}^{-2.4}\)
✓ 1
A1 \(0.909\)
✓ 1
Part (c)
M1 new mean for 90 minutes \(=3.6\) and \(P=\mathrm{e}^{-3.6}\)
✓ 1
A1 \(0.0273\)
✓ 1
Common error: When the time interval changes, the Poisson mean must be scaled to match it; keeping \(\lambda=2.4\) for 90 minutes gives a wrong probability.
56IBMI3H-4-04r38-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
Calculator
Seals use two haul-out sites, A and B. Each day a seal is at exactly one site. If a seal is at A today, the probability that it is at A tomorrow is 0.75. If it is at B today, the probability that it is at A tomorrow is 0.40. The transition matrix is \(\mathbf{T}=\begin{pmatrix}0.75&0.40\\0.25&0.60\end{pmatrix}\), where each column gives the probabilities for a seal at the site it heads (first column: from A; second column: from B).
(a)
A seal is at site A on Monday. Find the probability that it is at site A on Wednesday.
(2)
(b)
Find the long-term probability that a seal is at site A.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\mathbf{T}^2\) (two transitions) applied to \(\begin{pmatrix}1\\0\end{pmatrix}\), or \(0.75\times0.75+0.40\times0.25\)
✓ 1
A1 \(0.6625\)
✓ 1
Part (b)
M1 solve \(\mathbf{T}\mathbf{s}=\mathbf{s}\) with \(s_A+s_B=1\), e.g. \(0.25s_A=0.40s_B\)
✓ 1
M1 \(s_A=\frac{0.40}{0.25+0.40}\)
✓ 1
A1 \(0.615\)
✓ 1
Exam tip: The steady-state vector must satisfy \(\mathbf{T}\mathbf{s}=\mathbf{s}\) and also have entries that sum to 1; forgetting the second condition leaves the answer undetermined.
57IBMI3H-4-04r38-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
A courier company records whether 300 deliveries arrived on time or late, for three delivery methods.
Method
Courier
Post
Locker
On time
85
72
93
Late
15
28
7
The company tests, at the 1% significance level, whether delivery outcome is independent of delivery method.
(a)
State the null hypothesis.
(1)
(b)
Carry out a \(\chi^2\) test for independence and find the \(p\)-value.
(2)
(c)
State the conclusion of the test, giving a reason.
(1)
(d)
Find the expected frequency of late deliveries by locker if \(\mathrm{H}_0\) is true. Give your answer to three significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\mathrm{H}_0\): delivery outcome (on time or late) is independent of delivery method
✓ 1
Part (b)
M1 use of \(\chi^2\) test for independence with \(\nu=(2-1)(3-1)=2\) (\(\chi^2=16.2\))
✓ 1
A1 \(p=0.000307\)
✓ 1
Part (c)
R1 \(0.000307<0.01\), so reject \(\mathrm{H}_0\): there is evidence that outcome depends on delivery method
Common error: Expected frequencies are calculated from the row and column totals assuming independence, not from the observed value in the cell.
58IBMI3H-4-04r38-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
A new delivery app records its weekly orders. The table shows the number of orders, \(y\) hundred, in week \(x\) after launch.
\(x\)
1
2
3
4
5
6
7
\(y\)
12.4
16.8
24.9
34.6
48.1
71.0
97.3
The data are modelled by \(y=ab^x\).
(a)
Use exponential regression to find \(a\) and \(b\), giving each to four significant figures.
(2)
(b)
State the value of the coefficient of determination, \(R^2\), to three significant figures, and interpret the value of \(b\) in this context.
(2)
(c)
Use your values of \(a\) and \(b\) from part (a) to predict the number of orders, in hundreds, in week 10. Give your answer to three significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of exponential regression on the data
✓ 1
A1 \(a=8.634\), \(b=1.415\)
✓ 1
Part (b)
A1 \(R^2=0.999\) (very close to 1, so the model fits the data very well)
✓ 1
A1 \(b=1.415\): weekly orders are multiplied by about 1.415, an increase of about 41.5% each week
✓ 1
Part (c)
M1 \(8.634\times1.415^{10}\)
✓ 1
A1 \(278\) hundred
✓ 1
Common error: \(R^2\) near 1 shows a good fit within the data range, but it does not by itself make predictions far outside that range reliable.
59IBMI3H-4-04r45-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Eruptions of a volcano occur at random and independently, at a constant mean rate of 2.4 per year. Let X be the number of eruptions in one year.
(a)
Write down the distribution of X and find the probability of exactly 3 eruptions in one year.
(2)
(b)
Find the probability of at least one eruption in a six-month period.
(2)
(c)
Find the probability that there are exactly 2 eruptions in each of two consecutive years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ Po(2.4) and P(X = 3) = e−2.4 × 2.43/3!
✓ 1
A1 0.209
✓ 1
Part (b)
M1 mean for six months is 1.2, and 1 − P(Y = 0) = 1 − e−1.2
✓ 1
A1 0.699
✓ 1
Part (c)
M1 P(X = 2) = 0.26127 and squares it, using independence
✓ 1
A1 0.0683
✓ 1
Exam tip: Always rescale the Poisson mean to match the time period in the question before calculating any probability.
60IBMI3H-4-04r45-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
The running time of Act 1 of a play is a random variable A ~ N(52, 42) minutes and that of Act 2 is B ~ N(47, 52) minutes. The two running times are independent. The interval between the acts lasts exactly 20 minutes. Let T = A + B.
(a)
Find E(T) and Var(T).
(3)
(b)
Find the probability that the whole evening, including the interval, lasts more than 130 minutes.
(2)
(c)
Two independent performances of Act 1 have running times A and A′, each distributed as A. Find the standard deviation of A + A′, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 adds the variances, 16 + 25
✓ 1
A1 E(T) = 99 minutes
✓ 1
A1 Var(T) = 41
✓ 1
Part (b)
M1 recognises the condition T > 110, with T ~ N(99, 41)
✓ 1
A1 0.0429
✓ 1
Part (c)
M1 Var(A + A′) = 16 + 16 = 32
✓ 1
A1 standard deviation = 5.66 minutes
✓ 1
Common error: Adding standard deviations instead of variances: for independent variables it is the variances that add.
61IBMI3H-4-04r45-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
A ferry company compares the crossing times, in minutes, of two operators on the same route. Six sailings by operator A took 52, 55, 49, 58, 54 and 51 minutes. Seven sailings by operator B took 57, 60, 55, 62, 58, 56 and 59 minutes. Assume that both sets of times come from normal populations with the same variance. A test is carried out at the 5 % significance level to see whether the mean crossing times differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Carry out a two-sample t-test with pooled variance. State the test statistic to 3 significant figures and the p-value to 3 significant figures.
(3)
(c)
State your conclusion in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: the population mean crossing times are equal (μA = μB)
✓ 1
A1 H1: the population mean crossing times are different (μA ≠ μB)
✓ 1
Part (b)
M1 uses a two-sample pooled t-test, with 11 degrees of freedom
✓ 1
A1 t = −3.20
✓ 1
A1 p-value = 0.00839
✓ 1
Part (c)
A1 p < 0.05, so reject H0
✓ 1
R1 there is sufficient evidence that the mean crossing times of the two operators differ
✓ 1
Exam tip: Compare the p-value with the significance level directly, and write the conclusion in the context of the problem, not just 'reject H₀'.
62IBMI3H-4-04r45-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
The alert status of a volcano is recorded each week as Calm (C) or Alert (A). If the status is Calm one week, then it is Calm the next week with probability 0.85. If the status is Alert, then it is Calm the next week with probability 0.40. The status is Calm this week.
(a)
Find the probability that the status is Alert two weeks from now.
(2)
(b)
Find the long-term proportion of weeks for which the status is Alert. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 considers C → C → A and C → A → A: 0.85 × 0.15 + 0.15 × 0.60
✓ 1
A1 0.2175
✓ 1
Part (b)
M1 sets up steady-state equations, e.g. a = 0.15c + 0.60a with c + a = 1
✓ 1
M1 solves to get 0.40a = 0.15c
✓ 1
A1 a = 0.15/0.55 = 0.273
✓ 1
Exam tip: For a steady state, use the equations from Ts = s together with the condition that the probabilities add to 1.
63IBMI3H-4-06r45-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
The height of the ash plume produced by eruptions of a volcano is modelled by a normal distribution with mean 8.4 km and standard deviation 1.6 km.
(a)
Find the probability that an eruption produces a plume higher than 10.5 km. Give your answer to 3 significant figures.
(2)
(b)
The tallest 15% of plumes are classed as ‘major’. Find the minimum height of a major plume, giving your answer to 3 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 identifies H ~ N(8.4, 1.62) and attempts P(H > 10.5)
✓ 1
A1 0.0947
✓ 1
Part (b)
M1 P(H < h) = 0.85 (or P(H > h) = 0.15)
✓ 1
M1 attempts inverse normal, z = 1.036…
✓ 1
A1 h = 10.1 km
✓ 1
Exam tip: Sketching a quick normal curve and shading the required area shows whether you need the probability above or below the value.
64IBMI3H-4-06r45-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A theatre sells 40 tickets for a matinee. Independently, each ticket-holder fails to attend with probability 0.08. Let X be the number of ticket-holders who fail to attend.
(a)
Write down the expected value of X.
(1)
(b)
Find the probability that exactly 3 ticket-holders fail to attend. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that at least 5 ticket-holders fail to attend. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 3.2
✓ 1
Part (b)
M1 X ~ B(40, 0.08) and attempts P(X = 3)
✓ 1
A1 0.231
✓ 1
Part (c)
M1 P(X ≥ 5) = 1 − P(X ≤ 4)
✓ 1
A1 P(X ≤ 4) = 0.7868…
✓ 1
A1 0.213
✓ 1
Common error: P(X ≥ 5) is 1 − P(X ≤ 4), not 1 − P(X ≤ 5); check which value the cumulative function should be given.
65IBMI3H-4-06r45-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
At a theatre, the number of mobile phone interruptions in a performance is modelled by a Poisson distribution with mean 2.4. Let X be the number of interruptions in a performance.
(a)
State one condition that must hold for this Poisson model to be suitable.
(1)
(b)
Find P(X = 2). Give your answer to 3 significant figures.
(2)
(c)
Find the probability that there are at least 4 interruptions in a performance. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 interruptions occur independently of each other and at a constant average rate (accept singly / at random)
✓ 1
Part (b)
M1 X ~ Po(2.4) and attempts P(X = 2) = e−2.4 2.42/2!
✓ 1
A1 0.261
✓ 1
Part (c)
M1 1 − P(X ≤ 3)
✓ 1
A1 0.221
✓ 1
Common error: ‘At least 4’ means 4 or more, so subtract P(X ≤ 3) from 1, not P(X ≤ 4).
66IBMI3H-4-06r45-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
A cargo ship carries two types of container. The mass of a type A container is normally distributed with mean 18.5 tonnes and standard deviation 2.1 tonnes. The mass of a type B container is normally distributed with mean 12.4 tonnes and standard deviation 1.5 tonnes. The masses of all containers are independent. A crane lifts one type A container and two different type B containers together. Let L tonnes be the total mass lifted.
(a)
Find the mean and variance of L.
(2)
(b)
Find the probability that L is greater than 46 tonnes. Give your answer to 3 significant figures.
(2)
(c)
The crane operator wants a load limit that is exceeded by only 5% of lifts. Find this limit to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 E(L) = 18.5 + 2 × 12.4 = 43.3
✓ 1
A1 Var(L) = 2.12 + 2 × 1.52 = 8.91
✓ 1
Part (b)
M1 L ~ N(43.3, 8.91) and attempts P(L > 46)
✓ 1
A1 0.183
✓ 1
Part (c)
M1 attempts inverse normal, z = 1.645 for P(L > c) = 0.05
✓ 1
A1 48.2 tonnes
✓ 1
Common error: For two different type B containers, the variance is 2 × 1.5²; using 2² × 1.5² would wrongly treat them as the same container counted twice.
67IBMI3H-4-07r47-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
Calculator
A camera on a mountain-top observatory records meteors from a shower. The number of meteors, X, recorded in a 10-minute interval is modelled by a Poisson distribution with mean 3.5. Assume the meteors arrive independently of each other.
(a)
Find P(X = 5).
(2)
(b)
Find the probability that at least 2 meteors are recorded in a 10-minute interval.
(2)
(c)
The camera runs for 25 minutes. Find the probability that at most 6 meteors are recorded in this time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of Po(3.5) probability for X = 5
✓ 1
A1 0.132
✓ 1
Part (b)
M1 use of 1 − P(X ≤ 1)
✓ 1
A1 0.864
✓ 1
Part (c)
A1 mean for 25 minutes is 8.75
✓ 1
M1 use of Po(8.75) with P(Y ≤ 6)
✓ 1
A1 0.231
✓ 1
Common error: When the time interval changes, scale the mean first (25 min is 2.5 intervals) and only then use the Poisson distribution; do not scale the final probability.
68IBMI3H-4-07r47-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy8 marks
Calculator
A commuter train makes a journey in two legs. The time for the first leg, A minutes, is normally distributed with mean 42 and standard deviation 3.2. The time for the second leg, B minutes, is normally distributed with mean 28 and standard deviation 2.4. The two times are independent. Let T = A + B be the total journey time.
(a)
State one assumption made about the two legs that is needed to find the distribution of T.
(1)
(b)
Find E(T) and Var(T).
(3)
(c)
Find the probability that the total journey time exceeds 76 minutes.
(2)
(d)
Let D = A − B. Find the standard deviation of D.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the leg times are independent (accept: the delay on one leg does not affect the other)
✓ 1
Part (b)
A1 E(T) = 70
✓ 1
M1 Var(T) = 3.22 + 2.42 (variances added)
✓ 1
A1 Var(T) = 16
✓ 1
Part (c)
M1 use of T ~ N(70, 16) (or standard deviation 4)
✓ 1
A1 0.0668
✓ 1
Part (d)
M1 Var(D) = 3.22 + 2.42 (variances are added, not subtracted)
✓ 1
A1 standard deviation 4
✓ 1
Common error: Variances add for both a sum and a difference of independent variables; subtracting variances for A − B is a frequent slip.
69IBMI3H-4-07r47-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
A sports scientist records the weekly training time (hours) of seven club runners and their finishing positions in a race (1 = first). Runner: P, Q, R, S, T, U, V Training (h): 4, 9, 6, 12, 8, 10, 5 Position: 6, 3, 5, 1, 2, 4, 7
(a)
Rank the training times, giving rank 1 to the smallest value.
(2)
(b)
Calculate Spearman's rank correlation coefficient, rs, between training rank and finishing position. Give your answer to 3 significant figures.
(2)
(c)
Interpret your value of rs in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to rank the seven times in order
✓ 1
A1 ranks 1, 5, 3, 7, 4, 6, 2 for P to V
✓ 1
Part (b)
M1 use of rs with ranks (formula with Σd2 = 102, or GDC on the ranks)
✓ 1
A1 rs = −0.821
✓ 1
Part (c)
A1 strong negative (monotonic) association
✓ 1
R1 runners who train for more hours tend to have lower position numbers, that is, they finish nearer the front
✓ 1
Common error: The position scale runs the opposite way to the time scale (1 is best), so a negative coefficient here means more training goes with better finishes.
70IBMI3H-4-07r47-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy7 marks
Calculator
A sports scientist records the running speed, x km h−1, and the heart rate, y beats per minute, of a volunteer on a treadmill. x: 8, 9, 10, 12, 13, 14, 16 y: 128, 134, 141, 152, 160, 165, 176
(a)
Write down the product-moment correlation coefficient, r, and the equation of the regression line of y on x. Give values to 3 significant figures.
(3)
(b)
Interpret the gradient of the regression line in context.
(1)
(c)
Estimate the heart rate at a speed of 15 km h−1, to the nearest beat per minute.
(2)
(d)
Explain why the line should not be used to estimate the heart rate at 25 km h−1.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 r = 0.999
✓ 1
A1 gradient 6.06
✓ 1
A1 intercept 79.8, so y = 6.06x + 79.8
✓ 1
Part (b)
A1 heart rate increases by about 6.06 beats per minute for each 1 km h−1 increase in speed
✓ 1
Part (c)
M1 substitution of x = 15 into the regression line
✓ 1
A1 171
✓ 1
Part (d)
R1 25 is far outside the data range (extrapolation); the line gives about 231 beats per minute, which is unrealistic
✓ 1
Exam tip: Interpret a gradient with units and 'for each one unit increase in x'; interpolate only within the range of the data.
71IBMI3H-4-07r48-01Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
A volcano observatory records eruptions of a particular volcano. Eruptions occur independently and at random, at a constant average rate of 3.2 per year. Let X be the number of eruptions in one year, and assume X follows a Poisson distribution.
(a)
Find the probability that there are exactly 2 eruptions in a given year.
(2)
(b)
Find the probability that there are at least 5 eruptions in a given year.
(2)
(c)
Find the probability that there are no eruptions at all during a two-year period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of Po(3.2) with \(P(X=2)=\frac{e^{-3.2}3.2^2}{2!}\) or GDC Poisson pdf
✓ 1
A1 0.209 (0.2087…)
✓ 1
Part (b)
M1 \(1-P(X\le 4)\)
✓ 1
A1 0.219 (0.2193…)
✓ 1
Part (c)
M1 new mean \(\lambda=2\times3.2=6.4\) for two years
✓ 1
A1 \(e^{-6.4}=0.00166\) (0.0016615…)
✓ 1
Common error: When the time period changes, the Poisson mean must be scaled to match; reusing the yearly mean for a two-year period gives a wrong probability.
72IBMI3H-4-07r48-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy6 marks
Calculator
A theatre sells season subscriptions. Each year, 85% of those who held a subscription the year before renew it, and 40% of those who did not hold one the year before take one out. Let the states be S (holds a subscription) and N (does not). Use the convention that each column of the transition matrix T gives the probabilities of moving from a state, and \(\mathbf{s}_n=\mathbf{T}^n\mathbf{s}_0\).
(a)
Write down the transition matrix T, with states in the order S, N.
(1)
(b)
A person holds a subscription this year. Find the probability that they hold one in 3 years' time.
(2)
(c)
Find the long-term proportion of people who hold a subscription.
(2)
(d)
The theatre has 1200 people on its list. Estimate how many hold a subscription in the long term, to the nearest whole number.
M1 solving \(\mathbf T\mathbf s=\mathbf s\) with components summing to 1, or \(\mathbf T^n\) for large \(n\)
✓ 1
A1 \(\frac{8}{11}=0.727\)
✓ 1
Part (d)
A1 \(1200\times\frac{8}{11}=872.7\approx873\)
✓ 1
Common error: Check which way the matrix is set up: with columns as the "from" state, every column must sum to 1.
73IBMI3H-4-07r48-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Easy5 marks
Calculator
A theatre compares what patrons spend (£) on refreshments per visit. Matinee (10 patrons): 7.2, 6.8, 7.9, 6.5, 7.4, 8.1, 7.0, 6.9, 7.6, 7.3 Evening (9 patrons): 6.4, 7.1, 6.2, 6.9, 7.0, 6.5, 6.0, 6.8, 7.2 Assume the populations are normal with equal variances. The manager believes matinee patrons spend more on average and tests this at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Carry out a one-tailed pooled t-test and find the p-value.
(2)
(c)
State, with a reason, the conclusion of the test.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_M=\mu_E\) and \(H_1:\mu_M>\mu_E\)
✓ 1
Part (b)
M1 pooled two-sample t-test, one-tailed, with \(t=2.78\) and 17 degrees of freedom
✓ 1
A1 \(p=0.00643\) (0.006428…)
✓ 1
Part (c)
R1 \(p=0.00643<0.05\)
✓ 1
A1 reject \(H_0\): there is sufficient evidence that matinee patrons spend more on average
✓ 1
Common error: For a one-tailed alternative, use the one-tailed p-value; do not halve or double it unless the calculator gives the two-tailed value.
74IBMI-033-04-r2-02T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A café advertises that a large cup of its filter coffee contains a mean of 110 mg of caffeine. A consumer group buys 10 large cups at random and measures the caffeine content, in mg:
92, 105, 98, 110, 101, 95, 99, 107, 103, 96
Assume the caffeine content is normally distributed.
(a)
Find the sample mean and an unbiased estimate of the population variance.
(2)
(b)
Find a 95% confidence interval for the population mean caffeine content. Give the limits to one decimal place.
(3)
(c)
Comment on the café's advertised value, with a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\bar{x}\) = 100.6 mg
✓ 1
\(s_{n-1}^2\) = 32.3 mg2 (32.2667)
✓ 1
Part (b)
Use of t-distribution with 9 degrees of freedom (critical value 2.262) or GDC t-interval
✓ 1
Lower limit 96.5 mg
✓ 1
Upper limit 104.7 mg
✓ 1
Part (c)
110 mg lies outside (above) the 95% confidence interval
✓ 1
So the advertised mean is not supported; the true mean caffeine content is likely to be lower than 110 mg
✓ 1
Common error: The population standard deviation is unknown, so the interval uses the t-distribution with n − 1 degrees of freedom, not a z-interval.
75IBMI-033-04-r2-04T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
A running coach believes a six-week training programme reduces 5 km race times. Eight runners record their times, in minutes, before and after the programme.
Reject H0: there is significant evidence that the programme reduces the mean 5 km time
✓ 1
Common error: Using a two-sample test on paired data ignores the pairing and usually gives a much larger p-value. Always test the differences.
76IBMI-033-04-r1-04T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
A phone manufacturer claims that the mean battery life of its new model is 12.0 hours of continuous video playback. A reviewer suspects the true mean is lower. She tests a random sample of 10 phones and records a sample mean of 11.4 hours and an unbiased estimate of the standard deviation \(s_{n-1} = 0.85\) hours. Battery lives may be assumed to be normally distributed.
She carries out a one-sample t-test at the 5% significance level. The critical value for this test is −1.833.
(a)
State suitable hypotheses for the test.
(2)
(b)
Calculate the value of the test statistic.
(2)
(c)
State, with a reason, the conclusion of the test in context.
(2)
(d)
Using \(t_{9} = 1.833\) for 90% confidence, find a 90% confidence interval for the mean battery life.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: H0: \(\mu = 12.0\), where \(\mu\) is the population mean battery life in hours
✓ 1
A1: H1: \(\mu < 12.0\)
✓ 1
Part (b)
M1: \(t = \frac{11.4 - 12.0}{0.85/\sqrt{10}}\)
✓ 1
A1: \(t = -2.23\)
✓ 1
Part (c)
R1: \(-2.23 < -1.833\), so the test statistic lies in the critical region
✓ 1
A1: reject H0; there is sufficient evidence at the 5% level that the mean battery life is less than 12.0 hours
Common error: The population standard deviation is unknown, so a t-distribution with 9 degrees of freedom is used. Using the normal value 1.645 instead of 1.833 gives an interval that is too narrow.
77IBMI-033-01-r1-05Correlation and regressionPaper 1Medium7 marks
The number of public electric-vehicle charging points in a city, y hundred, is recorded x years after the start of 2019.
x: 1, 2, 3, 4, 5 y: 3.1, 4.8, 7.6, 11.9, 17.2
Two models are proposed: Model A: \(y = 3.5x - 1.4\) Model B: \(y = 0.6x^2 + 0.1x + 2.4\)
(a)
Calculate the sum of square residuals, SSres, for Model A.
(2)
(b)
Calculate SSres for Model B.
(2)
(c)
State, with a reason, which model fits the data better.
(1)
(d)
Use your chosen model to estimate the number of charging points 6 years after the start of 2019.
Model B, because its sum of square residuals is smaller (1.03 < 5.59)
✓ 1
Part (d)
M1: substitutes x = 6 into the chosen model, e.g. Model B: 0.6 × 36 + 0.1 × 6 + 2.4 = 24.6
✓ 1
A1: about 2460 charging points
✓ 1
(Follow through from (c): Model A gives 19.6 hundred = 1960 charging points, which earns M1 A1 FT.)
—
Common error: Forgetting the units of y: the model gives y in hundreds, so 24.6 must be converted to 2460 charging points to answer in context.
78IBMI-033-05-r2-06Sampling and descriptive statisticsPaper 1Medium6 marks
A machine fills bags of coffee labelled 250 g. A quality controller takes every 50th bag from the production line until she has 12 bags. She weighs each one and records the mass x grams. The results give \(\sum x = 3018\) and \(\sum x^2 = 759\,160\).
(a)
Find an unbiased estimate of the mean mass of all bags filled by the machine.
(1)
(b)
Find an unbiased estimate of the variance of the mass of all bags filled by the machine.
(3)
(c)
Name the sampling method used, and give one reason why it could produce a biased sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\bar{x}=\frac{3018}{12}\) = 251.5 g A1
✓ 1
Part (b)
Correct use of \(\sum x^2\) and \(\bar{x}\), e.g. \(759\,160-12(251.5)^2=133\) or \(\frac{759\,160}{12}-251.5^2=11.08...\) M1
✓ 1
Correct use of divisor \(n-1\), e.g. \(\frac{133}{11}\) or \(11.08...\times\frac{12}{11}\) M1
✓ 1
\(s_{n-1}^2\) = 12.1 g2 A1
✓ 1
Part (c)
Systematic sampling A1
✓ 1
If a fault or fill pattern in the machine repeats at an interval that matches the sampling interval (e.g. every 25 or 50 bags), the sample could keep selecting the faulty bags or always miss them R1
✓ 1
Common error: Σx²/n − mean² gives the variance with divisor n. For an unbiased estimate of the population variance, multiply by n/(n − 1), or divide the sum of squared deviations by n − 1.
79IBMI-033-04-r1-03T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
Priya travels to work each weekday either by bicycle (C) or by bus (B). If she cycles on one day, the probability that she cycles the next day is 0.7. If she takes the bus on one day, the probability that she takes the bus the next day is 0.6. Assume her choice depends only on the previous day's choice.
(a)
Write down the transition matrix T, with columns representing the current day's mode of travel (C then B) and rows the next day's.
(2)
(b)
Priya cycles on Monday. Find the probability that she cycles on Thursday of the same week.
(2)
(c)
Find the long-term proportion of days on which Priya cycles.
(3)
(d)
Priya works 210 days a year. Estimate the number of days in a year on which she cycles to work.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: one column fully correct, e.g. first column \(\begin{pmatrix}0.7\\0.3\end{pmatrix}\)
✓ 1
A1: \(\mathbf{T}=\begin{pmatrix}0.7 & 0.4\\ 0.3 & 0.6\end{pmatrix}\) fully correct, in the stated orientation
M1: forms \(\mathbf{T}\mathbf{s}=\mathbf{s}\) with \(\mathbf{s}=\begin{pmatrix}p\\q\end{pmatrix}\), e.g. \(0.7p+0.4q=p\), or finds an eigenvector of T for eigenvalue 1
✓ 1
M1: uses \(p+q=1\) (or normalises the eigenvector) to get one equation in one unknown, e.g. \(0.3p=0.4(1-p)\)
✓ 1
A1: \(p=\frac{4}{7}\approx 0.571\)
✓ 1
Part (d)
A1: \(210 \times \frac{4}{7} = 120\) days
✓ 1
Exam tip: Count transitions carefully: Monday to Thursday is three steps, so use \(\mathbf{T}^3\), not \(\mathbf{T}^4\).
80IBMI-033-01-r2-05Correlation and regressionPaper 1Medium8 marks
The number of downloads, N, of a new mobile game was recorded at the end of each of its first 9 weeks after launch, where t is the number of weeks. A developer suspects exponential growth, so she calculates ln N for each week. The regression line of ln N on t is \(\ln N = 0.231t + 5.12\) with correlation coefficient r = 0.994.
(a)
Show that the model can be written as \(N = Ae^{kt}\), and find the values of A and k.
(3)
(b)
Use the model to estimate the number of downloads at the end of week 10.
(2)
(c)
Find the value of t at which the model predicts that downloads first reach 5000.
(2)
(d)
Give one reason why this model is unlikely to be valid in the long term.
(1)
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Mark
Notes
Part (a)
Exponentiates both sides: \(N = e^{5.12}e^{0.231t}\) (M1)
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A = \(e^{5.12}\) ≈ 167 (A1)
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k = 0.231 (A1)
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Part (b)
Substitutes t = 10, e.g. \(N = e^{7.43}\) or \(167e^{2.31}\) (M1)
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N ≈ 1690 downloads (accept 1680 from rounded A) (A1)
Any one valid reason, e.g. exponential growth is unbounded but downloads are limited by the number of potential users or by falling interest; OR long-term predictions are extrapolation far beyond the 9 weeks of data (R1)
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Common error: The constant A is \(e^{5.12}\), not 5.12. The intercept of the linearised model is ln A.
81IBMI3H-4-08-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Faults occur at random along an underground cable at an average rate of 0.8 faults per kilometre.
(a)
Find the probability that a 1.5 km section of cable has no faults.
(2)
(b)
Five separate 1.5 km sections are inspected. Find the probability that exactly two of them have no faults.
(3)
(c)
Find the greatest length of cable for which the probability of no faults is at least 0.05.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 mean \(0.8\times1.5=1.2\), \(P(X=0)=\text{e}^{-1.2}\)
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A1 0.301
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Part (b)
M1 \(Y\sim B(5,0.3012)\)
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M1 \(P(Y=2)\)
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A1 0.310
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Part (c)
M1 \(\text{e}^{-0.8l}\ge0.05\)
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M1 take natural logarithms: \(l\le\frac{-\ln0.05}{0.8}\)
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A1 3.74 km
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Exam tip: If a Poisson question then asks about a count of sections, switch to a binomial model whose probability comes from the Poisson part.
82IBMI3H-4-08-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Customers choose one of three streaming services A, B or C each month. The monthly transition matrix is
\(T=\begin{pmatrix}0.7&0.2&0.1\\0.2&0.6&0.3\\0.1&0.2&0.6\end{pmatrix}\), with columns giving the probabilities of moving from one service (A, B, C).
Initially the market shares are A: 50%, B: 30%, C: 20%.
(a)
Find the market shares after three months.
(2)
(b)
Find the long-run market shares.
(3)
(c)
There are 12 000 customers in total. Estimate the number who use service A in the long run.
(2)
(d)
Explain why the long-run shares do not depend on the initial market shares.
R1 all transition probabilities are positive (the chain is regular), so it converges to the same steady state from any starting distribution
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Common error: For the steady state, use the unrounded fractions in later calculations; rounding 0.3448 to 0.34 gives 4080 customers, not 4140.
83IBMI3H-4-08b-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
In a triathlon relay, the swim leg time S and the run leg time R of a team, both in minutes, are independent with S ~ N(24, 22) and R ~ N(41, 32). Let T = S + R.
(a)
Write down E(T).
(1)
(b)
Find Var(T) and hence the standard deviation of T.
(2)
(c)
Find the probability that the team's total time is less than 60 minutes.
(2)
(d)
The team wants a total time that it beats with probability 0.9. Find this time.
(2)
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Notes
Part (a)
A1 24 + 41 = 65 minutes
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Part (b)
A1 Var(T) = 4 + 9 = 13
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A1 standard deviation = √13 = 3.61 minutes
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Part (c)
M1 T ~ N(65, 13) used
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A1 0.0828
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Part (d)
M1 inverse normal, z = 1.2816
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A1 65 + 1.2816 × 3.606 = 69.6 minutes
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Common error: For independent variables, variances add: Var(S + R) = Var(S) + Var(R). Standard deviations do not add.
84IBMI3H-4-04r7-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
An emergency department receives patients at random, at an average rate of 6 per hour. The number of arrivals in a given period is modelled by a Poisson distribution.
(a)
The manager wants to staff for the number of arrivals in a one-hour period with a probability of at least 0.95 that the number of arrivals does not exceed the staffing level \(k\). Find the smallest integer \(k\).
(3)
(b)
Find the probability that at least 10 patients arrive in a given two-hour period.
(3)
Show mark scheme
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Notes
Part (a)
M1 \(X\sim\mathrm{Po}(6)\) and attempt to compare \(P(X\le k)\) with 0.95
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A1 \(P(X\le9)=0.916<0.95\)
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A1 \(P(X\le10)=0.957\ge0.95\), so \(k=10\)
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Part (b)
A1 mean for two hours is 12, \(Y\sim\mathrm{Po}(12)\)
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M1 \(1-P(Y\le9)\)
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A1 0.758
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Exam tip: For 'at least' probabilities with a discrete distribution, use \(P(X\ge a)=1-P(X\le a-1)\); check that the value \(a-1\) is the one you enter.
85IBMI3H-4-04r7-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
In a Formula 1 qualifying session, the lap times of Mara are independent and normally distributed with mean 92.4 s and standard deviation 0.8 s. Her rival Teo's lap times are independent and normally distributed with mean 92.1 s and standard deviation 0.7 s, independent of Mara's.
(a)
Mara completes three laps. Let \(T\) be her total time. Find the probability that \(T\) is less than 276 s, to 3 significant figures.
(3)
(b)
On one lap each, let \(D=M-S\), where \(M\) is Mara's time and \(S\) is Teo's time. Find the probability that Mara is faster than Teo, that is \(P(D<0)\), to 3 significant figures.
(3)
(c)
A strategist suggests modelling Mara's time for three laps as \(3M\). Find \(\mathrm{Var}(3M)\) and explain why it differs from \(\mathrm{Var}(T)\) in (a).
R1 \(3M\) assumes the three laps are identical, whereas the real laps are independent so some variation cancels, giving a smaller variance
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Common error: \(\mathrm{Var}(X_1+X_2)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)\) even for a difference; the variances never subtract.
86IBMI3H-4-04r7-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A beekeeper records the honey yield, in kg, from a random sample of 12 hives: 28.4, 31.2, 26.9, 33.5, 29.8, 30.6, 27.3, 32.1, 28.8, 34.0, 29.5, 31.7. Assume the yields are normally distributed.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation.
(2)
(b)
Construct a 95% confidence interval for the population mean yield. Give the upper limit to 3 significant figures.
(3)
(c)
A supplier claims the mean yield of this type of hive is 32 kg. Comment on this claim using your interval.
(2)
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Notes
Part (a)
A1 mean \(=30.3\) kg (30.3167)
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A1 \(s_{n-1}=2.28\) kg
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Part (b)
M1 use of the \(t\)-distribution with \(11\) degrees of freedom
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A1 \(t=2.20\) and lower limit 28.9
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A1 upper limit 31.8
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Part (c)
R1 32 lies outside the interval (28.9, 31.8)
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A1 the data are not consistent with a mean of 32 kg at the 95% level; evidence the claim is too high
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Common error: With a small sample and an unknown population standard deviation, use the \(t\)-distribution with \(n-1\) degrees of freedom, not the normal distribution.
87IBMI3H-4-03r9-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Ships arrive at the port of Calvera at random and independently, at a mean rate of 5.5 per day. The number of arrivals per day is modelled by a Poisson distribution. Give probabilities to 3 significant figures.
(a)
Find the probability that more than 8 ships arrive on a given day.
(2)
(b)
Find the probability that the total number of arrivals in a three-day period is between 12 and 20 inclusive.
(3)
(c)
The port can handle at most 9 arrivals in one day. Over a 30-day month, find the expected number of days on which this limit is exceeded.
(3)
Show mark scheme
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Notes
Part (a)
M1 Let X ~ Po(5.5); P(X > 8) = 1 − P(X ≤ 8)
✓ 1
A1 = 0.106
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Part (b)
A1 Three-day mean = 16.5, so Y ~ Po(16.5)
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M1 P(12 ≤ Y ≤ 20) = P(Y ≤ 20) − P(Y ≤ 11)
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A1 = 0.734
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Part (c)
A1 P(X ≥ 10) = 1 − P(X ≤ 9) = 0.0538
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M1 Expected number = 30 × 0.0538
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A1 = 1.61 days
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Common error: 'Between 12 and 20 inclusive' needs P(Y ≤ 20) − P(Y ≤ 11): the lower bound is one below the first value you want.
88IBMI3H-4-03r9-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A theatre advertises that its performances of a play last 150 minutes on average. The running times, in minutes, of 10 randomly selected performances were: 152, 156, 149, 158, 154, 151, 160, 147, 155, 157. Assume running times are normally distributed. A one-sample t-test is carried out at the 5% significance level to decide whether the mean running time differs from the advertised time.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the sample mean and an unbiased estimate of the population variance, to 3 significant figures.
(2)
(c)
Calculate the value of the test statistic t, to 3 significant figures.
(2)
(d)
The p-value for this test is 0.0151. State the conclusion in context, with a reason.
(2)
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Notes
Part (a)
A1 H0: μ = 150
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A1 H1: μ ≠ 150
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Part (b)
A1 Sample mean = 154 (exactly 153.9)
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A1 Unbiased variance estimate = 17.0
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Part (c)
M1 Attempt to use t = (x̄ − 150)/(s/√10)
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A1 t = 2.99
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Part (d)
R1 0.0151 < 0.05, so reject H0
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A1 There is sufficient evidence that the mean running time differs from 150 minutes
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Exam tip: Compare the p-value directly with the significance level; for a two-tailed test the calculator's two-tailed p-value is used as it stands.
89IBMI3H-4-03r9-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A shipping company cleaned the hulls of 8 ships. Each ship's daily fuel consumption (tonnes) on the same route was recorded before and after cleaning. Ship 1 to 8, before: 38.2, 41.5, 36.8, 44.0, 39.7, 42.3, 37.5, 40.9 Ship 1 to 8, after: 36.9, 40.8, 37.0, 41.7, 38.1, 41.2, 36.2, 39.3 Differences may be assumed normally distributed. The company carries out a paired t-test at the 5% significance level, with d = before − after.
(a)
Explain why a paired test is appropriate here.
(1)
(b)
State suitable hypotheses for a test of whether cleaning reduces fuel consumption.
(2)
(c)
Find the mean and the unbiased estimate of the standard deviation of the 8 values of d, to 3 significant figures.
(2)
(d)
The critical value of t for this one-tailed test with 7 degrees of freedom is 1.895. Calculate the test statistic and state the conclusion in context.
(3)
Show mark scheme
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Notes
Part (a)
R1 The same ships are measured twice, so the two samples are not independent
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Part (b)
A1 H0: μd = 0
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A1 H1: μd > 0
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Part (c)
A1 Mean = 1.21
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A1 s = 0.734
✓ 1
Part (d)
M1 Attempt to use t = d̄/(s/√8)
✓ 1
A1 t = 4.67
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R1 4.67 > 1.895, so reject H0: there is evidence that cleaning reduces fuel consumption
✓ 1
Exam tip: Take differences first and then treat them as a single sample; do not run a two-sample test on paired data.
90IBMI3H-4-03r9-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
The alert level of the Thessa volcano is reviewed each day and is Green, Yellow or Red. The day-to-day changes are modelled by a Markov chain with transition matrix T, where the columns give the probabilities of moving from Green, Yellow and Red respectively, and the rows give the probabilities of moving to Green, Yellow and Red respectively: \(T=\begin{pmatrix}0.80&0.30&0.10\\0.15&0.50&0.30\\0.05&0.20&p\end{pmatrix}\) Use \(\mathbf{s}_{n}=T^{n}\mathbf{s}_{0}\) for the state vector after n days. Give answers to 3 significant figures where not exact.
(a)
Find the value of p.
(2)
(b)
The alert level is Green today. Find the probability that it is Red in two days' time.
(3)
(c)
Find the long-term proportion of days on which the alert level is Red.
(3)
Show mark scheme
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Notes
Part (a)
M1 Columns sum to 1, so 0.10 + 0.30 + p = 1
✓ 1
A1 p = 0.60
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Part (b)
M1 Attempt to use T2 or to sum the paths Green→Green→Red, Green→Yellow→Red, Green→Red→Red
✓ 1
A1 0.8 × 0.05 + 0.15 × 0.20 + 0.05 × 0.60
✓ 1
A1 = 0.100
✓ 1
Part (c)
M1 Set up Ts = s with the components summing to 1
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M1 Solve to obtain the invariant vector (14/27, 5/18, 11/54)
✓ 1
A1 Proportion Red = 11/54 = 0.204
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Exam tip: For the long-term distribution solve Ts = s together with the condition that the entries sum to 1; the equations alone have infinitely many solutions.
91IBMI3H-4-03r9-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A theatre compares two brands of stage-lighting bulb. The lifetimes, in hours, of random samples were: Brand X (8 bulbs): 1180, 1225, 1310, 1255, 1190, 1270, 1235, 1205 Brand Y (9 bulbs): 1290, 1345, 1260, 1380, 1315, 1330, 1275, 1360, 1300 Lifetimes for both brands are assumed to be normally distributed with the same population variance. A two-sample t-test (pooled) is used at the 5% significance level to test whether the mean lifetimes differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the sample standard deviation of each brand, using the unbiased estimate, to 3 significant figures.
(2)
(c)
Calculate the pooled estimate of the common variance, to 3 significant figures.
(2)
(d)
The p-value for the test is 0.000911. State the conclusion in context, and say which brand has the longer mean lifetime.
(2)
Show mark scheme
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Notes
Part (a)
A1 H0: μX = μY
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A1 H1: μX ≠ μY
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Part (b)
A1 Brand X: 43.6 hours
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A1 Brand Y: 40.0 hours
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Part (c)
M1 Use of (7sX2 + 8sY2)/15
✓ 1
A1 Pooled variance = 1740
✓ 1
Part (d)
R1 0.000911 < 0.05, so reject H0; there is evidence the mean lifetimes differ
✓ 1
A1 Brand Y, since its sample mean is larger
✓ 1
Exam tip: The pooled variance weights each sample variance by its degrees of freedom (n − 1), not by its sample size.
92IBMI3H-4-03r12-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A data centre has two independent server halls. In hall A the number of drive failures in a week follows a Poisson distribution with mean 1.4; in hall B it follows a Poisson distribution with mean 0.9. Let T be the total number of drive failures in the data centre in one week. Weekly totals in different weeks are independent.
(a)
State the distribution of T, including the value of its parameter.
(2)
(b)
Find \(P(T=3)\), giving your answer to 3 significant figures.
(2)
(c)
Find the probability that there are no drive failures at all in a period of 4 weeks. Give your answer in the form \(a\times10^{k}\) where \(1\leq a<10\) and \(a\) is given to 3 significant figures.
(3)
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Notes
Part (a)
A1 T follows a Poisson distribution
✓ 1
A1 with mean \(1.4+0.9=2.3\)
✓ 1
Part (b)
M1 \(e^{-2.3}\times\frac{2.3^3}{3!}\)
✓ 1
A1 0.203
✓ 1
Part (c)
M1 mean for 4 weeks is \(4\times2.3=9.2\)
✓ 1
M1 \(P(\text{no failures})=e^{-9.2}\)
✓ 1
A1 1.01×10−4
✓ 1
Common error: When the time interval changes, scale the Poisson mean by the same factor before using the formula; do not scale the probability.
93IBMI3H-4-03r12-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A web server handles two types of request. The response time of a type 1 request, \(R_1\) milliseconds, is normally distributed with mean 120 and standard deviation 15. The response time of a type 2 request, \(R_2\) milliseconds, is normally distributed with mean 95 and standard deviation 12. The two times are independent.
(a)
Let \(D=R_1-R_2\). State the distribution of D, giving its mean and variance.
(3)
(b)
Find the probability that a type 1 request takes longer than a type 2 request. Give your answer to 3 significant figures.
(2)
(c)
Three independent type 1 requests are made and their mean response time is \(M\). Find the standard deviation of \(M\), giving your answer to 3 significant figures.
(2)
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Notes
Part (a)
A1 D is normally distributed
✓ 1
A1 \(E(D)=120-95=25\)
✓ 1
A1 \(\text{Var}(D)=15^2+12^2=369\)
✓ 1
Part (b)
M1 \(P(D>0)\) with \(D\sim N(25,369)\) (standard deviation \(\sqrt{369}=19.2\))
Common error: \(\text{Var}(aX)=a^2\text{Var}(X)\), but the variance of a sum of n independent copies of X is \(n\,\text{Var}(X)\), not \(n^2\text{Var}(X)\).
94IBMI3H-4-03r12-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The transfer of a data file between two data centres takes place in two independent stages. Stage 1 takes X seconds, where \(X\sim N(\mu,4^2)\) and \(\mu\) is unknown. Stage 2 takes Y seconds, where \(Y\sim N(50,3^2)\). The total transfer time is \(T=X+Y\). It is known that 20% of transfers take longer than 90 seconds.
(a)
State the distribution of T in terms of \(\mu\).
(3)
(b)
Find the value of \(\mu\), giving your answer to 3 significant figures.
(4)
Show mark scheme
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Notes
Part (a)
A1 T is normally distributed with mean \(\mu+50\)
✓ 1
A1 variance \(4^2+3^2=25\) (standard deviation 5)
✓ 1
A1 \(T\sim N(\mu+50,\,25)\)
✓ 1
Part (b)
M1 \(P(T>90)=0.2\) so \(P(Z>z)=0.2\) giving \(z=0.8416\) (inverse normal)
✓ 1
M1 \(\frac{90-(\mu+50)}{5}=0.8416\)
✓ 1
M1 \(\mu=40-0.8416\times5=35.792\)
✓ 1
A1 35.8 seconds
✓ 1
Exam tip: Standardise using the standard deviation of the combination (here 5), not of one component.
95IBMI3H-4-03r12-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
The latency (in milliseconds) of a database query was measured on 10 randomly chosen occasions: 38.2, 41.5, 36.9, 44.1, 40.3, 39.8, 42.6, 37.4, 43.0, 40.9 Assume that latency is normally distributed.
(a)
Calculate the mean and the unbiased estimate of the standard deviation of the population, each to 3 significant figures.
(2)
(b)
Hence find a 95% confidence interval for the population mean latency. Give the limits to 3 significant figures, using the unrounded values from (a).
(4)
(c)
The system's specification states that the mean latency is 40 ms. State, with a reason, whether the confidence interval is consistent with this statement.
(2)
Show mark scheme
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Notes
Part (a)
A1 mean = 40.5 ms
✓ 1
A1 \(s_{n-1}\) = 2.43 ms
✓ 1
Part (b)
M1 uses a t-interval with 9 degrees of freedom (critical value 2.262)
✓ 1
M1 \(\bar x\pm 2.262\times\frac{s}{\sqrt{10}}\)
✓ 1
A1 lower limit 38.7
✓ 1
A1 upper limit 42.2 (ECF from (a))
✓ 1
Part (c)
R1 40 lies inside the interval
✓ 1
A1 so the data are consistent with a mean of 40 ms (no evidence against the specification)
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Common error: With an unknown population variance and a small sample, use the t-distribution with \(n-1\) degrees of freedom, not the normal distribution.
96IBMI3H-4-03r18-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The number of dropped calls at a mobile-phone mast in one hour is modelled by a Poisson distribution with unknown mean m. It is known that the probability of exactly 2 dropped calls in an hour is equal to the probability of exactly 3 dropped calls in an hour. Assume that dropped calls occur independently and at a constant rate.
(a)
Show that \(m=3\).
(3)
(b)
The mast is monitored for a shift of 8 hours. Find the probability that more than 20 calls are dropped during the shift.
(2)
(c)
Find the probability that at least one call is dropped during a particular 20-minute period.
M1 cancels \(e^{-m}m^2\) (valid as \(m>0\)) to obtain \(\frac12=\frac{m}{6}\)
✓ 1
A1 \(m=3\) (AG, correct working shown)
✓ 1
Part (b)
A1 shift count is Poisson with mean \(8\times3=24\)
✓ 1
A1 \(P(Y>20)=1-P(Y\le20)=0.757\) (3 s.f.)
✓ 1
Part (c)
A1 mean for 20 minutes is \(3\times\frac{20}{60}=1\)
✓ 1
A1 \(P(\ge1)=1-e^{-1}=0.632\) (3 s.f.)
✓ 1
Common error: Forgetting to rescale the mean when the time interval changes. The Poisson mean must always match the length of the period being asked about.
97IBMI3H-4-03r18-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A cable car on a mountain railway carries exactly four climbers at a time, each with one rucksack. The mass of a climber is normally distributed with mean 74 kg and standard deviation 6 kg. The mass of a rucksack is normally distributed with mean 12.5 kg and standard deviation 1.2 kg. All masses are independent. Let T be the total mass, in kg, of the four climbers and their four rucksacks.
(a)
State the distribution of T, giving its mean and variance.
(3)
(b)
The cable car is overloaded if the total mass exceeds 370 kg. Find the probability that it is overloaded.
(2)
(c)
The operator wishes to set a load limit c so that the probability of overload is 0.01. Find c to one decimal place.
(2)
Show mark scheme
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Notes
Part (a)
M1 \(T\) is a sum of eight independent normal variables, so normal
✓ 1
A1 mean \(=4(74+12.5)=346\)
✓ 1
A1 variance \(=4(6^2)+4(1.2^2)=149.76\)
✓ 1
Part (b)
M1 \(P(T>370)\) with \(T\sim N(346,149.76)\)
✓ 1
A1 \(0.0249\) (3 s.f.)
✓ 1
Part (c)
M1 \(P(T>c)=0.01\), uses inverse normal (\(z=2.3263\))
✓ 1
A1 \(c=374.5\) kg (1 d.p.)
✓ 1
Common error: Writing the variance of four climbers as \(4^2\times 36\) (treating four climbers as one climber multiplied by four). Independent masses add as \(n\sigma^2\), not \(n^2\sigma^2\).
98IBMI3H-4-03r18-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A food factory has two production lines. The mass of a dough portion from line A is normally distributed with mean 805 g and standard deviation 9 g. The mass of a dough portion from line B is normally distributed with mean 800 g and standard deviation 12 g. All portions are independent.
(a)
A portion is taken at random from each line. Find the probability that the portion from line A is heavier than the portion from line B.
(3)
(b)
Two portions are taken at random from line A and two from line B. Find the probability that the total mass of the two line A portions exceeds the total mass of the two line B portions.
(3)
(c)
A student says that the total mass of two line A portions has variance \(4\times 81\) because \"two portions is double one portion\". Explain why this is incorrect.
(1)
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Notes
Part (a)
M1 considers \(D=A-B\) and \(D\sim N(5,\,81+144)=N(5,225)\)
✓ 1
A1 SD of \(D\) is 15
✓ 1
A1 \(P(D>0)=0.631\) (3 s.f.)
✓ 1
Part (b)
M1 \(S=(A_1+A_2)-(B_1+B_2)\) has mean \(1610-1600=10\)
✓ 1
A1 variance \(2(81)+2(144)=450\)
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A1 \(P(S>0)=0.681\) (3 s.f.)
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Part (c)
R1 two different portions are independent random variables, so variances add: \(2\times 81\); doubling one portion (\(2A\)) would give \(4\times81\)
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Exam tip: The variance of a difference is the sum of the variances for independent variables: \(\mathrm{Var}(A-B)=\mathrm{Var}(A)+\mathrm{Var}(B)\).
99IBMI3H-4-03r18-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Small fragments of bone are occasionally found in boneless fish fillets. The number of fragments in a 500 g pack is modelled by a Poisson distribution with mean 3.5. Packs are independent.
(a)
Find the probability that a randomly chosen pack contains no bone fragments.
(2)
(b)
Find the smallest integer k such that the probability of a pack containing more than k fragments is less than 0.005.
(3)
(c)
Three packs are chosen at random. Find the probability that every one of them contains at least one bone fragment.
(2)
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Notes
Part (a)
M1 \(e^{-3.5}\)
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A1 \(0.0302\) (3 s.f.)
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Part (b)
M1 evaluates \(P(X>k)=1-P(X\le k)\) for successive values of k
✓ 1
A1 \(P(X>8)=0.00987>0.005\) and \(P(X>9)=0.00331<0.005\)
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A1 \(k=9\)
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Part (c)
M1 \(P(X\ge1)=1-e^{-3.5}=0.9698\) cubed
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A1 \(0.912\) (3 s.f.)
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Exam tip: "At least one" and "more than k" are both found through complements. When a threshold is asked for, show the two neighbouring cumulative values so that the choice of integer is justified.
100IBMI3H-4-03r18-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A telecoms engineer measures the download speed, in Mbps, at eight randomly chosen locations in a valley. The results are: 41.2, 38.5, 45.1, 43.8, 39.9, 42.6, 44.3, 40.7 The speeds can be modelled as a random sample from a normal distribution.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation, to 3 significant figures.
(2)
(b)
Find a 95% confidence interval for the population mean download speed, giving the limits to 3 significant figures. Use the unrounded mean and standard deviation.
(3)
(c)
State one assumption required for the interval in part (b) to be valid.
(1)
(d)
A network provider claims that the mean download speed in the valley is 45 Mbps. Comment on this claim.
(1)
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Notes
Part (a)
A1 mean \(=42.0\)
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A1 \(s_{n-1}=2.31\)
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Part (b)
M1 uses the t-distribution with \(n-1=7\) degrees of freedom, \(t=2.365\)
R1 the speeds in the valley are normally distributed (or the sample is random)
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Part (d)
R1 45 lies outside the interval, so the data do not support the claim
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Common error: Using \(z=1.96\) instead of the t-value when the population standard deviation is unknown and the sample is small. With 7 degrees of freedom the correct value is 2.365.
101IBMI3H-4-03r18-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Air pressure P (kPa) is measured at six altitudes h (km) on a mountain. A climber models the data by \(P=101.3e^{-0.1254h}\). h (km): 0, 1, 2, 3, 4, 5 Observed P: 101.3, 89.9, 79.5, 70.1, 61.6, 54.0 The coefficient of determination is \(R^2=1-\dfrac{SS_{res}}{SS_{tot}}\), where $SS_{res}=\sum (P_{obs}-P_{model})^2$ and \(SS_{tot}=\sum (P_{obs}-\bar P)^2\).
(a)
Calculate the model values of P for h = 1, 2, 3, 4 and 5, to one decimal place.
(2)
(b)
Using your rounded model values from part (a), find \(SS_{res}\) and \(SS_{tot}\).
(4)
(c)
Find \(R^2\) to 3 significant figures and interpret it.
(2)
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Notes
Part (a)
A1 89.4 and 78.8
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A1 69.5, 61.3 and 54.1
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Part (b)
M1 residuals \(0,\,0.5,\,0.7,\,0.6,\,0.3,\,-0.1\)
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A1 \(SS_{res}=1.20\) (ECF from (a))
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M1 uses \(\bar P=76.07\) in \(\sum(P_{obs}-\bar P)^2\)
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A1 \(SS_{tot}=1570\) (3 s.f.; exactly 1571.7)
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Part (c)
A1 \(R^2=1-\frac{1.20}{1571.7}=0.999\)
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R1 about 99.9% of the variation in observed pressure is explained by the model
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Common error: Describing \(R^2\) as the probability that the model is correct. It measures only the proportion of the variation in the data accounted for by the model.
102IBMI3H-4-05r18-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Weeds in a wheat field occur at random positions, at an average of 0.8 weeds per square metre. A farmer inspects square quadrats of area 2.5 m2. Let \(W\) be the number of weeds found in one quadrat.
(a)
Find the probability that a quadrat contains no weeds.
(2)
(b)
The farmer sprays a section of the field if its quadrat contains more than 4 weeds. Find the probability that a randomly placed quadrat leads to spraying.
(2)
(c)
Ten quadrats are placed independently in different parts of the field. Using the unrounded value from (b), find the probability that exactly 2 of the ten quadrats lead to spraying.
(2)
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Notes
Part (a)
A1 \(W \sim \mathrm{Po}(2.0)\) stated or used
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A1 \(P(W=0) = e^{-2} = 0.135\)
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Part (b)
M1 \(1 - P(W \le 4)\)
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A1 \(0.0527\)
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Part (c)
M1 recognise binomial \(B(10, 0.052653\ldots)\) and attempt \(P(=2)\)
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A1 \(0.0809\)
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Common error: Multiply the rate by the quadrat area to get the mean before using the Poisson formula. In part (c) the new random variable is binomial, because each quadrat is a trial with a fixed probability of spraying.
103IBMI3H-4-05r18-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Passengers on a charter flight each have one checked suitcase with mass \(S \sim N(18.4,\ 2.1^2)\) kg and one cabin bag with mass \(B \sim N(7.2,\ 0.9^2)\) kg. A family of three travels with three suitcases and one cabin bag. All masses are independent. Let \(T\) be the total mass of the three suitcases and the cabin bag, in kg.
(a)
Find \(\mathrm{E}(T)\) and \(\mathrm{Var}(T)\).
(2)
(b)
Find the probability that \(T\) exceeds 60 kg.
(2)
(c)
The airline will check the family's bags again if \(T\) is in the heaviest 5% of totals. Find the mass above which the bags are checked again, to 3 significant figures.
M1 \(T \sim N(62.4,\ 14.04)\) and attempt \(P(T > 60)\)
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A1 \(0.739\)
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Part (c)
M1 inverse normal with probability 0.95 (or \(z = 1.645\)): \(62.4 + 1.645\sqrt{14.04}\)
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A1 \(68.6\) kg
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Common error: Variances add for independent variables, but standard deviations do not. Three separate suitcases contribute \(3\sigma^2\), not \((3\sigma)^2\).
104IBMI3H-4-05r18-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A standard wheat variety has a mean yield of 5.50 tonnes per hectare. A new fertiliser is tested on 12 randomly chosen plots of this variety. The sample mean yield is 5.84 tonnes per hectare and the unbiased estimate of the population standard deviation is 0.55 tonnes per hectare. Yields are assumed to be normally distributed. The agronomist tests, at the 5% significance level, whether the fertiliser increases the mean yield.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the value of the \(t\) test statistic and the \(p\)-value.
(3)
(c)
State the conclusion of the test, giving a reason.
(1)
(d)
State why a \(t\)-test rather than a test using the normal distribution is used here.
(1)
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Notes
Part (a)
A1 \(H_0: \mu = 5.50\)
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A1 \(H_1: \mu > 5.50\)
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Part (b)
M1 \(t = \dfrac{5.84 - 5.50}{0.55/\sqrt{12}}\) with 11 degrees of freedom
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A1 \(t = 2.14\)
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A1 \(p = 0.0277\)
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Part (c)
R1 \(0.0277 < 0.05\), so reject \(H_0\): there is sufficient evidence that the fertiliser increases mean yield
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Part (d)
R1 the population standard deviation is unknown and the sample is small (estimated from the sample)
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Exam tip: Use the sign in \(H_1\) to decide between a one-tailed and a two-tailed \(p\)-value. Compare the \(p\)-value with the significance level, then state the conclusion in the context of the question.
105IBMI3H-4-05r18-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
The landing roll distance of a light aircraft on a grass strip is assumed to be normally distributed. A pilot records the roll distance, in metres, on 16 randomly chosen landings and calculates a 95% confidence interval for the population mean as \((412.7,\ 437.3)\). The interval was found using the \(t\)-distribution.
(a)
Find the sample mean.
(1)
(b)
Find the unbiased estimate of the population standard deviation, to 3 significant figures. (Use \(t = 2.131\) for 15 degrees of freedom.)
(3)
(c)
The aircraft manual states that the mean roll distance is 440 m. Comment on this claim, using the confidence interval.
(2)
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Notes
Part (a)
A1 \(\dfrac{412.7+437.3}{2} = 425.0\) m
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Part (b)
M1 half-width \(= 12.3\)
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M1 \(12.3 = 2.131\times\dfrac{s}{\sqrt{16}}\)
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A1 \(s = 23.1\) m
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Part (c)
R1 440 lies outside the interval \((412.7,\ 437.3)\)
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A1 evidence at the 5% level that the true mean differs from 440 m (the manual value looks too high)
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Exam tip: A confidence interval is centred on the sample mean, so the mean is the midpoint and the margin is half the width. Use the \(t\)-value for \(n-1\) degrees of freedom when the standard deviation is estimated.
106IBMI3H-4-05r18-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
The height \(h\) cm of a maize seedling is measured \(d\) days after planting. \(d\): 5, 10, 15, 20, 25, 30 \(h\): 9.8, 17.5, 24.1, 30.6, 35.2, 41.0 A student models the data with \(h = a\,d^{\,b}\).
(a)
Use power regression to find \(a\) and \(b\), each to 3 significant figures.
(2)
(b)
The software gives \(R^2 = 0.999\) for this model. State what this suggests about the model.
(1)
(c)
Use your values of \(a\) and \(b\) from (a) to predict the height of a seedling 35 days after planting. Give your answer to 3 significant figures.
(2)
(d)
Comment on the reliability of your prediction in (c).
(1)
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Notes
Part (a)
A1 \(a = 2.76\)
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A1 \(b = 0.796\)
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Part (b)
R1 very strong fit: almost all of the variation in the (transformed) height is explained by the model
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Part (c)
M1 \(2.76\times 35^{0.796}\)
✓ 1
A1 \(46.8\) cm
✓ 1
Part (d)
R1 35 days is outside the data range (extrapolation) so the prediction may be unreliable, as growth may slow down or stop
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Common error: A high \(R^2\) shows a good fit within the data range only; it does not make predictions outside that range reliable. Do not round \(a\) and \(b\) more than the question allows before predicting.
107IBMI3H-4-03r25-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A volcanologist measures the daily sulfur dioxide emission (in tonnes) from a volcano on ten randomly chosen days: 412, 388, 455, 430, 397, 441, 420, 405, 468, 434. Assume the emissions are normally distributed.
(a)
Calculate the mean and the unbiased estimate of the population standard deviation. Give the standard deviation to 3 significant figures.
(2)
(b)
Find a 95% confidence interval for the population mean daily emission. Give the limits to 1 decimal place.
(3)
(c)
A government report claims the mean daily emission is 450 tonnes. State whether your interval supports this claim, giving a reason.
(1)
(d)
Hence state the upper limit of the 95% confidence interval, to the nearest tonne.
(1)
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Notes
Part (a)
A1 mean = 425 tonnes
✓ 1
A1 sample standard deviation s = 25.5 (accept 25.5)
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Part (b)
M1 use of a t-interval with 9 degrees of freedom (not a normal interval)
✓ 1
A1 critical value 2.262
✓ 1
A1 interval from 406.7 to 443.3 (ECF from (a))
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Part (c)
R1 450 lies outside the interval [406.7, 443.3], so the data do not support the claim
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Part (d)
A1 443 tonnes
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Exam tip: With a small sample and an unknown population standard deviation use the t-distribution with n − 1 degrees of freedom.
108IBMI3H-4-03r25-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A theatre manager compares the amount spent on refreshments per person (in €) at matinee and evening performances. Random samples gave the results below. Assume both populations are normally distributed with equal variances.
Matinee: n = 15, mean = 6.80, sample standard deviation = 1.90 Evening: n = 18, mean = 7.90, sample standard deviation = 2.10
The manager carries out a pooled two-sample t-test at the 5% significance level to decide whether the mean spend is greater at evening performances.
(a)
State suitable null and alternative hypotheses.
(1)
(b)
Calculate the pooled estimate of the common variance. Give your answer to 3 significant figures.
(2)
(c)
Find the value of the test statistic. Give your answer to 3 significant figures; use the unrounded pooled variance.
(3)
(d)
The one-tailed p-value for this test is 0.064. State the conclusion in context.
(2)
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Part (a)
A1 H0: μevening = μmatinee and H1: μevening > μmatinee
A1 insufficient evidence that the mean spend at evening performances is greater than at matinees
✓ 1
Exam tip: Compare the p-value directly with the significance level; for a one-tailed test do not halve or double the value your GDC reports without checking which tail it gives.
109IBMI3H-4-03r25-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A shipping company cleans the hulls of eight vessels and records each vessel's fuel consumption (tonnes per 100 nautical miles) on a standard route before and after cleaning.
Vessel
1
2
3
4
5
6
7
8
Before
12.8
15.2
11.4
14.0
13.1
16.5
12.2
14.6
After
12.1
14.9
11.6
13.2
12.4
15.5
11.8
13.7
The company uses a paired t-test at the 5% significance level to test whether cleaning reduces mean fuel consumption. Assume the differences are normally distributed.
(a)
State the null and alternative hypotheses in terms of the mean difference μd = (before − after).
(2)
(b)
Calculate the mean and the sample standard deviation of the eight differences (before − after). Give the standard deviation to 3 significant figures.
(2)
(c)
Find the value of the test statistic. Give your answer to 3 significant figures.
(2)
(d)
The p-value for this test is 0.0021. State the conclusion in context.
A1 mean = 0.575, sample standard deviation = 0.392
✓ 1
Part (c)
M1 t = mean ÷ (s/√8) with 7 degrees of freedom
✓ 1
A1 t = 4.15
✓ 1
Part (d)
R1 0.0021 < 0.05, so reject H0
✓ 1
A1 there is sufficient evidence that hull cleaning reduces mean fuel consumption
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Exam tip: In a paired test, the data are the differences: find them first, then treat them as a single sample with n − 1 degrees of freedom.
110IBMI3H-4-03r25-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
The alert level at a volcano is classified each day as Low (L), Moderate (M) or High (H). The level on any day depends only on the level on the previous day. The transition matrix is
where the columns give today's level (L, M, H in order) and the rows give tomorrow's level (L, M, H in order).
(a)
The level is Low on Monday. Find the probability that it is High on Wednesday.
(2)
(b)
The level is Low on Monday. Find the probability distribution of the level on the following Saturday (5 days later). Give each probability to 3 significant figures.
(3)
(c)
Find the long-term proportion of days on which the alert level is High. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
M1 attempt to use T2 (or sum over the possible levels on Tuesday)
✓ 1
A1 0.095
✓ 1
Part (b)
M1 use of T5 multiplied by the initial state vector (1, 0, 0)
✓ 1
A1 P(L) = 0.565, P(M) = 0.280
✓ 1
A1 P(H) = 0.155
✓ 1
Part (c)
M1 set up Ts = s with the probabilities summing to 1 (or high power of T)
✓ 1
A1 correct steady-state values, e.g. L = 17/32, M = 19/64
✓ 1
A1 P(H) = 11/64 = 0.172
✓ 1
Common error: Multiplying the vector by the matrix in the wrong order: with columns as the 'from' state, use T^n times the state vector (matrix first).
111IBMI3H-4-03r25-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Volcanologists record the maximum height H (km) reached by the ash plume for six eruptions with different mass eruption rates m (in units of 104 kg s−1).
m
1
2
5
10
20
50
H
2.1
2.4
3.0
3.5
4.2
5.3
It is suggested that H = amb, where a and b are constants.
(a)
Use your GDC to find the power regression model for H in terms of m. Give a and b to 3 significant figures.
(2)
(b)
The coefficient of determination is R2 = 0.998. Interpret this value in context.
(2)
(c)
Using a = 2.06 and b = 0.238, estimate the plume height for an eruption with m = 30. Give your answer to 3 significant figures.
(2)
(d)
Using b = 0.238, find the percentage increase in predicted plume height when the mass eruption rate is doubled. Give your answer to 3 significant figures.
(2)
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Part (a)
A1 a = 2.06
✓ 1
A1 b = 0.238
✓ 1
Part (b)
A1 about 99.8% of the variation in (the logarithm of) plume height is explained by the model
✓ 1
R1 so the power model is a very good fit to these data
✓ 1
Part (c)
M1 substitute m = 30 into 2.06m0.238
✓ 1
A1 4.63 km
✓ 1
Part (d)
M1 ratio of heights = 20.238 (the factor a cancels)
✓ 1
A1 17.9%
✓ 1
Exam tip: In a power model y = a x^b, multiplying x by k multiplies y by k^b, whatever the value of a.
112IBMI3H-4-07r27-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
Larvae of a fruit fly are found in lemons at a packing plant. The number of larvae in a single lemon has a Poisson distribution with mean 0.35, and lemons are independent. Lemons are packed in boxes of 12. Let N be the total number of larvae in a box.
(a)
Write down the distribution of N and find the probability that a box contains at most 3 larvae.
(3)
(b)
A box is rejected if it contains more than 7 larvae. Find the probability that a randomly chosen box is rejected.
(2)
(c)
A delivery consists of 5 boxes chosen independently. Using the unrounded value of p from part (b), find the probability that at least one box in the delivery is rejected. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
A1 N ~ Po(4.2)
✓ 1
M1 use of P(N ≤ 3)
✓ 1
A1 0.395
✓ 1
Part (b)
M1 1 − P(N ≤ 7)
✓ 1
A1 p = 0.0639
✓ 1
Part (c)
M1 recognise binomial, B(5, p)
✓ 1
M1 1 − (1 − p)5 (or 1 − P(0 rejected))
✓ 1
A1 0.281 (ECF from (b))
✓ 1
Exam tip: When a Poisson mean is given per item, scale it by the number of items first (here 12 × 0.35) before reading probabilities from the calculator.
113IBMI3H-4-07r27-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A maintenance planner models the daily status of a regional aircraft as Flying (F), Inspection (I) or Repair (R). The transition matrix T, whose columns give the probabilities of moving from F, I, R respectively (in that order), is T = [[0.86, 0.70, 0.40], [0.09, b, 0.30], [a, 0.10, 0.30]] (rows listed in turn).
(a)
Find the values of a and b.
(2)
(b)
The aircraft is Flying on Monday. Find the probability that it is Flying on Wednesday.
(3)
(c)
Find the long-term proportion of days on which the aircraft is Flying.
(3)
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Notes
Part (a)
A1 a = 1 − 0.86 − 0.09 = 0.05
✓ 1
A1 b = 1 − 0.70 − 0.10 = 0.20
✓ 1
Part (b)
M1 two-step transition (T2 or sum of paths)
✓ 1
M1 0.86 × 0.86 + 0.70 × 0.09 + 0.40 × 0.05
✓ 1
A1 0.823
✓ 1
Part (c)
M1 solve Ts = s with entries summing to 1 (or high power of T)
✓ 1
A1 steady state (F, I, R) ≈ (0.807, 0.119, 0.0746)
✓ 1
A1 proportion of days Flying = 0.807
✓ 1
Common error: In a column-stochastic matrix each column sums to 1, not each row. Check this before multiplying, or the state vector will not stay a set of probabilities.
114IBMI3H-4-07r27-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
An airline records the ground roll (distance used to lift off) of 8 flights of the same aircraft type from the same runway, in metres: 1480, 1525, 1390, 1610, 1455, 1500, 1575, 1440. The ground roll may be assumed to be normally distributed.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation. Give the standard deviation to 3 significant figures.
(2)
(b)
Explain why a t-distribution, rather than the normal distribution, is used to construct a confidence interval for the mean here.
(1)
(c)
Find a 95% confidence interval for the mean ground roll, giving both limits to the nearest metre.
(3)
(d)
The airline's brochure claims that the mean ground roll is 1600 m. State, with a reason, whether the interval supports the claim.
(1)
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Notes
Part (a)
A1 mean = 1496.875 (≈ 1497) m
✓ 1
A1 sn−1 = 72.1 m
✓ 1
Part (b)
The population standard deviation is unknown and estimated from a small sample
✓ 1
Part (c)
M1 use of t with 7 degrees of freedom (critical value 2.365)
✓ 1
M1 mean ± 2.365 × 72.1 ÷ √8 (margin ≈ 60.3)
✓ 1
A1 (1437, 1557)
✓ 1
Part (d)
R1 No – 1600 lies outside the interval, so the data do not support a mean of 1600 m
✓ 1
Common error: Use the sample standard deviation with divisor n − 1 (sn−₁), not the population value σn that some calculators display first.
115IBMI3H-4-07r27-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A feed supplier claims that a new feed raises the mean milk yield of dairy cows above 28.0 litres per day. A farmer feeds 12 cows with the new feed and records the daily yields, which may be assumed to be normally distributed. The sample mean is 29.4 litres and the sample standard deviation (unbiased) is 2.6 litres. The farmer tests the claim at the 5% significance level.
(a)
State suitable null and alternative hypotheses.
(2)
(b)
Find the p-value for the test.
(2)
(c)
State the conclusion of the test in context.
(2)
(d)
Find the smallest sample mean, to 2 decimal places, that would lead to H0 being rejected, assuming the same sample size and standard deviation.
(2)
Show mark scheme
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Notes
Part (a)
A1 H0: μ = 28.0
✓ 1
A1 H1: μ > 28.0 (one-tailed)
✓ 1
Part (b)
M1 one-sample t-test with 11 degrees of freedom (t ≈ 1.87)
✓ 1
A1 p = 0.0445
✓ 1
Part (c)
R1 0.0445 < 0.05
✓ 1
A1 Reject H0: there is sufficient evidence that the new feed raises the mean yield above 28.0 litres per day
✓ 1
Part (d)
M1 28.0 + 1.796 × 2.6 ÷ √12 (critical t = 1.796 for 11 degrees of freedom)
✓ 1
A1 29.35 litres
✓ 1
Exam tip: State whether the test is one-tailed or two-tailed from the wording of the claim ("above" gives one-tailed) before finding the p-value; compare p directly with the significance level.
116IBMI3H-4-07r27-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A music technician measures the length, L cm, of a guitar string and the frequency, f Hz, of the note it produces at constant tension: L: 30, 36, 42, 48, 54, 60, 66 f: 441, 365, 316, 274, 246, 219, 201 A model of the form f = aLb is proposed.
(a)
Use power regression to find a and b, each to 3 significant figures.
(2)
(b)
The coefficient of determination is R2 = 0.9997. Comment on what this suggests about the model.
(1)
(c)
Use the model, with the 3 significant figure values of a and b, to predict the frequency when L = 45 cm. Give your answer to the nearest hertz.
(2)
(d)
The technician needs a note of frequency 440 Hz. Use the model, with the 3 significant figure values of a and b, to find the required length of string, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 a = 13 100 (1.31 × 104)
✓ 1
A1 b = −0.998
✓ 1
Part (b)
A1 The model fits the data very closely (very strong fit)
✓ 1
Part (c)
M1 13 100 × 45−0.998
✓ 1
A1 293 Hz
✓ 1
Part (d)
M1 440 = 13 100 × L−0.998
✓ 1
M1 L = (440 ÷ 13 100)1/(−0.998)
✓ 1
A1 30.0 cm
✓ 1
Exam tip: To rearrange a power model f = aLb for L, divide by a and then raise both sides to the power 1 ÷ b.
117IBMI3H-4-04r28-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
On a fishing vessel the number of unwanted fish (bycatch) in each trawl follows a Poisson distribution with mean 3.2. The numbers of bycatch fish in different trawls are independent.
(a)
Find the probability that a trawl contains at most 2 bycatch fish.
(2)
(b)
Find the probability that two trawls together contain at least 8 bycatch fish.
(2)
(c)
Five trawls are made. Find the probability that exactly 2 of them contain no bycatch at all.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of Poisson with mean 3.2, P(X ≤ 2)
✓ 1
A1 0.380
✓ 1
Part (b)
M1 total bycatch is Poisson with mean 6.4
✓ 1
A1 P(Y ≥ 8) = 1 − P(Y ≤ 7) = 0.313
✓ 1
Part (c)
M1 p = P(X = 0) = e−3.2 (= 0.0408)
✓ 1
M1 use of binomial B(5, p) with P(2 successes)
✓ 1
A1 0.0147
✓ 1
Exam tip: Sums of independent Poisson variables are Poisson with the means added; when a new variable is defined as a count of trawls, switch to a binomial model.
118IBMI3H-4-04r28-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Two tasks are carried out by an assembly robot. The time taken for task A is normally distributed with mean 12 s and standard deviation 1.5 s. The time taken for task B is normally distributed with mean 15 s and standard deviation 2 s. All task times are independent.
(a)
Find the probability that a single run of task B takes longer than a single run of task A.
(3)
(b)
In one cycle the robot performs task A twice (two separate runs) and then task B once. Let T be the total time for a cycle. Find E(T) and Var(T), and hence find the time t such that 90% of cycles take less than t seconds. Give t to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 D = B − A has E(D) = 3
✓ 1
A1 Var(D) = 22 + 1.52 = 6.25
✓ 1
A1 P(D > 0) = 0.885
✓ 1
Part (b)
A1 E(T) = 2(12) + 15 = 39
✓ 1
A1 Var(T) = 2(1.52) + 22 = 8.5
✓ 1
M1 uses z = 1.2816 with T ~ N(39, 8.5)
✓ 1
A1 t = 42.7 s
✓ 1
Common error: Variances add for both sums and differences of independent variables: Var(B − A) is Var(B) + Var(A), never a subtraction.
119IBMI3H-4-04r28-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A fisheries scientist measures the lengths (in cm) of a random sample of 12 mackerel from one catch:
34.1
36.8
31.5
35.2
33.9
38.0
32.4
35.7
34.6
30.8
37.1
33.5
The lengths are assumed to come from a normally distributed population.
(a)
Calculate the sample mean and the unbiased estimate of the population standard deviation.
(2)
(b)
Find a 95% confidence interval for the population mean length. Give the limits to 3 significant figures.
(3)
(c)
A fishing association claims that the population mean length is 35 cm. State, with a reason, whether the interval is consistent with this claim.
(1)
(d)
State why the t-distribution rather than the normal distribution is used.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 sample mean = 34.5 cm
✓ 1
A1 sn−1 = 2.23 cm
✓ 1
Part (b)
M1 uses the t-distribution with 11 degrees of freedom, t = 2.201
✓ 1
M1 interval uses x̄ ± t × s/√12
✓ 1
A1 (33.0, 35.9) cm
✓ 1
Part (c)
R1 35 lies within the interval, so the claim is consistent with the sample
✓ 1
Part (d)
R1 The population standard deviation is unknown and the sample is small (n = 12)
✓ 1
Exam tip: Report the interval limits as a pair and relate the claimed value to the interval; being inside the interval does not prove the claim true.
120IBMI3H-4-04r28-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A robotics team compares the battery life, in hours, of two models of mobile robot. A random sample of each model is tested under the same conditions.
Model X
14.2
15.1
13.8
16.0
14.9
15.5
13.5
14.6
15.8
Model Y
15.2
16.4
14.6
16.5
15.7
15.3
16.9
15.8
–
The team assumes that battery life is normally distributed for both models, with equal population variances. A two-sample pooled t-test is carried out at the 5% significance level to decide whether the population mean battery lives differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Use your calculator to find the p-value of the test. Give your answer to 3 significant figures.
(2)
(c)
State the conclusion of the test, giving a reason and writing your answer in context.
(2)
(d)
The team now repeats the test at the 1% significance level. State the conclusion, with a reason.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μX = μY (the population mean battery lives are equal)
✓ 1
A1 H1: μX ≠ μY (the population mean battery lives differ)
✓ 1
Part (b)
M1 uses a two-sample pooled t-test (two-tailed)
✓ 1
A1 p-value = 0.0278
✓ 1
Part (c)
R1 0.0278 < 0.05
✓ 1
A1 reject H0: there is sufficient evidence that the mean battery lives of the two models differ
✓ 1
Part (d)
R1 0.0278 > 0.01, so do not reject H0: there is not sufficient evidence at the 1% level that the mean battery lives differ
✓ 1
Exam tip: Compare the p-value with the significance level and write the conclusion in context; avoid saying the null hypothesis is "proved" or "accepted".
121IBMI3H-4-04r28-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Each day a delivery robot is in exactly one of three states: Working (W), Idle (I) or Under maintenance (M). The states are modelled by a Markov chain with transition matrix
from W
from I
from M
to W
0.80
0.30
0.50
to I
0.15
0.60
0.10
to M
0.05
0.10
0.40
Each column sums to 1. The robot is Working today.
(a)
Find the probability that the robot is Under maintenance two days from now.
(2)
(b)
Find the long-run proportion of days the robot is in each state. Give your answers to 3 significant figures.
(3)
(c)
Over a period of 200 days, estimate the number of days the robot is Under maintenance, assuming the long-run proportions apply. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses T2 applied to (1, 0, 0) or sums the three routes
✓ 1
A1 0.075
✓ 1
Part (b)
M1 solves Ts = s with the entries summing to 1
✓ 1
A1 W: 23/36 = 0.639
✓ 1
A1 I: 19/72 = 0.264 and M: 7/72 = 0.0972
✓ 1
Part (c)
M1 200 × (7/72)
✓ 1
A1 19.4 days
✓ 1
Common error: Check that every column (not every row) sums to 1 for this convention, and that the steady-state vector is normalised before it is used as a set of proportions.
122IBMI3H-4-04r28-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The monthly number of orders N received by an online fashion store is modelled by N = 120 × 1.18t, where t is the number of months since the store opened. The model was fitted to data for 0 ≤ t ≤ 10 and has a coefficient of determination R2 = 0.991.
(a)
Use the model to estimate the number of orders when t = 8. Give your answer to the nearest whole number.
(2)
(b)
Interpret the value 1.18 in the model.
(1)
(c)
Interpret the value of R2 in context.
(1)
(d)
Find the value of t at which the model predicts 1000 orders per month. Give your answer to 3 significant figures.
(2)
(e)
Comment on the reliability of using this model to predict orders when t = 36.
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 substitutes t = 8 into the model
✓ 1
A1 451 orders
✓ 1
Part (b)
R1 the number of orders increases by 18% each month
✓ 1
Part (c)
R1 about 99.1% of the variation in orders is explained by the model
✓ 1
Part (d)
M1 solves 120 × 1.18t = 1000
✓ 1
A1 t = 12.8 months
✓ 1
Part (e)
R1 unreliable: 36 is well beyond the data range (extrapolation), and growth of 18% per month cannot continue indefinitely
✓ 1
Exam tip: To solve an exponential equation, divide by the initial value first and then use logarithms or the calculator's equation solver.
123IBMI3H-4-05r29-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A beekeeper records the honey yield, in kg, of ten hives in one season: 21.4, 18.9, 24.2, 22.7, 19.6, 20.8, 23.1, 17.5, 22.0, 21.3. Assume the yields come from a normally distributed population.
(a)
Find the mean and the unbiased estimate of the population standard deviation.
(2)
(b)
Hence find a 95% confidence interval for the population mean yield.
(3)
(c)
A supplier claims that the mean yield is 23 kg. Comment on this claim using your interval.
(2)
(d)
Explain why the \(t\)-distribution is used rather than the normal distribution.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean 21.15 kg
✓ 1
A1 \(s_{n-1}=2.03\) kg
✓ 1
Part (b)
M1 use of the \(t\)-distribution with 9 degrees of freedom, \(t=2.262\)
✓ 1
M1 \(\bar x\pm t\frac{s}{\sqrt{10}}\)
✓ 1
A1 (19.7, 22.6) kg
✓ 1
Part (c)
R1 23 lies outside the interval
✓ 1
A1 evidence suggests the true mean is not 23 kg (claim not supported)
✓ 1
Part (d)
R1 the population standard deviation is unknown and the sample is small, so \(s\) is used as an estimate
✓ 1
Exam tip: Use the sample standard deviation with divisor \(n-1\) when the sample is used to estimate a population.
124IBMI3H-4-05r29-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A Formula 1 team compares two tyre compounds. Lap times (seconds) of eight laps on each compound, all set at the same stage of a stint, are: Soft: 91.82, 91.95, 92.10, 91.74, 92.03, 91.88, 92.15, 91.79 Medium: 92.08, 92.21, 92.34, 91.97, 92.29, 92.12, 92.40, 92.05 Assume both sets of lap times are normally distributed with the same population variance.
(a)
State suitable null and alternative hypotheses to test whether the mean lap time on soft tyres is lower than on medium tyres.
(2)
(b)
Carry out a two-sample pooled \(t\)-test and find the \(p\)-value.
(2)
(c)
State the conclusion at the 5% significance level, giving a reason in context.
(2)
(d)
Calculate the pooled variance estimate. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_S=\mu_M\)
✓ 1
A1 \(H_1:\mu_S<\mu_M\)
✓ 1
Part (b)
M1 attempt at pooled two-sample \(t\)-test, 14 degrees of freedom
✓ 1
A1 \(p=0.00258\)
✓ 1
Part (c)
R1 \(p<0.05\) so reject \(H_0\)
✓ 1
A1 sufficient evidence that mean lap time is lower on the soft compound
✓ 1
Part (d)
M1 \(s_p^2=\frac{7s_S^2+7s_M^2}{14}\)
✓ 1
A1 0.0228 s2
✓ 1
Exam tip: Check that the equal-variance assumption is stated before using the pooled test.
125IBMI3H-4-05r29-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
In a season, a driver either scores points (P) or does not (N) in each race. After a race in which the driver scored points, the probability of scoring in the next race is 0.7. After a race with no points, the probability of scoring in the next race is 0.4. The transition matrix is \(T=\begin{pmatrix}0.7&0.4\\0.3&0.6\end{pmatrix}\), with states in the order P, N, and each column giving the probabilities from the current state. The driver scores points in race 1.
(a)
Find the probability that the driver scores points in race 4.
(3)
(b)
Find the long-run probability that the driver scores points in a race.
(3)
(c)
In the long run, estimate the number of races in which the driver scores points in a 24-race season. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(T^3\begin{pmatrix}1\\0\end{pmatrix}\)
✓ 1
A1 \(\begin{pmatrix}0.583\\0.417\end{pmatrix}\)
✓ 1
A1 0.583
✓ 1
Part (b)
M1 \(T\pi=\pi\) with \(\pi_P+\pi_N=1\)
✓ 1
A1 \(0.3\pi_P=0.4\pi_N\)
✓ 1
A1 \(\pi_P=\frac{4}{7}=0.571\)
✓ 1
Part (c)
M1 \(24\times\frac{4}{7}\)
✓ 1
A1 13.7
✓ 1
Common error: In a steady-state calculation, use the probabilities from \(T\pi=\pi\) and the sum condition together; one equation alone cannot give a unique answer.
126IBMI3H-4-07r33-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Alarms from the gearbox monitoring system of an offshore wind turbine occur independently at a constant mean rate of 0.8 per week.
(a)
Find the probability that exactly 2 alarms occur in a period of 3 weeks.
(3)
(b)
Find the probability that at least one alarm occurs in each of 3 consecutive weeks.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mean for 3 weeks is 3 × 0.8 = 2.4, so Y ~ Po(2.4)
✓ 1
M1 P(Y = 2) = e−2.4 × 2.42 ÷ 2!
✓ 1
A1 0.261
✓ 1
Part (b)
M1 P(at least one alarm in a week) = 1 − e−0.8 (= 0.5507...)
✓ 1
M1 weeks are independent, so cube the probability: (1 − e−0.8)3
✓ 1
A1 0.167
✓ 1
Exam tip: When the time interval changes, scale the Poisson mean in proportion before calculating; here use 2.4, not 0.8.
127IBMI3H-4-07r33-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
The number of cases of a waterborne illness reported in a district in one month is modelled by a Poisson distribution with mean 6.5 when conditions are normal. The health authority issues an alert in any month when the number of cases is greater than c, where c is the smallest integer for which the probability of an alert in a normal month is less than 0.05.
(a)
Find the value of c.
(3)
(b)
Assuming normal conditions and independent months, find the probability that at least one alert is issued during a period of 12 months. Use the unrounded probability from the Poisson model.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ Po(6.5) and attempt to compare P(X > c) or P(X ≤ c) with 0.05 (0.95) for several values
M1 P(alert in a month) = P(X ≥ 12) = 0.03388 (accept 0.0339 or better)
✓ 1
M1 1 − (1 − 0.03388)12 (binomial, or complement of no alerts)
✓ 1
A1 0.339
✓ 1
Common error: Decide the strict inequality in P(X > c) carefully: an alert needs more than c cases, so the relevant tail starts at c + 1.
128IBMI3H-4-07r33-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The daily energy output, X kWh, of a solar array is normally distributed with mean 42 and standard deviation 6. Outputs on different days are independent. Let W be the total output over 7 days.
(a)
State the mean and variance of W.
(2)
(b)
Find P(W < 270).
(2)
(c)
A student claims that W = 7X. Using this claim, find P(7X < 270), and explain why it is different from your answer to part (b).
R1 the sum of 7 independent days varies less (variance 252 rather than 1764) because high and low days partly cancel, whereas 7X scales a single day's variation
✓ 1
Common error: A sum of n independent copies of X has variance nσ², but n times a single X has variance n²σ². Check which one the context describes.
129IBMI3H-4-07r33-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The times, in hours, between maintenance stops for a random sample of 9 turbine yaw motors are 412, 398, 431, 405, 389, 420, 415, 402, 407. Assume the times are normally distributed.
(a)
Find unbiased estimates of the population mean and population variance.
(2)
(b)
Construct a 95% confidence interval for the population mean.
(3)
(c)
The operator requires a mean time between maintenance stops of at least 420 hours. State whether the interval supports this requirement, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 408.8 hours (408.777...)
✓ 1
A1 variance = 155 hours2 (154.94..., from sn−1 = 12.45)
✓ 1
Part (b)
M1 use of the t-distribution with 8 degrees of freedom (critical value 2.306...)
✓ 1
M1 408.78 ± 2.306 × 12.45 ÷ √9
✓ 1
A1 (399.2, 418.3)
✓ 1
Part (c)
A1 it does not support the requirement
✓ 1
R1 420 lies above the upper limit of the interval, so it is not a plausible value of the mean at the 95% level
✓ 1
Exam tip: Because the population standard deviation is unknown and the sample is small, use the t-distribution with n − 1 degrees of freedom, not the normal distribution.
130IBMI3H-4-07r33-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
A manufacturer states that the mean efficiency of its solar panels is 19.2%. A buyer suspects the true mean is lower and tests 8 randomly chosen panels. Their efficiencies (%) are 19.1, 18.8, 19.4, 19.0, 18.6, 19.3, 18.9, 19.2. Assume efficiencies are normally distributed.
(a)
State suitable null and alternative hypotheses for a t-test.
(2)
(b)
Find the p-value for the test.
(2)
(c)
Using a 5% significance level, state the conclusion in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μ = 19.2
✓ 1
A1 H1: μ < 19.2
✓ 1
Part (b)
M1 one-tailed one-sample t-test with sample mean 19.04 and s = 0.267 (t = −1.72, 7 degrees of freedom)
✓ 1
A1 p = 0.0644
✓ 1
Part (c)
M1 0.0644 > 0.05, so do not reject H0
✓ 1
A1 there is insufficient evidence that the mean efficiency of the panels is below 19.2%
✓ 1
Common error: Write the conclusion in context and in terms of evidence: "do not reject H0" is not the same as proving the manufacturer's claim true.
131IBMI3H-4-07r33-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
A retailer compares the amounts spent (in €) by customers in two stores. Random samples give the following. Store A (8 customers): 24.5, 31.2, 28.7, 35.0, 26.4, 30.1, 29.8, 33.6 Store B (7 customers): 27.9, 22.4, 25.1, 29.0, 23.6, 26.2, 24.8 Assume both populations are normally distributed with equal variances.
(a)
State the null and alternative hypotheses for a test of whether the mean spend differs between the stores.
(2)
(b)
Carry out a two-sample pooled t-test and write down the p-value.
(2)
(c)
State the conclusion at the 5% significance level, in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μA = μB
✓ 1
A1 H1: μA ≠ μB (two-tailed)
✓ 1
Part (b)
M1 pooled two-sample t-test (means 29.91 and 25.57, t = 2.80, 13 degrees of freedom)
✓ 1
A1 p = 0.0150
✓ 1
Part (c)
M1 0.0150 < 0.05, so reject H0
✓ 1
A1 there is evidence that the mean amount spent differs between the two stores
✓ 1
Exam tip: Pooled tests assume equal population variances; state that assumption when you use the pooled option on your calculator.
132IBMI3H-4-05r34-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Two training programmes, X and Y, are compared for improving the standing jump height of athletes. Six athletes follow programme X and seven different athletes follow programme Y. The jump heights, in cm, after training are Programme X: 38, 41, 44, 39, 42, 45 Programme Y: 35, 37, 40, 34, 38, 36, 39 Assume both populations are normally distributed with equal variances. The test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses for a test of whether the mean jump heights differ.
(2)
(b)
Carry out a two-sample pooled t-test and find the p-value.
(2)
(c)
State the conclusion of the test, in context.
(2)
(d)
State why the data from the two groups are suitable for a two-sample test rather than a paired test.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: μX = μY
✓ 1
H1: μX ≠ μY
✓ 1
Part (b)
Use of a pooled two-sample t-test with 11 degrees of freedom
✓ 1
p-value = 0.00690 (3 s.f.)
✓ 1
Part (c)
0.00690 < 0.05, so reject H0
✓ 1
There is evidence that the mean jump heights differ between the two programmes
✓ 1
Part (d)
Different athletes in each group, so the samples are independent (and of different sizes)
✓ 1
Common error: A pooled test is only valid when the two variances can reasonably be assumed equal; state this assumption rather than leaving it out.
133IBMI3H-4-05r34-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
The brightness of a variable star, measured in magnitudes, is recorded on 10 independent nights. The sample mean is 12.46 and the unbiased estimate of the population standard deviation is 0.31. The measurements may be assumed to come from a normal population.
(a)
Find a 95% confidence interval for the population mean brightness. Give the interval to 3 significant figures.
(3)
(b)
A catalogue states that the true mean brightness is 12.8. Comment on whether the interval supports this value.
(2)
(c)
Explain why a t-distribution is used rather than the normal distribution.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Use of a t-interval with 9 degrees of freedom, t = 2.262
✓ 1
Margin = 2.262 × 0.31 ÷ √10 = 0.222
✓ 1
Interval (12.2, 12.7), the upper limit being 12.7
✓ 1
Part (b)
12.8 lies outside the interval
✓ 1
The data are not consistent with a true mean of 12.8 at the 95% confidence level
✓ 1
Part (c)
The population standard deviation is unknown and the sample is small
✓ 1
Exam tip: State the degrees of freedom (n − 1) when you use a t-interval; it shows the right distribution has been chosen.
134IBMI3H-4-05r34-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Rail passengers choose each month between three operators A, B and C. The monthly transition matrix, with columns giving the current operator and rows the next month's operator, is \(\mathbf{T}=\begin{pmatrix}0.7 & 0.1 & 0.2\\ 0.2 & 0.8 & 0.2\\ 0.1 & 0.1 & 0.6\end{pmatrix}\). Initially the proportions of passengers using A, B and C are 0.5, 0.3 and 0.2 respectively.
(a)
Find the proportion of passengers using B after two months.
(3)
(b)
Find the long-term proportions of passengers using each operator.
(3)
(c)
In the long term there are 12 000 passengers in total. Find the number using operator A.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
s2 = T2s0 (or two single-step calculations)
✓ 1
After one month: (0.42, 0.38, 0.2)
✓ 1
Proportion using B = 0.428
✓ 1
Part (b)
Solve Ts = s with the components summing to 1
✓ 1
Use of invariant vector from the eigenvector for eigenvalue 1
✓ 1
A: 0.3, B: 0.5, C: 0.2
✓ 1
Part (c)
12 000 × 0.3 = 3600
✓ 1
Exam tip: For a steady state, use the condition that the components add to 1 as well as Ts = s; otherwise the equations have infinitely many solutions.
135IBMI3H-4-05r34-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A rail company asks 310 commuters for their ticket type and the period in which they usually travel. The results are: Morning peak: Single 48, Day pass 35, Season 17 Off-peak: Single 30, Day pass 52, Season 28 Evening: Single 22, Day pass 33, Season 45 A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Find the expected frequency for commuters who travel in the morning peak with a single ticket. Give your answer to 3 significant figures.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 test statistic, to 3 significant figures.
(2)
(e)
The critical value at the 5% level for these degrees of freedom is 9.488. State the conclusion in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: ticket type and travel period are independent
✓ 1
Part (b)
Row total 100 and column total 100, total 310
✓ 1
Expected = 100 × 100 ÷ 310 = 32.3 (3 s.f.)
✓ 1
Part (c)
(3 − 1)(3 − 1) = 4
✓ 1
Part (d)
Use of Σ(O − E)2 ÷ E over the nine cells
✓ 1
χ2 = 29.3 (3 s.f.)
✓ 1
Part (e)
29.3 > 9.488, so reject H0
✓ 1
There is evidence that ticket type depends on the period of travel
✓ 1
Exam tip: Compare the statistic with the critical value (or the p-value with 0.05) and then state the conclusion in words about the context.
136IBMI3H-4-05r37-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium5 marks
Rescues by the lifeguards at a lake swimming area occur at random and independently, at a constant mean rate of 1.5 per week. Let R be the number of rescues in a period of 4 weeks.
(a)
Find the smallest integer k such that P(R > k) < 0.05. Show the distribution you use and the values that justify your answer.
(3)
(b)
Find the probability that there are fewer than 5 rescues in the 4-week period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 R ~ Po(6)
✓ 1
M1 evaluates P(R ≤ 9) or P(R > 9) (= 0.9161 or 0.0839) and P(R ≤ 10) or P(R > 10) (= 0.9574 or 0.0426)
✓ 1
A1 k = 10
✓ 1
Part (b)
M1 P(R ≤ 4) with Po(6)
✓ 1
A1 0.285
✓ 1
Common error: Using λ = 1.5 for a four-week period; the Poisson mean must be scaled to the length of the interval first.
137IBMI3H-4-05r37-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium8 marks
A library compares the loan period (days) for fiction and non-fiction books. A random sample of 15 fiction loans has mean 18.4 and standard deviation 3.2 (sn−1). A random sample of 12 non-fiction loans has mean 16.1 and standard deviation 3.6 (sn−1). Assume both populations are normal with equal variances. The library tests at the 5% level whether the mean loan periods differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Calculate the pooled estimate of the common variance, to 3 significant figures.
(2)
(c)
Calculate the value of the test statistic t, to 3 significant figures.
(2)
(d)
Using 25 degrees of freedom, state the conclusion in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μF = μN
✓ 1
A1 H1: μF ≠ μN
✓ 1
Part (b)
M1 (14 × 3.22 + 11 × 3.62)/25
✓ 1
A1 11.4
✓ 1
Part (c)
M1 t = (18.4 − 16.1)/√(11.44 × (1/15 + 1/12))
✓ 1
A1 1.76
✓ 1
Part (d)
M1 compares with the two-tailed critical value 2.060, or finds p = 0.0913 > 0.05
✓ 1
R1 do not reject H0: insufficient evidence that the mean loan periods differ
✓ 1
Exam tip: For a two-tailed test at 5%, compare the p-value with 0.05 (not 0.025), or use the critical value from the 97.5th percentile.
138IBMI3H-4-04r38-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
Steel beams used in a footbridge have lengths that are normally distributed with mean 5.40 m and standard deviation 0.03 m. Three beams, chosen independently, are joined end to end to form a span of total length \(T\) m.
(a)
Find the mean and the variance of \(T\).
(2)
(b)
Find the probability that the span is between 16.1 m and 16.3 m long.
(3)
(c)
A designer instead models the span as \(W=3B\), where \(B\) is the length of a single beam. Find the standard deviation of \(W\), and explain why it is larger than the standard deviation of \(T\).
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(\mathrm{E}(T)=3\times5.40=16.2\)
✓ 1
A1 \(\mathrm{Var}(T)=3\times0.03^2=0.0027\)
✓ 1
Part (b)
M1 \(T\sim N(16.2,\,0.0027)\) with standard deviation \(0.0520\)
✓ 1
M1 attempt to find \(P(16.1
✓ 1
A1 \(0.946\)
✓ 1
Part (c)
R1 the beams in \(T\) vary independently, so some errors cancel; for \(W\) the same error is tripled and cannot cancel (variances: \(\mathrm{Var}(3B)=9\sigma^2\) but \(\mathrm{Var}(B_1+B_2+B_3)=3\sigma^2\))
✓ 1
A1 \(\text{SD}(W)=3\times0.03=0.09\)
✓ 1
Common error: \(3B\) (one beam taken three times) and \(B_1+B_2+B_3\) (three different beams) have the same mean but different variances: \(9\sigma^2\) against \(3\sigma^2\).
139IBMI3H-4-04r38-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
Containers move between three depots X, Y and Z each day. The transition matrix, in which each column gives the probabilities for a container at the depot it heads (X, Y, Z in order), is \(\mathbf{T}=\begin{pmatrix}0.6&0.2&0.1\\0.3&0.5&0.3\\0.1&0.3&0.6\end{pmatrix}\). A container is at X, Y or Z at the start of day 1 with probabilities \(0.5\), \(0.3\) and \(0.2\) respectively.
(a)
Find the probability that the container is at each depot at the start of day 2.
(2)
(b)
Find the probability that the container is at depot Z at the start of day 4.
(2)
(c)
Find the long-term probability that the container is at depot Z.
Common error: From the start of day 1 to the start of day 4 there are three transitions, so the power of \(\mathbf{T}\) is 3, not 4.
140IBMI3H-4-04r38-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
A marine biologist measures the carapace length of 15 randomly selected crabs from a bay. The lengths are assumed to be normally distributed. The sample mean is 8.42 cm and the unbiased estimate of the population standard deviation is 0.63 cm.
(a)
Find a 95% confidence interval for the population mean carapace length. Give the limits to three significant figures.
(3)
(b)
A researcher claims that the population mean is 8.0 cm. Comment on this claim using your interval.
(1)
(c)
In a larger study of 60 crabs from the bay, the population standard deviation is known to be 0.60 cm. Find the width of the 95% confidence interval for the mean from a sample of this size. Give your answer to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(t\)-interval with \(\nu=14\), \(t=2.145\) (unknown population standard deviation)
Exam tip: Use a \(t\)-interval when only the sample standard deviation is available, and a normal (\(z\)) interval when the population standard deviation is known.
141IBMI3H-4-04r38-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
The manager of an apartment block states that the mean waiting time for a lift is 45 seconds. A resident believes it is longer. She records the waiting times of 12 randomly chosen occasions. The sample mean is 47.8 seconds and the unbiased estimate of the population standard deviation is 5.9 seconds. Waiting times are assumed to be normally distributed.
(a)
State suitable null and alternative hypotheses for a test of the resident's belief.
(1)
(b)
Find the \(p\)-value for the test.
(2)
(c)
State the conclusion of the test at the 5% significance level, giving a reason.
(2)
(d)
The resident repeats her study with 40 randomly chosen occasions and obtains the same sample mean and standard deviation. Find the value of the test statistic for the new sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\mathrm{H}_0:\mu=45\) and \(\mathrm{H}_1:\mu>45\)
✓ 1
Part (b)
M1 one-tailed \(t\)-test with \(\nu=11\), test statistic \(t=1.64\)
✓ 1
A1 \(p=0.0642\)
✓ 1
Part (c)
A1 \(0.0642>0.05\), so do not reject \(\mathrm{H}_0\)
✓ 1
A1 there is insufficient evidence that the mean waiting time exceeds 45 seconds
✓ 1
Part (d)
M1 \(t=\frac{47.8-45}{5.9/\sqrt{40}}\)
✓ 1
A1 \(3.00\)
✓ 1
Common error: Failing to reject \(\mathrm{H}_0\) does not prove it true; the conclusion should state that there is insufficient evidence against it.
142IBMI3H-4-04r45-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Cargo ships arrive at a harbour at random, independently of each other, with a mean of 3.5 arrivals per hour.
(a)
Find the probability that more than 6 ships arrive in a two-hour period.
(3)
(b)
Find the probability that, in exactly one of the next three separate one-hour periods, no ships arrive.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mean for two hours is 7, so X ~ Po(7)
✓ 1
M1 P(X > 6) = 1 − P(X ≤ 6)
✓ 1
A1 0.550
✓ 1
Part (b)
M1 P(no arrivals in one hour) = e−3.5 = 0.0302
✓ 1
M1 binomial with n = 3, p = 0.0302 and exactly one success: 3 × 0.0302 × (1 − 0.0302)2
✓ 1
A1 0.0852
✓ 1
Common error: Using 3.5 for a two-hour period: the mean must be doubled to 7 before using the Poisson formula or the calculator.
143IBMI3H-4-04r45-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Two independent thermal sensors measure the temperature of the same lava lake. Readings from sensor W are normally distributed with mean 850 °C and standard deviation 12 °C. Readings from sensor V are normally distributed with mean 842 °C and standard deviation 9 °C. Let D = W − V.
(a)
State the distribution of D.
(2)
(b)
Find the probability that a reading from W is higher than a simultaneous reading from V.
(2)
(c)
The mean, M, of four independent readings from sensor W is calculated. Find P(M > 856).
(2)
(d)
Find the standard deviation of the mean of 25 independent readings from sensor W.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 E(D) = 850 − 842 = 8 and Var(D) = 144 + 81 (variances add, even for a difference)
✓ 1
A1 D ~ N(8, 225)
✓ 1
Part (b)
M1 P(D > 0) with D ~ N(8, 152)
✓ 1
A1 0.703
✓ 1
Part (c)
M1 M ~ N(850, 122/4) = N(850, 36), so P(M > 856) with standard deviation 6
✓ 1
A1 0.159
✓ 1
Part (d)
A1 12/√25 = 2.4 °C
✓ 1
Common error: For a difference of independent variables, subtract the means but still add the variances.
144IBMI3H-4-04r45-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A theatre manager records the running time, in minutes, of ten performances of a play: 148, 152, 145, 150, 156, 147, 151, 149, 153, 146. Assume the running times are normally distributed.
(a)
Find a 95 % confidence interval for the mean running time. Give the end points to 1 decimal place.
(3)
(b)
The producer states that the mean running time is 150 minutes. Comment on this statement using your interval.
(2)
(c)
Find the width of a 99 % confidence interval for the mean running time. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses a t-interval with sample mean 149.7 and sample standard deviation 3.40
✓ 1
A1 margin of error 2.43
✓ 1
A1 (147.3, 152.1)
✓ 1
Part (b)
A1 150 lies inside the interval
✓ 1
R1 so the data are consistent with the producer's claim; the interval does not provide evidence against it
✓ 1
Part (c)
M1 2 × 3.2498 × 3.40098/√10
✓ 1
A1 6.99 minutes
✓ 1
Common error: Using the normal distribution instead of the t-distribution when the population standard deviation is unknown and the sample is small.
145IBMI3H-4-04r45-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
The mass of cargo in a sample of 16 containers is used to construct a 95 % confidence interval for the population mean mass, giving (62.4, 69.2) tonnes. Assume the masses are normally distributed.
(a)
Find the sample mean and the margin of error.
(2)
(b)
Find the sample standard deviation. Use the unrounded critical value from your calculator and give your answer to 3 significant figures.
(3)
(c)
A 90 % confidence interval is calculated from the same sample. Find its margin of error to 3 significant figures, and state without calculation whether this interval is wider or narrower than the 95 % interval.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 sample mean = (62.4 + 69.2)/2 = 65.8 tonnes
✓ 1
A1 margin of error = 3.4 tonnes
✓ 1
Part (b)
M1 margin of error = t × s/√16
✓ 1
M1 t = 2.131 for 15 degrees of freedom, so s = 3.4 × 4/2.131
✓ 1
A1 s = 6.38 tonnes
✓ 1
Part (c)
M1 1.753 × 6.38/4 (t = 1.753 for 15 degrees of freedom)
✓ 1
A1 2.80 tonnes, so the 90 % interval is narrower
✓ 1
Exam tip: The sample mean is always the midpoint of a confidence interval, and half the width is the margin of error.
146IBMI3H-4-06r45-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A shipping company records whether container vessels were delayed on departure from three ports, Alpha, Beta and Gamma, during one year.
Alpha
Beta
Gamma
Delayed
18
31
26
Not delayed
62
49
54
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Show that the expected frequency of delayed vessels at Alpha is 25.
(2)
(c)
Write down the number of degrees of freedom and calculate the χ2 statistic. Give the statistic to 3 significant figures.
(2)
(d)
The p-value for the test is 0.0819. State, with a reason, the conclusion of the test.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: whether a vessel is delayed is independent of the port
✓ 1
Part (b)
M1 (row total × column total)/grand total = 80 × 75/240
✓ 1
A1 = 25
✓ 1
Part (c)
A1 degrees of freedom = (2 − 1)(3 − 1) = 2
✓ 1
A1 χ2 = 5.00
✓ 1
Part (d)
R1 0.0819 > 0.05
✓ 1
A1 do not reject H0; there is insufficient evidence that the chance of delay depends on the port
✓ 1
Exam tip: Always compare the p-value with the stated significance level and give the conclusion in the context of the question, not just ‘accept’ or ‘reject’.
147IBMI3H-4-06r45-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
A volcanologist records, on 8 days, the number of small earthquakes x beneath a volcano and the mass y of sulfur dioxide released, in hundreds of tonnes.
x
2
5
7
9
12
14
17
20
y
16
18
25
29
38
36
50
53
(a)
Find the Pearson’s product-moment correlation coefficient r and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x, giving the coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Estimate the mass of sulfur dioxide, in hundreds of tonnes, on a day with 10 small earthquakes. Give your answer to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.983
✓ 1
A1 strong positive linear correlation
✓ 1
Part (b)
M1 uses regression of y on x
✓ 1
A1 y = 2.20x + 9.49
✓ 1
Part (c)
R1 each additional small earthquake is associated with about 220 more tonnes of sulfur dioxide released
✓ 1
Part (d)
A1 31.5 hundred tonnes (accept 31.5 from the unrounded line or from 2.20 × 10 + 9.49 = 31.49)
✓ 1
Common error: Describe correlation with both strength and direction (for example ‘strong positive’); and remember that correlation alone does not prove one variable causes the other.
148IBMI3H-4-06r45-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
A seismic alarm is installed at a volcano. In any given month the probability that an eruption occurs is 0.04. If an eruption occurs, the alarm sounds with probability 0.90. If no eruption occurs, the alarm sounds (falsely) with probability 0.07.
(a)
Find the probability that the alarm sounds in a given month. Give your answer to 4 decimal places.
(3)
(b)
Given that the alarm sounds, find the probability that an eruption occurs. Give your answer to 3 significant figures.
M1 uses P(eruption | alarm) = P(eruption and alarm)/P(alarm)
✓ 1
A1 0.036/0.1032
✓ 1
A1 0.349
✓ 1
Exam tip: For a conditional probability, write the formula first and identify the numerator and denominator from your tree diagram.
149IBMI3H-4-06r45-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
A volcanologist measures the temperature of gas at a vent, in °C, on 10 randomly chosen occasions: 512, 498, 531, 505, 520, 489, 526, 515, 502, 509. The temperatures can be modelled as a random sample from a normal distribution.
(a)
Calculate the sample mean and the unbiased estimate of the population variance. Give the variance to 3 significant figures.
(2)
(b)
Construct a 95% confidence interval for the population mean temperature. Give the end-points to 3 significant figures.
(3)
(c)
A previous study stated that the mean gas temperature is 525 °C. Comment on this claim using your interval.
(1)
(d)
Explain why a t-distribution is used instead of the normal distribution.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 mean = 510.7 °C
✓ 1
A1 unbiased variance = 166
✓ 1
Part (b)
M1 uses the t-distribution with 9 degrees of freedom (critical value 2.262…)
✓ 1
M1 mean ± 2.262… × 12.89…/√10
✓ 1
A1 (501, 520)
✓ 1
Part (c)
R1 525 lies outside the interval, so the data do not support the claim that the mean is 525 °C
✓ 1
Part (d)
R1 the population variance is unknown and estimated from a small sample
✓ 1
Common error: Use the unbiased standard deviation (divisor n − 1) from the calculator’s sample statistics, not the population standard deviation, in a confidence interval for a mean.
150IBMI3H-4-07r47-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Calculator
A sports scientist models the weekly fitness of a footballer using two states: fit (F) or injured (I). If the player is fit at the start of a week, the probability of being injured at the start of the next week is 0.06. If injured, the probability of being fit at the start of the next week is r. Over many seasons, the player is fit for 85% of weeks.
(a)
Find the value of r.
(3)
(b)
Use r = 0.34. The player is injured at the start of week 1. Find the probability that the player is fit at the start of week 3.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 long-term proportions fit 0.85 and injured 0.15
✓ 1
M1 balance equation 0.06 × 0.85 = r × 0.15 (or 0.85 = 0.94 × 0.85 + r × 0.15)
✓ 1
A1 r = 0.34
✓ 1
Part (b)
A1 transition matrix with columns (0.94, 0.06) and (0.34, 0.66)
✓ 1
M1 use of T2 with initial state (0, 1)T (or two successive steps)
✓ 1
A1 0.544
✓ 1
Common error: Make sure each column of a transition matrix sums to 1 before multiplying; mixing up rows and columns changes the answer.
151IBMI3H-4-07r48-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
Fog delays at a harbour occur independently and at random, at a constant average rate of \(\lambda\) per week, so that the number of delays in a week follows a Poisson distribution. The harbour master finds that the probability of a week with no fog delays is 0.2.
(a)
Find the value of \(\lambda\), correct to 3 significant figures.
(2)
(b)
Find the probability that there are at least 3 fog delays in a two-week period. Use the unrounded value of \(\lambda\).
(3)
(c)
Over the next 5 weeks, find the probability that exactly 2 of the weeks have no fog delays.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(e^{-\lambda}=0.2\)
✓ 1
A1 \(\lambda=\ln 5=1.61\) (1.6094…)
✓ 1
Part (b)
M1 mean for two weeks \(=2\ln5=3.22\) (3.2188…)
✓ 1
M1 \(1-P(X\le2)\)
✓ 1
A1 0.624 (0.6240…)
✓ 1
Part (c)
M1 recognises binomial \(B(5,0.2)\) with \(P=\binom52(0.2)^2(0.8)^3\)
✓ 1
A1 0.2048 = 0.205
✓ 1
Exam tip: Read what the question asks for: once a Poisson probability has been found for one interval, counting such intervals is a separate binomial model.
152IBMI3H-4-07r48-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Calculator
Sulfur dioxide concentrations (parts per billion) were measured at two monitoring stations on the flank of a volcano. Station A (8 readings): 41.2, 39.8, 43.5, 40.1, 42.7, 44.0, 38.9, 41.6 Station B (9 readings): 44.3, 42.9, 45.8, 43.1, 46.2, 44.7, 41.8, 45.0, 43.6 Assume both samples come from normal populations with equal variances. A two-tailed test at the 5% significance level is used to decide whether the mean concentrations differ.
(a)
State the null and alternative hypotheses.
(1)
(b)
Perform a pooled two-sample t-test and find the p-value.
(3)
(c)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_A=\mu_B\) and \(H_1:\mu_A\neq\mu_B\)
✓ 1
Part (b)
M1 pooled two-sample t-test used
✓ 1
A1 \(t=-3.38\) with 15 degrees of freedom (accept \(+3.38\) if samples entered in the other order)
✓ 1
A1 \(p=0.00409\) (0.004088…)
✓ 1
Part (c)
R1 \(p=0.00409<0.05\)
✓ 1
A1 reject \(H_0\): there is sufficient evidence that the mean concentrations at the two stations differ
✓ 1
Exam tip: State the comparison of the p-value with the significance level, then give the conclusion in the context of the question.
153IBMI3H-4-07r48-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium7 marks
Calculator
The temperature of gas from a volcanic vent was measured at 10 randomly chosen times. The readings, in °C, can be assumed to come from a normal population with unknown variance. A 95% confidence interval for the population mean, based on these 10 readings, is [41.2, 46.8].
(a)
Write down the sample mean.
(1)
(b)
Find the sample standard deviation \(s\), to 3 significant figures.
(3)
(c)
Find the 99% confidence interval for the population mean. Give both end points to 3 significant figures.
(3)
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Notes
Part (a)
A1 \(\bar x=\frac{41.2+46.8}{2}=44.0\)
✓ 1
Part (b)
M1 margin of error \(=\frac{46.8-41.2}{2}=2.8\)
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M1 \(2.8=t\times\frac{s}{\sqrt{10}}\) with \(t=2.262\) (9 degrees of freedom)
Common error: With the standard deviation unknown the critical value comes from the t-distribution with \(n-1\) degrees of freedom, not from the normal distribution.
154IBMI3H-4-07r48-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Medium6 marks
Calculator
During the first seven days of a theatre run, the number of tickets sold each day, \(y\) (hundreds), on day \(x\) was:
Day \(x\)
1
2
3
4
5
6
7
Tickets \(y\)
14.2
21.5
26.1
28.3
28.9
26.4
22.0
The manager models the data by \(y=ax^2+bx+c\).
(a)
Use quadratic regression to find the values of \(a\), \(b\) and \(c\), to 3 significant figures.
(2)
(b)
Write down the coefficient of determination \(R^2\) and state what it suggests about the model.
(2)
(c)
Use the unrounded regression coefficients to predict the number of tickets sold, in hundreds, on day 8. Give your answer to 3 significant figures.
(1)
(d)
Explain why the model should not be used to predict sales on day 10.
R1 about 99.9% of the variation in sales is explained by the model, so it is a very good fit for these days
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Part (c)
A1 15.2 (15.171…)
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Part (d)
R1 the model predicts a negative value (−5.40) for day 10, which is impossible; day 10 is outside the data range (extrapolation)
✓ 1
Common error: Using the rounded coefficients in a prediction can change the third significant figure; keep the full GDC values for calculations.
155IBMI3H-4-04r28-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Hard7 marks
Small flaws occur at random along a roll of woven fabric. The number of flaws in a 10 m length of fabric follows a Poisson distribution with mean λ. The probability that a 10 m length has no flaws is 0.2.
(a)
Find the value of λ, giving your answer to 3 significant figures.
(2)
(b)
Find the probability that a 10 m length has at least 3 flaws. Use the unrounded value of λ.
(2)
(c)
A 10 m length is known to have at least one flaw. Find the probability that it has exactly 2 flaws. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
M1 sets e−λ = 0.2
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A1 λ = ln 5 = 1.61
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Part (b)
M1 1 − P(X ≤ 2) with λ = ln 5
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A1 0.219
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Part (c)
M1 P(X = 2 | X ≥ 1) = P(X = 2) / P(X ≥ 1)
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A1 P(X = 2) = 0.259 and P(X ≥ 1) = 0.8
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A1 0.324
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Common error: For a conditional probability the denominator is the probability of the condition (here at least one flaw, which is 0.8), not 1.
156IBMI3H-4-07r33-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Hard8 marks
A small island's daily electricity demand, D kWh, is normally distributed with mean 310 and standard deviation 20. Two identical, independent wind turbines each generate an amount of energy per day, G1 and G2 kWh, normally distributed with mean 150 and standard deviation 15. D, G1 and G2 are independent. A shortfall occurs on a day when demand exceeds the total generated by the two turbines. Let Y = D − (G1 + G2).
(a)
Find the distribution of Y, giving its mean and variance.
(3)
(b)
Find the probability that a shortfall occurs on a given day.
(2)
(c)
Over 5 independent days, find the probability that a shortfall occurs on exactly 2 days. Use the unrounded answer to part (b).
(3)
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Notes
Part (a)
A1 E(Y) = 310 − (150 + 150) = 10
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M1 Var(Y) = Var(D) + Var(G1) + Var(G2) (variances add, even for a difference)
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A1 Y ~ N(10, 850) (variance 400 + 225 + 225)
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Part (b)
M1 P(Y > 0) with Y ~ N(10, 850), standard deviation 29.15...
Common error: For a difference of independent variables the variances still add: Var(D − G) = Var(D) + Var(G).
157IBMI3H-4-04r45-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 1Hard8 marks
At a theatre, the number of late arrivals in a ten-minute interval at door A is X ~ Po(1.8) and at door B is Y ~ Po(2.7). The two doors operate independently. Let T = X + Y be the total number of late arrivals.
(a)
State the distribution of T and find P(T = 5).
(3)
(b)
Find the probability that exactly 2 people arrive late at door A and exactly 3 arrive late at door B.
(2)
(c)
Given that the total number of late arrivals is 5, find the probability that exactly 2 of them arrived at door A.
Exam tip: For conditional probability, write the formula P(A | B) = P(A and B)/P(B) first, then identify which values come from which distribution.
158IBMI3H-4-03r9-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Easy9 marks
In a theatre, the running time of Act 1, X minutes, is normally distributed with mean 48 and standard deviation 3. The running time of Act 2, Y minutes, is normally distributed with mean 55 and standard deviation 4. The two acts are independent. There is a fixed interval of 20 minutes between the acts, so the total time is T = X + Y + 20.
(a)
Find E(T) and Var(T).
(2)
(b)
Find the probability that the performance, including the interval, lasts longer than 130 minutes.
(2)
(c)
Find the probability that Act 2 is longer than Act 1.
(3)
(d)
Only 5% of performances last longer than k minutes in total. Find k, to the nearest minute.
(2)
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Notes
Part (a)
A1 E(T) = 48 + 55 + 20 = 123 minutes
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A1 Var(T) = 9 + 16 = 25 (the constant adds no variance)
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Part (b)
M1 T ~ N(123, 25), so standard deviation 5
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A1 P(T > 130) = 0.0808
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Part (c)
A1 Let D = Y − X; E(D) = 7
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A1 Var(D) = 16 + 9 = 25
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A1 P(D > 0) = 0.919
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Part (d)
M1 z-value for upper 5% is 1.645
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A1 k = 123 + 1.645 × 5 = 131 minutes
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Common error: Variances add for both the sum and the difference of independent variables; only the means subtract.
159IBMI-033-03-r1-05Normal distribution and chi-squared testsPaper 2Medium10 marks
The masses of adults using a lift are normally distributed with mean 78 kg and standard deviation 12 kg. The masses of children using the lift are normally distributed with mean 32 kg and standard deviation 6 kg. All masses are independent.
(a)
Twelve adults enter the lift. Find the probability that their total mass exceeds 1000 kg.
(4)
(b)
An adult and a child are chosen at random. Find the probability that the adult's mass is more than twice the child's mass.
(3)
(c)
The manufacturer wants the probability that the total mass of twelve adults exceeds the lift's rated load L to be 0.01. Find L, giving your answer to the nearest kg.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1: mean of total = 12 × 78 = 936 kg
✓ 1
M1: variance of total = 12 × 122 = 1728
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M1: P(T > 1000) where T ~ N(936, 1728)
✓ 1
A1: 0.0618
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Part (b)
M1: A − 2C has mean 78 − 64 = 14
✓ 1
M1: variance = 122 + 22 × 62 = 288
✓ 1
A1: P(A − 2C > 0) = 0.795
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Part (c)
M1: uses T ~ N(936, 1728)
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M1: P(T > L) = 0.01, using inverse normal with upper tail
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A1: L = 1033 kg (1032.7…)
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Common error: Var(2C) = 4Var(C), whereas Var(C₁ + C₂) = 2Var(C). Twice one child's mass is not the same as the total mass of two children.
160IBMI-033-04-r1-05T-tests, confidence intervals and Markov chainsPaper 2Medium9 marks
A language school claims that a one-week intensive course improves vocabulary test scores. Eight students take a vocabulary test before and after the course. Their scores are:
Differences may be assumed to be normally distributed. A test is carried out at the 5% significance level, for which the critical value is 1.895.
(a)
Explain why a paired t-test is appropriate here.
(1)
(b)
State suitable hypotheses, defining d = after − before.
(2)
(c)
Find the mean and the unbiased standard deviation \(s_{n-1}\) of the differences.
(2)
(d)
Calculate the value of the test statistic.
(2)
(e)
State, with a reason, the conclusion of the test in context.
(2)
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Notes
Part (a)
R1: each before score and after score belong to the same student, so the two samples are not independent
✓ 1
Part (b)
A1: H0: \(\mu_d = 0\), where \(\mu_d\) is the population mean improvement
✓ 1
A1: H1: \(\mu_d > 0\)
✓ 1
Part (c)
A1: \(\bar{d} = 3\)
✓ 1
A1: \(s_{n-1} = 2.51\)
✓ 1
Part (d)
M1: \(t = \frac{3}{2.507/\sqrt{8}}\)
✓ 1
A1: \(t = 3.38\)
✓ 1
Part (e)
R1: \(3.38 > 1.895\)
✓ 1
A1: reject H0; there is sufficient evidence at the 5% level that the course improves mean vocabulary scores
✓ 1
Exam tip: Define the difference direction (after − before) before writing hypotheses; it fixes whether H₁ uses > or <.
161IBMI-033-08-r2-05Normal distribution and chi-squared testsPaper 2Medium10 marks
The times, T seconds, that a swimmer takes to complete a 100 m race can be modelled by a normal distribution with mean μ and standard deviation σ. Records show that 12% of her times are under 62.0 s and 30% of her times are over 66.5 s.
(a)
Find the value of μ and the value of σ.
(5)
(b)
To qualify for a national event, the swimmer must record a time under 64.0 s. Find the probability that she qualifies in a single race.
(2)
(c)
During a season she swims 15 independent races. Find the probability that she records a qualifying time in at least 4 of these races.
(3)
Show mark scheme
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Notes
Part (a)
z = −1.175 (−1.17498…) for the lower 12%
✓ 1
z = 0.5244 (0.52440…) for the upper 30%
✓ 1
Forms two equations, e.g. μ − 1.175σ = 62.0 and μ + 0.5244σ = 66.5
✓ 1
μ = 65.1 s (65.111…)
✓ 1
σ = 2.65 s (2.6480…)
✓ 1
Part (b)
Finds P(T < 64.0) using the values of μ and σ found in the previous part
✓ 1
0.337 (0.33735…)
✓ 1
Part (c)
Recognises X ~ B(15, 0.337…)
✓ 1
Uses 1 − P(X ≤ 3)
✓ 1
0.800
✓ 1
Exam tip: Carry the unrounded values of μ and σ into the qualifying-time probability. Using the rounded values 65.1 and 2.65 gives 0.339, which is not accurate enough.
162IBMI3H-4-08-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A box of biscuits contains 12 biscuits and a cardboard tray. The mass of one biscuit, X grams, is normally distributed with mean 25 and variance 0.64. The mass of the tray, Y grams, is normally distributed with mean 40 and variance 4. All masses are independent. The total mass of a box is W = X1 + X2 + … + X12 + Y.
(a)
Find E(W) and Var(W).
(4)
(b)
Find the probability that a box has mass less than 335 g.
(2)
(c)
A student suggests modelling the biscuit total as \(12X\) instead. Find Var(12X) and compare it with the variance of the sum of 12 independent biscuits.
(2)
(d)
Explain why the two variances in part (c) differ.
(1)
Show mark scheme
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Notes
Part (a)
M1 \(E(W)=12\times25+40\)
✓ 1
A1 340 g
✓ 1
M1 \(\text{Var}(W)=12\times0.64+4\)
✓ 1
A1 11.68
✓ 1
Part (b)
M1 \(W\sim N(340,11.68)\)
✓ 1
A1 0.0717
✓ 1
Part (c)
A1 Var(12X) = \(12^2\times0.64=92.16\)
✓ 1
A1 the sum of 12 independent biscuits has variance \(12\times0.64=7.68\), much smaller
✓ 1
Part (d)
R1 independent biscuits vary in different directions, so heavy and light ones partly cancel; 12X assumes all 12 are equally heavy or light
✓ 1
Exam tip: Distinguish between 'n copies of one variable' (\(nX\), variance \(n^2\sigma^2\)) and 'n independent variables added' (variance \(n\sigma^2\)).
163IBMI3H-4-08-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium11 marks
The resting metabolic rate, R watts, of mammals of body mass m kg is recorded for six species.
A model of the form \(R=am^b\) is fitted to the data.
(a)
Use your calculator to find the values of a and b.
(2)
(b)
The calculator gives R2 = 0.9997 for this model. Interpret this value.
(2)
(c)
Estimate the metabolic rate of a mammal of mass 30 kg.
(2)
(d)
Find the factor by which the metabolic rate is multiplied when the body mass doubles, and comment on its size.
(2)
(e)
Estimate the body mass of a mammal with a metabolic rate of 150 W.
(3)
Show mark scheme
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Notes
Part (a)
A1 a = 4.20
✓ 1
A1 b = 0.751
✓ 1
Part (b)
A1 about 99.97% of the variation in (the transformed) metabolic rate is explained by the model
✓ 1
A1 the model fits the data very closely
✓ 1
Part (c)
M1 substitute m = 30 into \(R=am^b\)
✓ 1
A1 53.9 W (accept 54.0 from rounded parameters)
✓ 1
Part (d)
A1 \(2^{0.751}=1.68\)
✓ 1
R1 doubling the mass increases the rate by less than a factor of 2, so large animals use less energy per kilogram
✓ 1
Part (e)
M1 \(4.198m^{0.7507}=150\)
✓ 1
M1 \(m=\left(\frac{150}{a}\right)^{1/b}\)
✓ 1
A1 117 kg
✓ 1
Common error: To solve for the base of a power, raise both sides to the power \(1/b\); taking logarithms of only one side of the equation does not work.
164IBMI3H-4-08-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A manufacturer claims that the mean battery life of its tablets is 18 hours. A consumer group tests ten tablets and records the battery lives, in hours: 17.2, 16.8, 18.1, 16.5, 17.9, 17.4, 16.9, 17.7, 16.2, 17.3
The group suspects that the true mean is lower than claimed and carries out a one-tailed t-test at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the sample mean and the unbiased estimate of the population standard deviation.
(2)
(c)
Find the number of degrees of freedom, the t-statistic and the p-value.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
State one assumption needed for this test to be valid.
(1)
Show mark scheme
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Notes
Part (a)
A1 H0: μ = 18; H1: μ < 18
✓ 1
Part (b)
A1 mean = 17.2 hours
✓ 1
A1 sn−1 = 0.609 hours
✓ 1
Part (c)
A1 degrees of freedom = 9
✓ 1
A1 t = −4.15
✓ 1
A1 p-value = 0.00124
✓ 1
Part (d)
R1 0.00124 < 0.05
✓ 1
A1 reject H0: there is evidence that the mean battery life is less than 18 hours
✓ 1
Part (e)
R1 battery lives are (approximately) normally distributed (and the sample is random)
✓ 1
Common error: Use the sample standard deviation with divisor \(n-1\) (not the population standard deviation of the sample) when calculating the t-statistic by hand.
165IBMI3H-4-08-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
Eight athletes run a 100 m sprint before and after a six-week training programme. Their times, in seconds, are:
The coach claims that the programme reduces sprint times. Assume the differences in times are normally distributed.
(a)
State the null and alternative hypotheses for a paired t-test, where μd is the mean of (before − after).
(1)
(b)
Find the mean and the standard deviation (with divisor n − 1) of the eight differences.
(2)
(c)
Carry out the test at the 5% significance level, stating the t-statistic, the degrees of freedom and the p-value.
(3)
(d)
State your conclusion in context.
(2)
(e)
Explain why a paired test is more appropriate than a two-sample test here.
(1)
Show mark scheme
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Notes
Part (a)
A1 H0: μd = 0; H1: μd > 0
✓ 1
Part (b)
A1 mean = 0.300 s
✓ 1
A1 s = 0.207 s
✓ 1
Part (c)
A1 degrees of freedom = 7
✓ 1
A1 t = 4.10
✓ 1
A1 p-value = 0.00229
✓ 1
Part (d)
R1 0.00229 < 0.05
✓ 1
A1 reject H0: there is evidence that the training reduces mean sprint time
✓ 1
Part (e)
R1 the same athletes appear in both samples, so the data are matched pairs, not independent samples
✓ 1
Exam tip: For a paired test, work with the single list of differences and keep the sign convention you stated in your hypotheses.
166IBMI3H-4-08-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A gardener compares the heights of seedlings, in cm, after four weeks with fertiliser A (8 seedlings) and fertiliser B (8 seedlings).
Assume both populations are normal with equal variances. A two-tailed, two-sample pooled t-test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Write down the mean height for each fertiliser.
(2)
(c)
Find the number of degrees of freedom, the t-statistic and the p-value.
(3)
(d)
State the conclusion of the test in context.
(2)
(e)
State why the pooled test needs the assumption of equal variances.
(1)
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Notes
Part (a)
A1 H0: μA = μB; H1: μA ≠ μB
✓ 1
Part (b)
A1 A: 22.2 cm
✓ 1
A1 B: 20.5 cm
✓ 1
Part (c)
A1 degrees of freedom = 14
✓ 1
A1 t = 2.68
✓ 1
A1 p-value = 0.0180
✓ 1
Part (d)
R1 0.0180 < 0.05
✓ 1
A1 reject H0: there is evidence that the mean seedling heights differ between the fertilisers
✓ 1
Part (e)
R1 the pooled estimate combines the two sample variances into a single estimate of a common population variance
✓ 1
Common error: In a two-tailed test, compare the two-tailed p-value directly with 0.05; do not halve or double it a second time.
167IBMI3H-4-08b-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
The number of goals scored in a match by a football club follows a Poisson distribution with mean 2.6, independently from match to match.
(a)
Find the probability that the club scores at least 4 goals in a match.
(3)
(b)
Find the probability that the club scores exactly 5 goals in a two-match period.
(2)
(c)
Find the probability that, in a run of 10 matches, there are exactly 3 matches in which the club scores no goals.
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 X ~ Po(2.6)
✓ 1
M1 1 − P(X ≤ 3)
✓ 1
A1 0.264
✓ 1
Part (b)
A1 mean for two matches is 5.2
✓ 1
A1 0.175
✓ 1
Part (c)
M1 P(no goals) = e−2.6
✓ 1
A1 0.07427
✓ 1
M1 number of scoreless matches ~ B(10, 0.07427)
✓ 1
A1 0.0286
✓ 1
Common error: Two matches means Po(5.2), not 2 × the probability from one match. For the final part, combine a Poisson probability with a binomial model.
168IBMI3H-4-08b-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A delivery van carries 6 parcels and one trolley. The mass of each parcel is X ~ N(18, 2.52) kg and the mass of the trolley is W ~ N(35, 42) kg. All masses are independent. Let S be the total mass of the parcels and the trolley.
(a)
Find E(S) and Var(S).
(2)
(b)
Find the probability that S is greater than 150 kg.
(3)
(c)
A second model assumes the six parcels all have the same mass X, so that the total is 6X + W. Find the standard deviation of 6X + W.
(3)
(d)
Explain why the standard deviation in (c) is larger than in (a).
(1)
Show mark scheme
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Notes
Part (a)
A1 E(S) = 6 × 18 + 35 = 143 kg
✓ 1
A1 Var(S) = 6 × 6.25 + 16 = 53.5
✓ 1
Part (b)
M1 S ~ N(143, 53.5)
✓ 1
M1 standardise or use normal cdf with limit 150
✓ 1
A1 0.169
✓ 1
Part (c)
M1 Var(6X) = 36 Var(X)
✓ 1
A1 Var = 36 × 6.25 + 16 = 241
✓ 1
A1 standard deviation = 15.5 kg
✓ 1
Part (d)
R1 the same parcel mass is repeated, so errors reinforce each other instead of partly cancelling as for independent parcels
✓ 1
Common error: 6X (one variable counted six times) is not the same as X₁ + … + X₆. Var(6X) = 36 Var(X), but the sum of six independent copies has variance 6 Var(X).
169IBMI3H-4-08b-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium11 marks
Each night an observatory classes the sky as Clear (C), Partly cloudy (P) or Cloudy (D). The weather changes from night to night according to a Markov chain with transition matrix (rows = tonight, columns = tomorrow, in the order C, P, D): C: 0.6, 0.3, 0.1 P: 0.3, 0.4, y D: 0.1, x, 0.5 Tonight is Clear.
(a)
Find the values of x and y.
(2)
(b)
Find the probability that the night after tomorrow is Clear.
(2)
(c)
Find the long-term probabilities of Clear, Partly cloudy and Cloudy nights.
(3)
(d)
Estimate the number of Clear nights in a long 40-night observing season.
(2)
(e)
Explain why the long-term probabilities do not depend on tonight's state.
(2)
Show mark scheme
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Notes
Part (a)
A1 x = 0.4
✓ 1
A1 y = 0.3
✓ 1
Part (b)
M1 first row of T2, or sum over tomorrow's states
✓ 1
A1 0.6 × 0.6 + 0.3 × 0.3 + 0.1 × 0.1 = 0.46
✓ 1
Part (c)
M1 πT = π with π1 + π2 + π3 = 1
✓ 1
A1 πC = 9/26 = 0.346
✓ 1
A1 πP = 0.365, πD = 0.288
✓ 1
Part (d)
M1 40 × πC = 40 × 0.3462
✓ 1
A1 13.8 nights
✓ 1
Part (e)
R1 every entry of T is positive, so any state can be reached from any other
✓ 1
R1 so the distribution converges to the same steady state from every starting state
✓ 1
Common error: The steady state needs the normalising condition (probabilities add to 1) as well as πT = π, which alone has infinitely many solutions.
170IBMI3H-4-08b-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Aircraft arrive at a regional airport at random at a mean rate of 3.4 per 10-minute period. Let N be the number of arrivals in a 10-minute period, and assume N ~ Po(3.4).
(a)
State two conditions that must hold for a Poisson model to be suitable.
(2)
(b)
Find the probability that no aircraft arrive in a 10-minute period.
(2)
(c)
The airport can handle at most c arrivals in a period. Find the smallest c such that P(N ≤ c) ≥ 0.99.
(3)
(d)
With this value of c, find the probability that capacity is exceeded in at least one of four independent 10-minute periods.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 arrivals occur independently of one another
✓ 1
R1 arrivals occur at a constant average rate
✓ 1
Part (b)
M1 e−3.4
✓ 1
A1 0.0334
✓ 1
Part (c)
A1 P(N ≤ 7) = 0.977, which is below 0.99
✓ 1
A1 P(N ≤ 8) = 0.9917
✓ 1
A1 c = 8
✓ 1
Part (d)
M1 P(capacity not exceeded in all four) = 0.99174
✓ 1
A1 0.9672
✓ 1
A1 1 − 0.9672 = 0.0328
✓ 1
Common error: 'At least one period exceeds' is the complement of 'no period exceeds', so use 1 − (P(not exceeded))⁴, not 4 × P(exceeded).
171IBMI3H-4-08b-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A road-safety study measures the stopping distance d metres of a car at speed v km h−1. Speed v: 20, 40, 60, 80, 100, 120 Distance d: 11, 27, 50, 81, 118, 163
(a)
Find the quadratic regression model d = av2 + bv + c.
(3)
(b)
For the quadratic model, R2 = 0.99999. For a linear model, R2 = 0.971. Interpret these values and state which model fits better.
(2)
(c)
Use the quadratic model to estimate the stopping distance at 70 km h−1.
(2)
(d)
Comment on using the model to estimate the stopping distance at 250 km h−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 a = 0.00897
✓ 1
A1 b = 0.264
✓ 1
A1 c = 2.1
✓ 1
Part (b)
R1 the quadratic model explains about 99.999% of the variation in stopping distance, the linear about 97.1%
✓ 1
R1 the quadratic model fits better because its R2 is closer to 1
✓ 1
Part (c)
M1 substitute v = 70
✓ 1
A1 64.5 m
✓ 1
Part (d)
R1 250 is far outside the data range, so this is extrapolation
✓ 1
R1 the quadratic form may not hold, e.g. at high speed, so the estimate is unreliable
✓ 1
Common error: A high R² shows a good fit within the data range only; it does not make predictions outside that range reliable.
172IBMI3H-4-04r7-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A beekeeper classifies each hive at the start of each month as Strong (S), Weak (W) or Queenless (Q). From records, the monthly changes are modelled by a Markov chain. The transition matrix \(\mathbf{T}\) has columns that sum to 1: the entry in row \(i\), column \(j\) is the probability of moving from state \(j\) to state \(i\), and \(\mathbf{s}_{n+1}=\mathbf{T}\mathbf{s}_n\). The states are in the order S, W, Q and \(\mathbf{T}=\begin{pmatrix}0.8&0.3&0.1\\0.15&0.5&0.4\\0.05&0.2&0.5\end{pmatrix}\).
(a)
A hive is Weak now. Find the probability that it is Queenless in two months' time.
(3)
(b)
The apiary has 100 hives, all Strong now. Find the expected number of hives in each state after 3 months.
(3)
(c)
Find the long-term proportion of hives that are Queenless, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(\mathbf{T}^2\) or add the possible paths
Common error: Check which convention the matrix uses: if columns sum to 1, multiply the matrix on the left of the state vector.
173IBMI3H-4-04r7-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium11 marks
A beekeeper compares two hive designs, A and B. The honey yields (kg) from random samples are: Design A (8 hives): 42.1, 38.5, 45.3, 40.2, 43.8, 39.6, 44.7, 41.5 Design B (9 hives): 39.4, 41.0, 38.2, 40.1, 37.9, 40.6, 39.0, 40.3, 38.5 Assume the yields are normally distributed with equal population variances.
(a)
Find the mean and the unbiased estimate of the standard deviation for each design. Give your answers to 3 significant figures.
(2)
(b)
State the hypotheses for a test of whether design A has a different mean yield from design B.
(2)
(c)
Carry out a two-sample \(t\)-test using the pooled estimate of variance, and state the \(p\)-value.
(3)
(d)
State the conclusion at the 5% significance level, in context.
(2)
(e)
Calculate the pooled estimate of the common variance, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Design A: mean 42.0, \(s_{n-1}=2.48\)
✓ 1
A1 Design B: mean 39.4, \(s_{n-1}=1.11\)
✓ 1
Part (b)
A1 \(H_0:\mu_A=\mu_B\)
✓ 1
A1 \(H_1:\mu_A\ne\mu_B\)
✓ 1
Part (c)
M1 attempt to use a pooled two-sample \(t\)-test (15 degrees of freedom)
✓ 1
A1 \(t=2.76\)
✓ 1
A1 \(p=0.0146\)
✓ 1
Part (d)
R1 \(0.0146<0.05\)
✓ 1
A1 reject \(H_0\): there is evidence of a difference in mean yield between the two designs
✓ 1
Part (e)
M1 \(\frac{7s_A^2+8s_B^2}{15}\) with \(s_A^2=6.13\) and \(s_B^2=1.24\)
✓ 1
A1 3.52 kg2
✓ 1
Common error: In a pooled test the degrees of freedom are \(n_1+n_2-2\), and a two-tailed test compares the \(p\)-value (from both tails) with the full significance level.
174IBMI3H-4-04r7-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
A hospital compares the recovery times (in days) of patients given drug A and drug B. Summary data are given for independent random samples: Drug A: \(n=15\), mean 6.2, unbiased standard deviation 1.4. Drug B: \(n=18\), mean 5.4, unbiased standard deviation 1.6. Assume recovery times are normally distributed with equal population variances. The hospital tests, at the 5% level, whether drug A has a longer mean recovery time than drug B.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the pooled estimate of variance, the test statistic and the \(p\)-value.
(3)
(c)
State the conclusion of the test, in context.
(2)
(d)
Find the smallest difference between the sample means, \(\bar x_A-\bar x_B\), that would lead to \(H_0\) being rejected at the 5% level. Use the critical value \(t=1.6955\) with 31 degrees of freedom and give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_A=\mu_B\)
✓ 1
A1 \(H_1:\mu_A>\mu_B\)
✓ 1
Part (b)
A1 pooled variance \(=\frac{14\times1.4^2+17\times1.6^2}{31}=2.29\)
✓ 1
A1 \(t=1.51\)
✓ 1
A1 \(p=0.0703\)
✓ 1
Part (c)
R1 \(0.0703>0.05\)
✓ 1
A1 do not reject \(H_0\): insufficient evidence that drug A gives a longer mean recovery time
✓ 1
Part (d)
M1 reject when \(\frac{d}{\sqrt{s_p^2(\frac1{15}+\frac1{18})}}>1.6955\)
Exam tip: A one-tailed alternative hypothesis (\(>\) or \(<\)) uses the area in one tail only; do not double the \(p\)-value.
175IBMI3H-4-03r9-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A port authority inspects ships and records the outcome as No deficiency, Minor deficiency or Major deficiency. The results for 180 inspections are: Bulk carriers (90): 48 No deficiency, 30 Minor, 12 Major Container ships (90): 62 No deficiency, 21 Minor, 7 Major A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for bulk carriers with a Minor deficiency is 25.5.
(2)
(c)
Write down the number of degrees of freedom and calculate the χ2 test statistic, to 3 significant figures.
(3)
(d)
The critical value at the 5% level with this number of degrees of freedom is 5.991. State the conclusion in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: inspection outcome is independent of ship type
✓ 1
A1 H1: inspection outcome is not independent of ship type
✓ 1
Part (b)
A1 Row total 90 and column total 51 (30 + 21)
✓ 1
A1 Expected = 90 × 51/180 = 25.5
✓ 1
Part (c)
A1 Degrees of freedom = (2 − 1)(3 − 1) = 2
✓ 1
M1 Attempt to use Σ(O − E)2/E
✓ 1
A1 χ2 = 4.69
✓ 1
Part (d)
R1 4.69 < 5.991, so do not reject H0
✓ 1
A1 There is insufficient evidence of an association between ship type and inspection outcome
✓ 1
Exam tip: Always check that each expected frequency is at least 5 before using the χ² test; here the smallest is 9.5.
176IBMI3H-4-03r12-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
Each weekend, members of a cycling club choose one of three routes: Coast (C), Hills (H) or Forest (F). The choices are modelled by a Markov chain with transition matrix \(T=\begin{pmatrix}0.5&0.2&0.3\\0.3&0.6&0.2\\0.2&0.2&0.5\end{pmatrix}\) where the rows and columns are in the order C, H, F, and the entry in row i, column j is the probability of moving from route j to route i at the next weekend.
(a)
A member chooses Coast this weekend. Find the probability that the member chooses Hills in 3 weekends' time. Give your answer to 3 significant figures.
(3)
(b)
This weekend, 40% of members choose Coast, 40% choose Hills and 20% choose Forest. Find the percentage choosing each route in 2 weekends' time, giving each answer to 3 significant figures.
(3)
(c)
Find the long-term proportion of members choosing Hills, giving your answer to 3 significant figures.
Exam tip: On a calculator, raise T to a large power (such as 50) to check a steady-state vector, but show the algebraic equations in your working.
177IBMI3H-4-03r12-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A cyclist records her steady climbing speed v (km/h) on six hills with average gradient g (%).
Gradient g (%)
2
4
6
8
10
12
Speed v (km/h)
24.5
16.6
14.2
11.7
10.9
9.6
She proposes the model \(v=34.0\,g^{-0.5}\).
(a)
Use the model to predict her speed on a hill of gradient 9%. Give your answer to 3 significant figures.
(2)
(b)
Calculate the sum of the squares of the residuals, \(SS_{res}=\sum(v-\hat{v})^2\), for the six data points. Give your answer to 3 significant figures.
(3)
(c)
The total sum of squares for the observed speeds is \(SS_{tot}=\sum(v-\bar v)^2=149.3\). Calculate the coefficient of determination \(R^2\) for the model, using the unrounded value of \(SS_{res}\) from (b). Give your answer to 3 significant figures.
(2)
(d)
Interpret the value of \(R^2\) in the context of the question.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(v=34.0\times9^{-0.5}\)
✓ 1
A1 11.3 km/h
✓ 1
Part (b)
M1 predicted speeds from the model: 24.04, 17.00, 13.88, 12.02, 10.75, 9.81
✓ 1
M1 residuals squared and summed: attempt at \(\sum(v-\hat v)^2\)
✓ 1
A1 0.643
✓ 1
Part (c)
M1 \(R^2=1-\frac{SS_{res}}{SS_{tot}}\) (ECF from (b))
✓ 1
A1 0.996
✓ 1
Part (d)
A1 about 99.6% of the variation in her speed is explained by the model (by gradient)
✓ 1
A1 the remaining variation is due to other factors (wind, fatigue, road surface) not included in the model
✓ 1
Exam tip: Always interpret \(R^2\) as a proportion of the variation in the dependent variable that the model accounts for, naming that variable.
178IBMI3H-4-03r12-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
A manufacturer advertises that its road bikes have a mean mass of 8.4 kg. A cyclist suspects that the true mean mass is higher. She weighs 9 randomly chosen bikes and finds a sample mean of 8.70 kg and a sample standard deviation \(s_{n-1}=0.52\) kg. Assume that masses are normally distributed.
(a)
State suitable null and alternative hypotheses.
(2)
(b)
Carry out a one-tailed t-test and find the p-value, giving your answer to 3 significant figures.
(3)
(c)
State the conclusion of the test at the 5% significance level, giving a reason.
(2)
(d)
Find the largest sample mean, to 3 significant figures, that would lead to \(H_0\) not being rejected at the 5% level in a test using the same sample size and the same sample standard deviation. The critical t-value for 8 degrees of freedom (one-tailed, 5%) is 1.860.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu=8.4\)
✓ 1
A1 \(H_1:\mu>8.4\)
✓ 1
Part (b)
M1 \(t=\frac{8.70-8.4}{0.52/\sqrt9}\)
✓ 1
A1 \(t=1.731\) with 8 degrees of freedom
✓ 1
A1 p-value = 0.0609
✓ 1
Part (c)
R1 0.0609 > 0.05
✓ 1
A1 do not reject \(H_0\): insufficient evidence that the mean mass exceeds 8.4 kg
✓ 1
Part (d)
M1 critical value 1.860: reject when \(\frac{\bar x-8.4}{0.52/3}>1.860\)
✓ 1
M1 $\bar x=8.4+t_{crit}\times\frac{0.52}{3}$
✓ 1
A1 8.722, so 8.72 kg
✓ 1
Exam tip: For "greater than" alternatives use the upper tail only; do not double the tail probability.
179IBMI3H-4-05r18-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium12 marks
A music teacher compares two ways of learning a new piece. Method A is repeated physical practice. Method B combines practice with mental rehearsal. Eight students use method A and nine different students use method B. The number of minutes each student needed before playing the piece without error is recorded. Method A: 48, 52, 45, 60, 55, 49, 57, 51 Method B: 42, 47, 39, 50, 44, 41, 53, 46, 43 Assume both populations are normally distributed with equal variances. The teacher tests, at the 5% significance level, whether method B reduces the mean learning time.
(a)
Find the sample mean and the unbiased estimate of the variance for each method. Give answers to 3 significant figures.
(2)
(b)
State the null and alternative hypotheses.
(2)
(c)
Find the pooled estimate of the common variance, to 3 significant figures.
(2)
(d)
Find the \(t\) test statistic and the \(p\)-value for this test.
(3)
(e)
State the conclusion of the test.
(1)
(f)
Find a 95% confidence interval for \(\mu_A - \mu_B\), with limits to 3 significant figures.
Common error: Pooled variance is a weighted average using degrees of freedom \(n-1\), not a simple average of the two variances. State the equal-variance assumption whenever a pooled test is used.
180IBMI3H-4-05r18-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
A concert hall surveys 260 customers about their age and the type of concert they prefer. Under 25: Rock 34, Jazz 22, Orchestral 9 Age 25 to 50: Rock 41, Jazz 38, Orchestral 31 Over 50: Rock 16, Jazz 27, Orchestral 42 A \(\chi^2\) test for independence is carried out at the 1% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Show that the expected frequency for customers over 50 who prefer orchestral concerts is 26.8, correct to 3 significant figures.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the \(\chi^2\) test statistic, to 3 significant figures, and the \(p\)-value.
(3)
(e)
State the conclusion of the test, giving a reason.
(2)
(f)
State the smallest expected frequency and explain why the conditions for the test are met.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): age group and preferred concert type are independent; \(H_1\): they are not independent (associated)
✓ 1
Part (b)
M1 \(\dfrac{85\times 82}{260}\) (row total \(\times\) column total \(\div\) grand total)
A1 there is evidence of an association between age group and preferred type of concert
✓ 1
Part (f)
R1 smallest expected frequency is 20.5, which is at least 5, so the test is valid
✓ 1
Common error: Expected frequencies come from row total times column total divided by the grand total, not from the observed counts. Degrees of freedom are \((r-1)(c-1)\), not the number of cells minus 1.
181IBMI3H-4-03r25-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
In a theatre, the first act of a play lasts X minutes and the second act lasts Y minutes, where X ~ N(54, 3.52) and Y ~ N(46, 2.82). The two durations are independent. The interval between the acts is always exactly 20 minutes. Let T = X + Y.
(a)
Find the mean and the standard deviation of T. Give the standard deviation to 3 significant figures.
(3)
(b)
The performance, including the interval, is advertised to last 125 minutes at most. Find the probability that a performance lasts longer than 125 minutes. Give your answer to 3 significant figures.
(3)
(c)
The theatre wants to state a time t minutes such that there is a 90% probability that the whole performance, including the interval, lasts no more than t minutes. Find t correct to 1 decimal place.
A1 Var(T) = 20.09, standard deviation = 4.48 (accept 4.48)
✓ 1
Part (b)
M1 recognise the condition is T > 105
✓ 1
M1 normal probability with mean 100 and standard deviation 4.48 (use the unrounded value)
✓ 1
A1 P(T > 105) = 0.132
✓ 1
Part (c)
M1 total duration = T + 20 has mean 120 minutes (standard deviation unchanged)
✓ 1
M1 inverse normal: P(T + 20 ≤ t) = 0.9, with z = 1.28155
✓ 1
A1 t = 125.7
✓ 1
Common error: Adding standard deviations instead of variances: for independent variables, Var(X + Y) = Var(X) + Var(Y), and a fixed time such as the interval changes the mean but not the variance.
182IBMI3H-4-07r27-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A pianist's performance of piece A has a duration (in minutes) that is modelled as N(4.8, 0.32). Piece B has a duration modelled as N(6.1, 0.42). All performances are independent. Let D = (duration of B) − (duration of A).
(a)
Find E(D) and Var(D).
(3)
(b)
Find the probability that piece B is longer than piece A by less than 1 minute (that is, 0 < D < 1).
(3)
(c)
A recital consists of two separate performances of piece A followed by one performance of piece B. Find the expected total duration of the recital and its standard deviation.
M1 variance = 2 × 0.32 + 0.42 (two independent performances of A)
✓ 1
A1 standard deviation = 0.583 minutes
✓ 1
Common error: Two separate performances of the same piece are independent, so their variances add (2σ²); this differs from doubling one performance (2A), which has variance 4σ².
183IBMI3H-4-07r27-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
The mass of an apple from an orchard is normally distributed with mean 180 g and standard deviation 22 g. A packer selects 6 apples independently and calculates their mean mass, X̄, and their total mass, T.
(a)
State the distribution of X̄.
(2)
(b)
Find P(X̄ > 190).
(2)
(c)
The packer wants 97% of packs to have a mean mass greater than m grams. Find the value of m, to the nearest gram.
Exam tip: For a mean of n independent observations, the standard deviation is σ ÷ √n; for the total, it is σ × √n. Check which of the two the question needs.
184IBMI3H-4-04r28-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium12 marks
A research vessel records the mean catch rate y (kg per hour) of a trawler at different sea-surface temperatures x (°C).
x (°C)
8
10
12
14
16
18
20
22
y (kg per hour)
5.1
7.4
9.0
9.9
10.2
9.6
8.1
6.2
A quadratic model y = ax2 + bx + c is fitted to the data using least-squares regression.
(a)
Find the values of a, b and c for the quadratic model, giving each to 3 significant figures.
(3)
(b)
Write down the coefficient of determination R2 for the quadratic model and interpret it in context.
(2)
(c)
Use your model to find the temperature at which the catch rate is greatest, and the greatest catch rate. Give your answers to 3 significant figures.
(3)
(d)
Use your model, with unrounded coefficients, to estimate the catch rate at 13 °C. Give your answer to 3 significant figures.
(2)
(e)
A linear model fitted to the same data gives R2 = 0.0443. State which model is more appropriate and give a reason.
(1)
(f)
The quadratic model predicts a negative catch rate at 40 °C. State what this shows about using the model at 40 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses quadratic regression on a calculator
✓ 1
A1 a = −0.0918
✓ 1
A1 b = 2.83 and c = −11.7
✓ 1
Part (b)
A1 R2 = 0.998
✓ 1
R1 about 99.8% of the variation in catch rate is explained by the quadratic model in temperature
✓ 1
Part (c)
M1 uses x = −b/(2a) or the calculator maximum
✓ 1
A1 15.4 °C
✓ 1
A1 10.1 kg per hour
✓ 1
Part (d)
M1 substitutes x = 13 into the fitted quadratic
✓ 1
A1 9.59 kg per hour
✓ 1
Part (e)
R1 the quadratic model, because its R2 is much closer to 1 and the data rise then fall
✓ 1
Part (f)
R1 40 °C lies far outside the data range (extrapolation) and the prediction is not physically possible, so the model is unreliable there
✓ 1
Common error: R² near 1 shows a good fit within the data range only; it does not make predictions outside the range reliable.
185IBMI3H-4-05r29-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Patients arrive at the emergency department of a hospital at random with a mean of 6.5 per hour. The number of arrivals in an hour is modelled by a Poisson distribution, and different hours are independent.
(a)
Find the probability that exactly four patients arrive in one hour.
(2)
(b)
Find the probability that more than 15 patients arrive during a two-hour period.
(3)
(c)
An extra triage nurse is called in whenever 12 or more patients arrive in one hour. Find the probability that this happens in a given hour. Store the unrounded value for use in part (d).
(2)
(d)
During a five-hour shift, find the probability that an extra nurse is called in during exactly one of the hours. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{e^{-6.5}6.5^4}{4!}\)
✓ 1
A1 0.112
✓ 1
Part (b)
M1 mean 13 for two hours
✓ 1
M1 \(P(Y\ge16)=1-P(Y\le15)\)
✓ 1
A1 0.236
✓ 1
Part (c)
M1 \(1-P(X\le11)\) with mean 6.5
✓ 1
A1 0.0339
✓ 1
Part (d)
M1 recognise binomial with \(n=5\) and \(p=0.03388\) (unrounded from (c))
✓ 1
M1 \(\binom{5}{1}p(1-p)^4\)
✓ 1
A1 0.148
✓ 1
Exam tip: For 'more than 15', use \(P(Y\ge16)\); check whether the inequality is strict before using the cumulative function.
186IBMI3H-4-05r29-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
At a honey-packing site, the mass of an empty jar is \(J\sim N(142,\,3^2)\) grams and the mass of honey put into a jar is \(H\sim N(454,\,6^2)\) grams. The two masses are independent. The total mass of a filled jar is \(T=J+H\). Filled jars are independent of each other.
(a)
Write down \(E(T)\) and \(\mathrm{Var}(T)\).
(2)
(b)
Find the probability that a filled jar has total mass less than 585 g.
(3)
(c)
Three filled jars are placed in a gift box. Find the probability that their total mass exceeds 1800 g.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(E(T)=596\) g
✓ 1
A1 \(\mathrm{Var}(T)=45\) g2
✓ 1
Part (b)
M1 standardise: \(z=\frac{585-596}{\sqrt{45}}\)
✓ 1
A1 \(z=-1.64\) (or equivalent normal calculation)
✓ 1
A1 0.0505
✓ 1
Part (c)
M1 mean of total \(=3\times596=1788\)
✓ 1
M1 variance of total \(=3\times45=135\) (independent jars)
✓ 1
M1 \(z=\frac{1800-1788}{\sqrt{135}}\)
✓ 1
A1 0.151
✓ 1
Common error: Variances add for both sums and differences of independent variables; subtracting variances is a frequent slip.
187IBMI3H-4-05r29-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium12 marks
A hospital's standard recovery time after a minor operation has a mean of 6.0 days. A new care protocol is tried on twelve patients, with recovery times (days): 5.4, 6.1, 4.8, 5.2, 5.9, 4.6, 5.5, 6.3, 5.0, 4.9, 5.7, 5.3. Assume recovery times under the new protocol are normally distributed.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation.
(2)
(b)
State suitable null and alternative hypotheses to test whether the new protocol reduces the mean recovery time.
(2)
(c)
Carry out a one-tailed \(t\)-test and find the \(p\)-value.
(3)
(d)
State the conclusion at the 5% significance level, giving a reason.
(2)
(e)
Using \(n=12\) and \(s_{n-1}=0.532\), find the critical value of the sample mean below which \(H_0\) would be rejected at the 5% level. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
A1 mean 5.39 days
✓ 1
A1 \(s_{n-1}=0.532\) days
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Part (b)
A1 \(H_0:\mu=6.0\)
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A1 \(H_1:\mu<6.0\)
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Part (c)
M1 attempt at a one-sample \(t\)-test with 11 degrees of freedom
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A1 \(t=-3.96\)
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A1 \(p=0.00111\)
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Part (d)
R1 \(p<0.05\), so reject \(H_0\)
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A1 there is sufficient evidence that the mean recovery time is lower under the new protocol
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Part (e)
M1 critical \(t=-1.796\) (11 degrees of freedom, 5% one-tail)
Exam tip: State the number of degrees of freedom and compare the \(p\)-value with the significance level in a full sentence in context.
188IBMI3H-4-05r29-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
The table shows the lap time \(y\) seconds of a car on lap number \(x\) of a stint, as the tyres wear. Lap \(x\): 2, 5, 8, 11, 14, 17 Time \(y\) (s): 86.9, 86.4, 86.7, 87.1, 88.4, 89.8
(a)
Use quadratic regression to find a model of the form \(y=ax^2+bx+c\). Give each constant to 3 significant figures.
(3)
(b)
Write down the coefficient of determination, \(R^2\), and interpret it.
(2)
(c)
Use the model, with unrounded constants, to predict the lap time on lap 20.
(2)
(d)
The team changes tyres when the model predicts a lap time above 91 s. Find the value of \(x\), to 3 significant figures, at which the model gives 91 s, and hence state the first lap on which the predicted time exceeds 91 s.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempt at quadratic regression
✓ 1
A1 \(a=0.0268\) and \(b=-0.310\)
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A1 \(c=87.4\)
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Part (b)
A1 \(R^2=0.995\)
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A1 about 99.5% of the variation in lap time is explained by the model
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Part (c)
M1 substitute \(x=20\)
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A1 91.9 s
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Part (d)
M1 solve \(ax^2+bx+c=91\) (unrounded constants)
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A1 \(x=18.8\)
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A1 lap 19
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Common error: Do not use the model to predict outside the range of data without comment; here lap 20 is only a short extrapolation.
189IBMI3H-4-07r33-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
In a town of 1000 people, each person is in one of three states each week: susceptible (S), infected (I) or recovered (R). The weekly transition probabilities are: S to S 0.9, S to I 0.1; I to I 0.5, I to R 0.5; R to R 0.8, R to S 0.2. Initially 940 people are susceptible, 60 are infected and none are recovered.
(a)
Write down the transition matrix T, with columns in the order S, I, R, so that each column sums to 1.
(2)
(b)
Find the expected number of infected people after 2 weeks, to the nearest whole number.
(2)
(c)
Show that in the long term the proportion of the population that is infected is 2/17.
(3)
(d)
Hence find the expected long-term number of infected people in a town of 4500, to the nearest whole number.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 first column (0.9, 0.1, 0) and second column (0, 0.5, 0.5)
✓ 1
A1 third column (0.2, 0, 0.8)
✓ 1
Part (b)
M1 use of T2 × (940, 60, 0)T (week 1: 846, 124, 30)
✓ 1
A1 147 (146.6)
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Part (c)
M1 steady state Ts = s, giving 0.1sS = 0.5sI and 0.5sI = 0.2sR
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M1 sI = 0.2sS, sR = 2.5sI = 0.5sS, with total 1
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A1 sS = 1/1.7 so sI = 0.2/1.7 = 2/17 (AG)
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Part (d)
M1 4500 × 2/17
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A1 529
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Exam tip: Set up the matrix so each column sums to 1 when state vectors are columns; a quick column-sum check catches most input errors.
190IBMI3H-4-05r34-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
A commuter's train journey has two stages. The time spent on the first stage, A minutes, is normally distributed with mean 42 and standard deviation 3. The time on the second stage, B minutes, is normally distributed with mean 35 and standard deviation 4. The two times are independent. Let T = A + B.
(a)
Find E(T) and Var(T).
(2)
(b)
Find the probability that the whole journey takes more than 85 minutes.
(3)
(c)
Find the probability that the first stage takes longer than the second stage.
(3)
(d)
The commuter wants to leave early enough to be on time on 95% of days. Find the journey time t such that P(T < t) = 0.95. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
E(T) = 42 + 35 = 77
✓ 1
Var(T) = 9 + 16 = 25
✓ 1
Part (b)
T ~ N(77, 25), standard deviation 5
✓ 1
P(T > 85) = P(Z > 1.6)
✓ 1
= 0.0548 (3 s.f.)
✓ 1
Part (c)
Consider D = A − B; E(D) = 7 and Var(D) = 9 + 16 = 25 (variances add)
✓ 1
P(D > 0) = P(Z > −1.4)
✓ 1
= 0.919 (3 s.f.)
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Part (d)
z = 1.645 (inverse normal for 0.95)
✓ 1
t = 77 + 1.645 × 5 = 85.2 minutes (3 s.f.)
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Common error: When subtracting independent variables the variances still add; Var(A − B) is not 9 − 16.
191IBMI3H-4-05r34-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A sports scientist claims that a training programme reduces the mean reaction time of sprinters to below 0.165 s, the mean before training. After the programme, the reaction times (in seconds) of a random sample of eight sprinters were 0.152, 0.168, 0.145, 0.159, 0.171, 0.149, 0.162, 0.155. Reaction times may be assumed to be normally distributed.
(a)
State suitable null and alternative hypotheses.
(2)
(b)
Find the sample mean and the sample standard deviation (unbiased estimate). Give each to 3 significant figures.
(2)
(c)
Perform a one-tailed t-test and find the p-value.
(2)
(d)
Using a 5% significance level, state the conclusion in context.
(2)
(e)
State why a t-test is used here instead of a test based on the normal distribution with a known standard deviation.
(1)
Show mark scheme
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Notes
Part (a)
H0: μ = 0.165
✓ 1
H1: μ < 0.165
✓ 1
Part (b)
Mean = 0.158 s
✓ 1
Standard deviation s = 0.00910 s
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Part (c)
Use of a one-sample t-test with 7 degrees of freedom
✓ 1
p-value = 0.0278 (3 s.f.)
✓ 1
Part (d)
0.0278 < 0.05, so reject H0
✓ 1
There is sufficient evidence that the mean reaction time is below 0.165 s after the programme
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Part (e)
The population standard deviation is unknown and estimated from a small sample
✓ 1
Exam tip: Write the conclusion by comparing the p-value with the significance level and then answer in the words of the context.
192IBMI3H-4-05r34-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
A shot-putter records the distance d metres of throws made at different launch angles θ degrees: θ: 20, 25, 30, 35, 40, 45, 50 d: 12.1, 12.9, 13.5, 13.9, 14.0, 13.8, 13.3
(a)
Use quadratic regression to find the model d = aθ2 + bθ + c, giving a, b and c to 4 significant figures.
(3)
(b)
The coefficient of determination is R2 = 0.997 (3 s.f.). Interpret this value.
(2)
(c)
Use the model (with the coefficients from part (a)) to find the angle that gives the greatest distance. Give your answer to 3 significant figures.
(2)
(d)
Use the model to estimate the distance for a launch angle of 60°, and comment on the reliability of your estimate. Give your estimate to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
Use of quadratic regression on a GDC
✓ 1
a = −0.005286 and b = 0.4121
✓ 1
c = 5.936
✓ 1
Part (b)
About 99.7% of the variation in d is explained by the model (variation in angle)
✓ 1
The model is a very good fit to the data
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Part (c)
θ = −b ÷ 2a (or use GDC maximum)
✓ 1
θ = 39.0° (3 s.f.)
✓ 1
Part (d)
Substitute θ = 60 into the model
✓ 1
d = 11.6 m (3 s.f.)
✓ 1
The estimate is an extrapolation outside the data range (20° to 50°), so it may be unreliable
✓ 1
Common error: A high R² does not make predictions outside the observed range reliable; always comment on extrapolation.
193IBMI3H-4-05r37-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
The daily energy output of two independent wind turbines is modelled by normal distributions. Turbine A: X ~ N(48, 62) MWh. Turbine B: Y ~ N(40, 52) MWh. Let T = X + Y.
(a)
Write down E(T) and Var(T).
(2)
(b)
Find P(T > 100).
(2)
(c)
Find the probability that turbine A produces more energy than turbine B on a given day.
(3)
(d)
The farm operator wants a value c such that P(T > c) = 0.01. Find c to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
A1 E(T) = 88
✓ 1
A1 Var(T) = 61
✓ 1
Part (b)
M1 T ~ N(88, 61), standardise or use normal cdf
✓ 1
A1 0.0622
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Part (c)
M1 considers D = X − Y
✓ 1
A1 D ~ N(8, 61)
✓ 1
A1 P(D > 0) = 0.847
✓ 1
Part (d)
M1 inverse normal, z = 2.3263 for upper tail 0.01
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M1 c = 88 + 2.3263 × √61
✓ 1
A1 c = 106 MWh
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Common error: For a difference of independent variables the variances are added, not subtracted: Var(X − Y) = Var(X) + Var(Y).
194IBMI3H-4-05r37-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A library packs books into boxes for a mobile service. The mass of a book is modelled by B ~ N(0.45, 0.122) kg, and masses of different books are independent. A box holds 20 books, and the empty box has a fixed mass of 1.2 kg. Let M be the total mass of a full box.
(a)
Find the mean and standard deviation of M, giving the standard deviation to 3 significant figures.
(3)
(b)
A box is rejected if its total mass exceeds 11 kg. Find the probability that a box is rejected.
(3)
(c)
A colleague models the total book mass as 20B. Find Var(20B) and explain why it differs from the variance of the sum of 20 independent books.
(3)
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Notes
Part (a)
A1 E(M) = 20 × 0.45 + 1.2 = 10.2 kg
✓ 1
M1 Var = 20 × 0.122 (= 0.288), fixed box mass adds no variance
✓ 1
A1 standard deviation 0.537 kg
✓ 1
Part (b)
M1 M ~ N(10.2, 0.288) (ECF from (a))
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M1 standardise or use normal cdf with 11
✓ 1
A1 0.0680
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Part (c)
A1 Var(20B) = 202 × 0.122 = 5.76
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R1 20B treats all 20 books as the same book repeated, so their deviations from the mean are perfectly correlated
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R1 independent books partly cancel (some heavier, some lighter), giving a much smaller variance of 0.288
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Common error: Var(2X) = 4Var(X), but Var(X₁ + X₂) = Var(X₁) + Var(X₂) for independent copies; these are different models.
195IBMI3H-4-05r37-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium11 marks
The wind at a coastal wind farm is classified each day as Low (L), Medium (M) or High (H). The day-to-day changes are modelled by a Markov chain with transition matrix T below. Each column gives the probabilities of the next day's state, given the current state, in the order L, M, H. Distributions are column vectors, so the distribution on day n + 1 is T times the distribution on day n.
T = [[0.5, 0.3, 0.1], [0.4, 0.5, 0.4], [0.1, 0.2, 0.5]] (rows listed in order: L, M, H)
On Monday (day 0) the wind is Medium.
(a)
Find the probability distribution for the wind on Wednesday (day 2).
(3)
(b)
Given that Monday is Low, find the probability that Thursday (day 3) is also Low.
(2)
(c)
Find the long-term proportion of days in each state.
(4)
(d)
In the long run, estimate the number of High-wind days in a 365-day year. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
M1 attempts T2 applied to (0, 1, 0)T, or two successive multiplications
✓ 1
A1 distribution after day 1 is (0.3, 0.5, 0.2)
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A1 day 2: P(L) = 0.32, P(M) = 0.45, P(H) = 0.23
✓ 1
Part (b)
M1 element in row 1, column 1 of T3
✓ 1
A1 0.34
✓ 1
Part (c)
M1 sets Ts = s with the components summing to 1
✓ 1
M1 solves the resulting equations (or uses the eigenvector for eigenvalue 1)
✓ 1
A1 P(L) = 0.315 (17/54), P(M) = 0.444 (24/54)
✓ 1
A1 P(H) = 0.241 (13/54)
✓ 1
Part (d)
M1 365 × 13/54 (or 365 × 0.2407)
✓ 1
A1 87.9
✓ 1
Exam tip: State clearly whether your transition matrix has columns or rows summing to 1, and multiply in the matching order.
196IBMI3H-4-05r37-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A site engineer records the mean wind speed (m s−1) at hub height on 10 randomly chosen days: 8.1, 6.9, 7.4, 9.0, 5.8, 7.7, 8.4, 6.5, 7.9, 7.1. Assume the speeds come from a normal distribution with unknown variance.
(a)
Find the sample mean and the unbiased estimate of the population standard deviation, to 3 significant figures.
(2)
(b)
Find a 95% confidence interval for the population mean wind speed.
(3)
(c)
The turbine supplier states that the mean wind speed at the site is 8.0 m s−1. Comment on whether the sample contradicts this.
(2)
(d)
Find the width of a 99% confidence interval for the mean, to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
A1 mean = 7.48 m s−1
✓ 1
A1 sn−1 = 0.945 m s−1
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Part (b)
M1 uses a t-interval with 9 degrees of freedom (critical value 2.262)
✓ 1
M1 7.48 ± 2.262 × 0.9449/√10
✓ 1
A1 (6.80, 8.16)
✓ 1
Part (c)
M1 compares 8.0 with the interval (ECF from (b))
✓ 1
R1 8.0 lies within (6.80, 8.16), so the data do not contradict the claim
✓ 1
Part (d)
M1 width = 2 × 3.250 × 0.9449/√10
✓ 1
A1 1.94 m s−1
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Common error: With an unknown population variance use the t-distribution with n − 1 degrees of freedom, not the normal distribution.
197IBMI3H-4-04r45-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium13 marks
A shipping company moves a container between three ports A, B and C once each week. The probabilities are given by the transition matrix T, whose columns give the current port (A, B, C from left to right) and whose rows give the next port (A, B, C from top to bottom): T = [0.6, 0.2, 0.1; 0.3, 0.5, 0.2; 0.1, 0.3, 0.7] (rows separated by semicolons). A particular container is at port A at the start of week 0.
(a)
Explain why each column of T sums to 1.
(1)
(b)
Find the probability that the container is at port C at the start of week 3.
(3)
(c)
Find the long-term proportion of weeks for which a container is at each port. Give the exact fractions.
(4)
(d)
A fleet of 200 containers is distributed in the long term according to the steady state. Estimate the number of containers at port B, to the nearest whole number.
(2)
(e)
Find the probability that the container is at port B at the start of week 2 and at port C at the start of week 3.
(3)
Show mark scheme
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Notes
Part (a)
R1 from any port the container must go to exactly one of A, B or C (including staying), so the probabilities of all possible moves add to 1
✓ 1
Part (b)
M1 initial state vector (1, 0, 0)
✓ 1
M1 calculates T3 multiplied by the initial vector
✓ 1
A1 0.302
✓ 1
Part (c)
M1 Tπ = π with the entries of π adding to 1
✓ 1
A1 correct equations in two unknowns
✓ 1
A1 πA = 9/34
✓ 1
A1 πB = 11/34 and πC = 14/34 = 7/17
✓ 1
Part (d)
M1 200 × 11/34
✓ 1
A1 65
✓ 1
Part (e)
M1 state after two weeks (0.43, 0.35, 0.22) so P(at B in week 2) = 0.35
✓ 1
M1 multiplies by P(B → C) = 0.3
✓ 1
A1 0.105
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Common error: Mixing up rows and columns of the transition matrix: with columns as the current state, the probability of moving from B to C is the entry in row C, column B.
198IBMI3H-4-04r45-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium12 marks
The hourly running cost, C thousand pounds, of a cargo ship depends on its speed, v knots. Measurements are: v: 8, 10, 12, 14, 16, 18 C: 6.1, 5.2, 5.0, 5.5, 6.6, 8.3
(a)
Use quadratic regression to find the model C = av2 + bv + c, giving a, b and c to 3 significant figures.
(4)
(b)
Find the coefficient of determination, R2, to 4 significant figures, for (i) a linear model and (ii) the quadratic model.
(2)
(c)
State, with a reason, which model is more suitable.
(2)
(d)
Use the quadratic model to find the speed that minimises the running cost, and state the minimum cost.
(2)
(e)
Use the unrounded quadratic model to predict the cost at 24 knots, to 3 significant figures.
(1)
(f)
Comment on the reliability of the prediction in part (e).
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 uses quadratic regression on the data
✓ 1
A1 a = 0.0813
✓ 1
A1 b = −1.89
✓ 1
A1 c = 16.0
✓ 1
Part (b)
A1 (i) 0.4715
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A1 (ii) 0.9995
✓ 1
Part (c)
A1 the quadratic model
✓ 1
R1 its R2 is much closer to 1: about 99.95 % of the variation in cost is explained, compared with about 47 % for the linear model
✓ 1
Part (d)
A1 11.6 knots
✓ 1
A1 5.01 thousand pounds per hour
✓ 1
Part (e)
A1 17.5 thousand pounds per hour
✓ 1
Part (f)
R1 24 knots is outside the range of the data (8 to 18 knots), so this is extrapolation and may be unreliable
✓ 1
Exam tip: A high R² shows a good fit to the data given but does not make predictions outside the data range reliable.
199IBMI3H-4-06r45-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
In a theatre production, Act 1 lasts A minutes and Act 2 lasts B minutes, where A ~ N(52, 42) and B ~ N(47, 3.52), independent of each other. There is a fixed interval of 20 minutes between the acts. Let T be the total duration of the evening, in minutes.
(a)
Find the mean and variance of T.
(3)
(b)
Find the probability that the evening lasts more than 130 minutes. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that Act 2 is longer than Act 1. Give your answer to 3 significant figures.
(3)
(d)
The theatre wants a closing time such that the evening is longer than this time with probability 0.01. Find this duration, in minutes, to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
M1 T = A + B + 20
✓ 1
A1 E(T) = 119
✓ 1
A1 Var(T) = 16 + 12.25 = 28.25 (the constant 20 adds nothing to the variance)
✓ 1
Part (b)
M1 T ~ N(119, 28.25) and attempts P(T > 130)
✓ 1
A1 0.0192
✓ 1
Part (c)
M1 D = A − B and requires P(D < 0)
✓ 1
A1 D ~ N(5, 28.25)
✓ 1
A1 0.173
✓ 1
Part (d)
M1 z = 2.326… for P(T > t) = 0.01
✓ 1
A1 131 minutes
✓ 1
Common error: Var(A − B) = Var(A) + Var(B): variances add for both sums and differences of independent variables.
200IBMI3H-4-06r45-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
A shipping company claims that the mean turnaround time of a container vessel at a port is 36 hours. A port agent believes that the mean is higher. The turnaround times, in hours, for 9 randomly chosen vessels are 38.2, 35.9, 40.1, 37.4, 36.8, 39.5, 41.2, 34.7, 38.9. The agent assumes that turnaround times follow a normal distribution and tests at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the sample mean and the unbiased estimate of the population standard deviation, each to 3 significant figures.
(2)
(c)
Carry out a one-sample t-test to find the value of the test statistic (to 3 significant figures) and the p-value (to 3 significant figures).
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
Explain why the agent used a t-test rather than a test based on the normal distribution.
(1)
Show mark scheme
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Notes
Part (a)
A1 H0: μ = 36, H1: μ > 36
✓ 1
Part (b)
A1 sample mean = 38.1 (38.08)
✓ 1
A1 unbiased standard deviation = 2.09
✓ 1
Part (c)
M1 uses a one-sample t-test with 8 degrees of freedom
✓ 1
A1 t = 2.99
✓ 1
A1 p = 0.00868
✓ 1
Part (d)
R1 0.00868 < 0.05
✓ 1
A1 reject H0; there is sufficient evidence that the mean turnaround time is greater than 36 hours
✓ 1
Part (e)
R1 the population standard deviation is unknown and the sample is small
✓ 1
Exam tip: A ‘greater than’ alternative hypothesis gives a one-tailed test, so use the upper-tail p-value and do not double it.
201IBMI3H-4-06r45-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
The daily fuel consumption F tonnes of a cargo ship at speed v knots is recorded.
v
10
12
14
16
18
20
22
F
21
26
34
45
57
72
89
A linear regression model F = 5.70v − 42.0 has coefficient of determination R2 = 0.970. A quadratic regression model of the form F = av2 + bv + c is also fitted.
(a)
Find the values of a, b and c in the quadratic model, to 3 significant figures.
(2)
(b)
Write down R2 for the quadratic model to 4 decimal places, and explain which model fits the data better and why the quadratic form is sensible in this context.
(3)
(c)
Use the quadratic model, with the unrounded values of a, b and c from your calculator, to estimate the daily fuel consumption at 15 knots. Give your answer to 3 significant figures.
(2)
(d)
Explain why the quadratic model should not be used to estimate the fuel consumption at 28 knots.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses quadratic regression on the data
✓ 1
A1 a = 0.289, b = −3.54, c = 27.3
✓ 1
Part (b)
A1 R2 = 0.9999
✓ 1
R1 0.9999 is closer to 1 than 0.970, so the quadratic model fits better
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R1 fuel use rises faster than speed because resistance grows more than proportionally with speed (accept any sensible context reason)
✓ 1
Part (c)
M1 substitutes v = 15 into the quadratic
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A1 39.1 tonnes
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Part (d)
R1 28 knots is outside the range of the data (10 to 22 knots), so this would be extrapolation
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R1 the relationship may not continue to follow the same curve beyond the range, so the estimate is unreliable
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Exam tip: Comparing R² values only tells you which model fits these data better; it does not guarantee good predictions outside the data range.
202IBMI3H-4-07r47-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Calculator
A suburban tram line is classed each morning as on time (O), late (L) or cancelled (X). The state on one morning depends only on the state on the previous morning. The transition matrix T is shown, where each column gives the probabilities of moving from that state, in the order O, L, X: \(\mathbf{T}=\begin{pmatrix}0.80&0.50&0.30\\0.15&0.35&0.40\\0.05&0.15&0.30\end{pmatrix}\)
(a)
Write down the probability that the tram is on time the morning after it was cancelled.
(1)
(b)
On Monday the tram is on time. Find the probability that it is late on Thursday.
(3)
(c)
Find the long-term proportion of mornings on which the tram is cancelled.
(3)
(d)
Use your answer to part (c) to estimate the number of cancelled mornings in a year of 365 mornings. Give your answer to 3 significant figures.
(2)
(e)
State one reason why a Markov chain may not be a perfect model for the tram line.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.30
✓ 1
Part (b)
M1 use of T3 with initial state (1, 0, 0)T
✓ 1
A1 state vector (0.7035, 0.207875, 0.088625)
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A1 0.208
✓ 1
Part (c)
M1 use of Tπ = π (or an eigenvector for eigenvalue 1)
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M1 use of the components of π summing to 1
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A1 0.0957 (= 22/230)
✓ 1
Part (d)
M1 365 × 0.0957 (or 365 × 22/230)
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A1 34.9 (34.91 from the unrounded value)
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Part (e)
R1 e.g. the state may depend on more than the previous morning (engineering faults lasting several days, seasons, strikes) or the probabilities may change over time
✓ 1
Exam tip: For a long-term proportion, solve for the invariant vector and check that its components sum to 1; then multiply by the number of trials for an expected count.
203IBMI3H-4-07r47-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Calculator
A sports scientist tests whether compression sleeves reduce the time taken to sprint 20 m. Sixteen athletes are split at random into two groups of eight. Group 1 wears sleeves and group 2 does not. The times, in seconds, are: Group 1: 2.91, 3.02, 2.88, 2.95, 2.99, 2.84, 2.93, 2.96 Group 2: 3.05, 3.12, 2.98, 3.09, 3.01, 3.15, 3.04, 3.10 Assume that both sets of times come from normal populations with the same variance. Let μ1 and μ2 be the population mean times for groups 1 and 2.
(a)
State the null and alternative hypotheses.
(2)
(b)
Perform a one-tailed two-sample t-test using a pooled variance. Write down the value of t and the p-value, giving the p-value to 3 significant figures.
(3)
(c)
Using a 5% significance level, state the conclusion in context.
(2)
(d)
Explain why random allocation to the two groups is important.
(1)
(e)
Find the pooled estimate of the common variance, giving your answer to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μ1 = μ2
✓ 1
A1 H1: μ1 < μ2
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Part (b)
M1 use of a pooled two-sample t-test with 14 degrees of freedom
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A1 t = −4.58
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A1 p = 0.000216
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Part (c)
R1 p < 0.05
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A1 reject H0: there is evidence that sleeves reduce the mean sprint time
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Part (d)
R1 it reduces bias, so differences in times are less likely to be caused by differences in the athletes' ability
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Part (e)
M1 (7s12 + 7s22)/14 with s1 = 0.0583 and s2 = 0.0575
✓ 1
A1 0.00335
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Exam tip: State which test and which tail you are using, then compare the p-value with the significance level in a sentence written in the context of the question.
204IBMI3H-4-07r47-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
Calculator
The dome of a telescope is in one of three states each night: open (O), closed for weather (W) or closed for maintenance (M). The state on one night depends only on the state on the previous night. The transition matrix is shown, with each column giving the probabilities of moving from that state, in the order O, W, M: \(\mathbf{T}=\begin{pmatrix}0.7&0.4&0.5\\0.2&0.5&0\\0.1&0.1&0.5\end{pmatrix}\)
(a)
The dome is closed for maintenance on night 1. Find the probability that it is open on night 4.
(3)
(b)
Find the probability that the dome is closed for weather on night 4.
(1)
(c)
Find the long-term proportion of nights on which the dome is open.
(3)
(d)
Estimate the number of nights in a year of 365 nights on which the dome is open. Give your answer to the nearest whole number.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 initial state vector (0, 0, 1)T
✓ 1
M1 use of T3 applied to it
✓ 1
A1 0.61
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Part (b)
A1 0.17
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Part (c)
M1 Tπ = π (or eigenvector for eigenvalue 1)
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M1 components sum to 1
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A1 0.595 (= 25/42)
✓ 1
Part (d)
M1 365 × 25/42 (or 365 × 0.595)
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A1 217 (217.26 from the exact fraction)
✓ 1
Exam tip: For 'night n', apply T exactly n − 1 times to the night-1 state vector.
205IBMI3H-4-07r47-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
Calculator
A manufacturer states that the mean mass of its match footballs is 410 g. A quality-control scientist weighs a random sample of 10 footballs. The masses, in grams, are: 408.2, 411.5, 409.8, 412.6, 410.9, 413.1, 409.4, 411.8, 412.2, 410.3 Assume the masses are normally distributed.
(a)
State suitable null and alternative hypotheses to test whether the mean mass differs from 410 g.
(1)
(b)
Carry out a one-sample t-test. Write down the p-value to 3 significant figures.
(2)
(c)
State the conclusion at the 5% significance level, in context.
(2)
(d)
Find a 95% confidence interval for the mean mass, giving the end points to 1 decimal place, and state how it supports your conclusion.
(4)
Show mark scheme
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Mark
Notes
Part (a)
A1 H0: μ = 410 and H1: μ ≠ 410
✓ 1
Part (b)
M1 use of one-sample t-test with 9 degrees of freedom (sample mean 410.98, s = 1.55)
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A1 p = 0.0767
✓ 1
Part (c)
R1 p > 0.05, so do not reject H0
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A1 there is insufficient evidence that the mean mass differs from 410 g
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Part (d)
M1 critical value t = 2.262 with 9 degrees of freedom
✓ 1
M1 410.98 ± 2.262 × 1.550/√10
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A1 (409.9, 412.1)
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R1 410 lies inside the interval, consistent with not rejecting H0
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Common error: 'Do not reject H₀' is not the same as 'prove the mean is 410 g'; say only that there is insufficient evidence of a difference.
206IBMI3H-4-07r48-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Calculator
A ferry company receives cancellations of cabin bookings each night from two independent sources. The number of cancellations made online, O, follows a Poisson distribution with mean 2.4 per night. The number made by telephone, H, follows a Poisson distribution with mean 1.1 per night. Let T = O + H be the total number of cancellations in a night.
(a)
State the distribution of T, including its parameter.
(1)
(b)
Find \(P(T\ge 6)\).
(2)
(c)
Given that exactly 3 cancellations are received in one night, find the probability that exactly 2 of them were made online.
(4)
(d)
Using the unrounded value of \(P(T\ge6)\), find the probability that, in a week of 7 independent nights, at least one night has 6 or more cancellations.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(T\sim\text{Po}(3.5)\)
✓ 1
Part (b)
M1 \(1-P(T\le5)\)
✓ 1
A1 0.142 (0.14238…)
✓ 1
Part (c)
M1 conditional probability \(\frac{P(O=2)\times P(H=1)}{P(T=3)}\)
✓ 1
A1 \(P(O=2)=0.2613\) and \(P(H=1)=0.3662\)
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A1 \(P(T=3)=0.2158\)
✓ 1
A1 0.443 (0.4433…)
✓ 1
Part (d)
M1 \(P(\text{no night with }T\ge6)=(1-0.14238…)^7=0.3412\)
✓ 1
M1 \(1-(\ldots)^7\)
✓ 1
A1 0.659 (0.6587…)
✓ 1
Common error: The sum of independent Poisson variables is Poisson with the means added; do not add the probabilities or the standard deviations.
207IBMI3H-4-07r48-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium9 marks
Calculator
A theatre production has two acts separated by an interval. The durations, in minutes, are modelled as independent random variables: first act \(A\sim N(52,4^2)\), second act \(B\sim N(48,5^2)\), interval \(I\sim N(20,3^2)\). Let \(T=A+B+I\) be the total running time, including the interval.
(a)
Find \(E(T)\) and \(\text{Var}(T)\).
(3)
(b)
Find the probability that the total time exceeds 130 minutes.
(2)
(c)
Find the probability that the first act is longer than the second act.
(2)
(d)
The production runs for 60 independent performances. Using your answer to part (b), estimate the number of performances in which the total time exceeds 130 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(E(T)=120\)
✓ 1
M1 adding the variances (not the standard deviations): \(16+25+9\)
Common error: For \(A-B\) the variances still add (\(16+25\)), because \(\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)\) for independent variables.
208IBMI3H-4-07r48-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Medium10 marks
Calculator
A volcano is classified each month as Dormant (D), Restless (R) or Erupting (E). The transition probabilities between months are modelled by the matrix T, in which each column gives the probabilities of moving from a state, with states in the order D, R, E: $\(\mathbf T=\begin{pmatrix}0.90&0.30&0.50\\0.08&0.55&0.30\\0.02&0.15&0.20\end{pmatrix}\)$
(a)
The volcano is Dormant this month. Find the probability that it is Erupting in two months' time.
(2)
(b)
This month the volcano is Dormant with probability 0.5 and Restless with probability 0.5. Find the probability that it is Erupting in 3 months' time.
(2)
(c)
Find the long-term proportion of months in each state.
(3)
(d)
Over 240 months, estimate the number of months in which the volcano is Erupting.
(2)
(e)
Explain why the long-term proportions do not depend on the state at the start.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathbf T^2\) or paths D→D→E, D→R→E, D→E→E: \(0.9\times0.02+0.08\times0.15+0.02\times0.2\)
✓ 1
A1 0.034
✓ 1
Part (b)
M1 \(\mathbf T^3(0.5,0.5,0)^T\)
✓ 1
A1 0.0658 (0.065803)
✓ 1
Part (c)
M1 \(\mathbf T\mathbf s=\mathbf s\) with sum of components 1
R1 all entries of T are positive (the chain is regular), so \(\mathbf T^n\) converges to the same matrix whatever the initial state
✓ 1
Exam tip: For a long-term state, solve \(\mathbf T\mathbf s=\mathbf s\) together with the condition that the probabilities sum to 1; the equations alone are dependent.
209IBMI-033-07-r2-06Probability and the binomial distributionPaper 2Hard13 marks
An airline states that 8% of its flights are delayed by more than 30 minutes. A consumer group believes the true proportion is higher. It selects a random sample of 60 of the airline's flights and finds that 9 are delayed by more than 30 minutes. The group carries out a hypothesis test at the 5% significance level. Let X be the number of delayed flights in a sample of 60.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the p-value for this test.
(2)
(c)
State the conclusion of the test in context, giving a reason.
(2)
(d)
Find the critical region for this test.
(3)
(e)
Write down the probability of a Type I error for this test.
(1)
(f)
The true proportion of delayed flights is 0.12. Find the probability of a Type II error.
(3)
Show mark scheme
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Mark
Notes
Part (a)
H0: p = 0.08 A1
✓ 1
H1: p > 0.08 A1
✓ 1
Part (b)
Under H0, X ~ B(60, 0.08); p-value = P(X ≥ 9) = 1 − P(X ≤ 8) M1
✓ 1
= 0.0482 A1
✓ 1
Part (c)
0.0482 < 0.05 R1
✓ 1
Reject H0: there is sufficient evidence that more than 8% of the airline's flights are delayed by over 30 minutes A1
✓ 1
Part (d)
Considers upper-tail probabilities, e.g. P(X ≥ 8) = 0.104 M1
✓ 1
and P(X ≥ 9) = 0.0482 A1
✓ 1
Critical region: X ≥ 9 A1
✓ 1
Part (e)
P(X ≥ 9 | p = 0.08) = 0.0482 A1
✓ 1
Part (f)
Type II error: accepting H0, i.e. X ≤ 8 M1
✓ 1
using X ~ B(60, 0.12) M1
✓ 1
P(X ≤ 8) = 0.710 A1
✓ 1
Exam tip: For a discrete test the Type I error probability is the actual probability of the critical region (0.0482), not the nominal 5%. A Type II error uses the non-rejection region with the true value of p.
210IBMI-033-05-r1-06Sampling and descriptive statisticsPaper 2Hard10 marks
A sports-science company develops an app that measures reaction time, in milliseconds (ms). To assess the app, 8 volunteers each take the test twice, one week apart, under the same conditions. The results are: Volunteer: A, B, C, D, E, F, G, H First test (x ms): 210, 245, 230, 198, 260, 225, 240, 215 Second test (y ms): 215, 240, 236, 205, 255, 220, 248, 212
(a)
State what is meant by the reliability of a test, and name the type of reliability test being carried out here.
(2)
(b)
Calculate the Pearson product-moment correlation coefficient, r, between the first and second test results.
(2)
(c)
Interpret your value of r in terms of the reliability of the app.
(2)
(d)
Explain why a high value of r does not show that the app is valid, and describe how criterion-related validity could be tested.
(2)
(e)
The company later groups reaction times into categories for a χ2 goodness-of-fit test with a larger sample. State one consideration when choosing the categories, and why it matters.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: Reliability is the extent to which the test produces consistent results when repeated under the same conditions
✓ 1
A1: Test-retest reliability
✓ 1
Part (b)
M1: Correct use of GDC with both lists entered as paired data
✓ 1
A1: r = 0.957
✓ 1
Part (c)
A1: Very strong positive linear correlation between first and second results
✓ 1
R1: Volunteers obtain similar times on both occasions, so the app appears to be reliable
✓ 1
Part (d)
R1: Validity concerns whether the app measures true reaction time; it could be consistently wrong (e.g. adding a fixed screen delay) and still give a high r
✓ 1
A1: Compare app results with those from an established, accurate laboratory reaction-time device on the same people and check for strong correlation/agreement
✓ 1
Part (e)
A1: Categories should be chosen so that each expected frequency is at least 5 (combine sparse classes if necessary)
✓ 1
R1: Otherwise the χ2 distribution is a poor approximation to the test statistic, making the conclusion unreliable
✓ 1
Common error: Reliability and validity are different: a strong test-retest correlation shows consistency only, not that the measurement is accurate.
211IBMI-033-02-r2-06Probability and the binomial distributionPaper 2Hard11 marks
Faults in a type of woven fabric occur randomly at a mean rate of 0.6 faults per metre. The fabric is sold in rolls of length 5 metres. Let \(Y\) be the number of faults in a roll.
(a)
Find the probability that a roll contains at most 2 faults.
(2)
(b)
A shop receives a batch of 12 rolls. Find the probability that at least 8 of these rolls contain at most 2 faults.
(3)
(c)
The manufacturer also sells short offcuts of length \(L\) metres. Find the greatest value of \(L\) for which the probability that an offcut has no faults is at least 0.2.
(3)
(d)
Given that a 5 metre roll contains at least one fault, find the probability that it contains exactly 3 faults.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(Y \sim Po(3)\)
✓ 1
A1: \(P(Y \le 2) = 0.423\)
✓ 1
Part (b)
M1: \(R \sim B(12, 0.42319…)\)
✓ 1
M1: \(P(R \ge 8) = 1 - P(R \le 7)\)
✓ 1
A1: \(0.0794\)
✓ 1
Part (c)
M1: \(e^{-0.6L} \ge 0.2\)
✓ 1
M1: \(L \le \frac{\ln 5}{0.6}\) or solves graphically
Exam tip: Scale the Poisson mean to the interval in the question: 0.6 per metre becomes 3 per 5 m roll and 0.6L for an offcut of length L.
212IBMI-033-07-r1-06Probability and the binomial distributionPaper 2Hard14 marks
The number of calls received by a customer helpline can be modelled by a Poisson distribution with mean 3.2 calls per 10-minute interval. The numbers of calls in non-overlapping intervals are independent.
(a)
Find the probability that exactly 5 calls are received in a 10-minute interval.
(2)
(b)
Each operator can handle at most one call per 10-minute interval. Find the least number of operators needed so that the probability that all calls in a 10-minute interval can be handled is at least 0.95.
(2)
(c)
Find the probability that more than 6 calls are received in a 20-minute period.
(3)
(d)
Given that exactly 6 calls are received in a 20-minute period, find the probability that exactly 3 of them arrive in the first 10 minutes.
(3)
(e)
An hour is divided into six 10-minute intervals. An interval is classed as “busy” if more than 4 calls are received in it. Find the probability that exactly 2 of the six intervals in an hour are busy.
She believes the data can be modelled by \(N = ae^{kt}\), where a and k are positive constants.
(a)
Show that, if the model is correct, the graph of ln N against t is a straight line.
(2)
(b)
Find the equation of the regression line of ln N on t.
(2)
(c)
Find the value of Pearson's correlation coefficient for ln N and t, and state what it indicates about the model.
(2)
(d)
Hence find the values of a and k.
(2)
(e)
Use the model to estimate the number of bacteria after 7 hours.
(2)
(f)
Find the time at which the model predicts that the number of bacteria first reaches 5000.
(3)
(g)
Comment on the reliability of your answer to the previous part.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: takes natural logs of both sides, ln N = ln a + ln(ekt)
✓ 1
R1: ln N = kt + ln a, which is of the form Y = mt + c with gradient k and intercept ln a, so linear
✓ 1
Part (b)
Gradient 0.435 (0.434604…)
✓ 1
Intercept 4.79 (4.78977…), giving ln N = 0.435t + 4.79
✓ 1
Part (c)
r = 1.00 (0.99995…)
✓ 1
Almost perfect positive linear relationship between ln N and t, so the exponential model is highly appropriate
✓ 1
Part (d)
M1: k = gradient = 0.435 and a = e4.78977…
✓ 1
A1: a = 120 (120.27…)
✓ 1
Part (e)
M1: N = 120.27e0.434604 × 7
✓ 1
A1: N ≈ 2520
✓ 1
Part (f)
M1: sets up equation 120.27e0.434604t = 5000
✓ 1
M1: solves, e.g. t = ln(5000 ÷ 120.27) ÷ 0.434604 or by GDC
✓ 1
A1: t = 8.58 hours (accept 8.57 from rounded values)
✓ 1
Part (g)
Unreliable: t = 8.58 is outside the data range (0–5 hours), and growth is likely to slow as nutrients or space run out
✓ 1
Exam tip: Keep the full GDC values of the gradient and intercept stored in memory. Rounding a to 120 and k to 0.435 before solving can shift the final answer in the third significant figure.
214IBMI-033-04-r2-05T-tests, confidence intervals and Markov chainsPaper 2Hard16 marks
In a city, every household subscribes to exactly one of two streaming services, X or Y. Each month, 85% of X subscribers stay with X and the rest switch to Y. Of Y subscribers, 75% stay with Y and the rest switch to X. Initially, 40% of households subscribe to X and 60% to Y.
(a)
Write down the transition matrix T, with columns representing the current service (X, Y).
(2)
(b)
Find the proportion of households subscribing to X after 3 months.
(2)
(c)
Find the eigenvalues of T and a corresponding eigenvector for each.
(4)
(d)
Hence, by writing \(\mathbf{T}=\mathbf{PDP}^{-1}\), show that the proportion subscribing to X after n months is \(0.625-0.225(0.6)^n\).
(3)
(e)
Find the least number of months after which more than 62% of households subscribe to X.
(3)
(f)
State the long-term proportion of households subscribing to X, justifying your answer.
Exam tip: Eigenvalues of a transition matrix always include 1; the other eigenvalue (here 0.6) controls how quickly the chain approaches its steady state.
215IBMI-033-06-r2-05Correlation and regressionPaper 2Hard11 marks
A cup of coffee is left to cool in a room kept at 20 °C. The excess temperature, D °C, is the coffee's temperature minus the room temperature. It is recorded every 5 minutes.
t (minutes): 0, 5, 10, 15, 20, 25 D (°C): 62.0, 47.5, 37.0, 28.6, 22.0, 17.1
It is suggested that the data can be modelled by \(D = ab^t\).
(a)
By considering ratios of consecutive values of D, justify the use of an exponential model.
(2)
(b)
Use exponential regression to find the values of a and b.
(3)
(c)
Interpret the value of b in context.
(1)
(d)
A linear model for D against t has coefficient of determination R2 = 0.965. Despite this high value, explain why the linear model is not suitable for predicting the coffee's temperature over a longer period.
(2)
(e)
Use your exponential model to estimate the time at which the temperature of the coffee is 45 °C.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: calculates ratios such as 47.5/62.0 = 0.766, 37.0/47.5 = 0.779, 28.6/37.0 = 0.773, 22.0/28.6 = 0.769, 17.1/22.0 = 0.777
✓ 1
R1: the ratios are approximately constant (about 0.77) over equal time intervals, which is the property of an exponential model
✓ 1
Part (b)
M1: exponential regression on the GDC, or a linear regression of ln D on t
✓ 1
A1: a = 61.8
✓ 1
A1: b = 0.950
✓ 1
Part (c)
Each minute the excess temperature is multiplied by 0.950, i.e. it falls by about 5% per minute
✓ 1
Part (d)
R1: the linear model predicts that D becomes zero and then negative (at about 33 minutes), so the coffee would cool below room temperature, which is unrealistic
✓ 1
R1: the data show a curved, decreasing rate of cooling. The exponential model approaches D = 0 asymptotically and fits this behaviour, so a high R2 alone does not make a model appropriate
✓ 1
Part (e)
M1: recognises that the excess temperature is D = 45 − 20 = 25
✓ 1
M1: solves 61.8(0.950)t = 25, e.g. graphically or with logarithms
✓ 1
A1: t = 17.6 minutes
✓ 1
Common error: The model is for the excess temperature, not the actual temperature. Subtract the room temperature before substituting into the model.
216IBMI-033-03-r1-06Normal distribution and chi-squared testsPaper 2Hard12 marks
A marine biologist measures the lengths, L cm, of 200 fish of one species. From the raw data, the mean is estimated as 30.0 cm and the standard deviation as 5.0 cm. She tests, at the 5% significance level, whether the lengths can be modelled by a normal distribution. The observed frequencies are:
L < 22: 15 22 ≤ L < 26: 25 26 ≤ L < 30: 62 30 ≤ L < 34: 50 34 ≤ L < 38: 36 L ≥ 38: 12
Under the model N(30, 52), the expected frequencies for 22 ≤ L < 26, 30 ≤ L < 34, 34 ≤ L < 38 and L ≥ 38 are 31.41, 57.63, 31.41 and 10.96 respectively.
(a)
Show that the expected frequency for 26 ≤ L < 30 is 57.6 to 3 significant figures, and find the expected frequency for L < 22.
(3)
(b)
State the null hypothesis and explain why the test has 3 degrees of freedom.
(3)
(c)
Find the χ2 statistic.
(2)
(d)
The critical value is 7.815. State the conclusion of the test in context.
(2)
(e)
A colleague suggests using twelve classes, each 2 cm wide, to make the test more detailed. Explain one problem with this suggestion.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1: P(26 ≤ L < 30) = 0.28814…
✓ 1
A1: 200 × 0.28814… = 57.63 ≈ 57.6
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A1: 200 × P(L < 22) = 200 × 0.054799 = 10.96
✓ 1
Part (b)
A1: H0: the fish lengths can be modelled by a normal distribution
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R1: two parameters (the mean and the standard deviation) were estimated from the data, and the expected frequencies must total 200
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A1: 6 − 1 − 2 = 3
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Part (c)
M1: \(\sum \frac{(O-E)^2}{E}\) over all six classes
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A1: χ2 = 4.91 (4.908…)
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Part (d)
R1: 4.91 < 7.815
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A1: do not reject H0; there is insufficient evidence that the lengths are not normally distributed, so the normal model may be used
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Part (e)
R1: the expected frequencies in the tail classes would fall below 5
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R1: those classes would have to be combined, which reduces the degrees of freedom and removes the extra detail gained
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Exam tip: Subtract one degree of freedom for every parameter estimated from the sample, as well as one for the fixed total. For a normal model with both μ and σ estimated, that is 3 in total.
217IBMI-033-04-r1-06T-tests, confidence intervals and Markov chainsPaper 2Hard14 marks
A car-rental company has depots at the Airport (A), the Beach (B) and the City (C). Each car is rented once a day and returned to one of the depots. For a car rented from A, the probabilities of return to A, B and C are 0.6, 0.3 and 0.1 respectively. For a car rented from B they are 0.2, 0.7 and 0.1. For a car rented from C they are 0.2, 0.2 and 0.6. Assume the return depot depends only on the depot of rental.
(a)
Write down the transition matrix T, with columns representing the depot of rental (A, B, C) and rows the depot of return.
(2)
(b)
A car starts at A. Find the probability that it is at C after two rentals.
(2)
(c)
A car starts at A. Find the probability that it is not returned to C at any time during its next three rentals, justifying your method.
(3)
(d)
Find the steady-state probability vector for the location of a car.
(4)
(e)
The company has 150 cars. The City depot has 25 parking spaces. Find the expected number of cars at C in the long term and comment on whether the depot is large enough.
R1: 30 > 25, so in the long term the City depot is expected to be too small
✓ 1
Exam tip: Before multiplying out matrices, look for structure: equal probabilities of moving to C from both A and B make part (c) a one-line calculation.
218IBMI-033-08-r1-05Normal distribution and chi-squared testsPaper 2Hard14 marks
At an airport, the masses of adult passengers, M kg, are modelled by N(78, 122), and the masses of their cabin luggage, B kg, are modelled by N(15, 42). All masses are independent.
(a)
Find the probability that a randomly chosen passenger is more than 70 kg heavier than a randomly chosen piece of luggage.
(3)
(b)
A shuttle bus has a maximum safe load of 1100 kg. Eleven passengers, each with one piece of luggage, board the bus. Find the probability that the total mass exceeds the safe load.
(3)
(c)
A student models the total mass of the eleven passengers as 11M. Explain why this model is not appropriate.
(2)
(d)
The operator requires the probability that the total mass of passengers and luggage exceeds 1100 kg to be less than 0.01. Find the largest number of passengers, each with luggage, who may board.
(3)
(e)
A random sample of 25 passengers is weighed. Find the value of k such that the probability that the sample mean mass exceeds k kg is 0.05.
(3)
Show mark scheme
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Notes
Part (a)
E(M − B) = 78 − 15 = 63
✓ 1
Var(M − B) = 144 + 16 = 160
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P(M − B > 70) = 0.290
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Part (b)
Mean of total = 11 × (78 + 15) = 1023
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Variance of total = 11 × 160 = 1760
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P(T > 1100) = 0.0332
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Part (c)
11M assumes all eleven passengers have the same mass, but they are different, independent people, so the total is M1 + … + M11
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Var(11M) = 121 × 144 = 17424, whereas Var(M1 + … + M11) = 11 × 144 = 1584, so the model greatly overstates the spread
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Part (d)
For n passengers, total ~ N(93n, 160n)
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Testing values or solving 1100 − 93n ≥ 2.326√(160n): n = 11 gives 0.0332 > 0.01, while n = 10 gives about 1.1×10−5 < 0.01
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n = 10
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Part (e)
\(\bar{M}\) ~ N(78, \(\frac{144}{25}\)), so the standard deviation is 2.4
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Use of inverse normal: k = 78 + 1.645 × 2.4
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k = 81.9 kg
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Common error: For a difference of independent variables, the variances add: Var(M − B) = 144 + 16, not 144 − 16. Also distinguish the sum of n independent copies (variance nσ²) from a multiple nM (variance n²σ²).
219IBMI-033-06-r1-05Correlation and regressionPaper 2Hard10 marks
A pharmacologist measures the concentration, C mg L−1, of a drug in a patient's blood t hours after it was given.
It is believed that the data can be modelled by an equation of the form C = aekt, where a and k are constants.
(a)
Find the equation of the regression line of ln C on t.
(3)
(b)
Hence find the values of a and k.
(3)
(c)
Interpret the value of a in context.
(1)
(d)
The drug is no longer effective when its concentration falls below 5 mg L−1. Use the model to estimate the time after which the drug is no longer effective.
(2)
(e)
Comment on the reliability of your answer to the previous part.
(1)
Show mark scheme
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Mark
Notes
Part (a)
Calculates ln C for each value, e.g. 4.086, 3.784, 3.490, 3.182, 2.890, 2.580
✓ 1
Gradient = −0.301 (−0.30053…)
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ln C = −0.301t + 4.39 (intercept 4.3874…)
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Part (b)
Uses a = eintercept with the unrounded intercept, e4.3874…
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a = 80.4
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k = −0.301
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Part (c)
The model estimates the concentration immediately after the drug is given (t = 0) as about 80.4 mg L−1
✓ 1
Part (d)
Sets up 80.43e−0.3005t = 5 and solves it by graph or by logarithms
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t = 9.24 hours
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Part (e)
9.24 hours is beyond the last measurement at 6 hours, so it is extrapolation and may be unreliable
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Exam tip: Keep the unrounded intercept from the ln C regression when you calculate a = eiⁿtercept. Rounding to 4.39 first gives a = 80.6, which loses accuracy.
220IBMI-033-01-r2-06Correlation and regressionPaper 2Hard12 marks
An oceanographer measures the speed, v m s−1, of waves travelling over water of depth d m at five coastal sites.
d: 1, 2, 4, 8, 16 v: 3.0, 4.3, 6.1, 8.4, 12.2
She proposes a model of the form \(v = ad^{\,b}\).
(a)
By considering ln v and ln d, find the equation of the regression line of ln v on ln d.
(3)
(b)
Hence find the values of a and b.
(2)
(c)
A colleague fits a linear model, v = md + c, to the original data. Find the coefficient of determination, R2, for this linear model.
(2)
(d)
The regression of ln v on ln d has R2 = 0.9996. Give two reasons why the power model is more appropriate than the linear model.
(2)
(e)
Use the power model to estimate the wave speed where the water depth is 12 m.
(2)
(f)
Comment on the reliability of using the model to estimate the wave speed where the depth is 100 m.
(1)
Show mark scheme
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Notes
Part (a)
Takes natural logs of both variables and uses the GDC (M1)
✓ 1
gradient = 0.501 (A1)
✓ 1
intercept = 1.10, so ln v = 0.501 ln d + 1.10 (A1)
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Part (b)
a = e1.10… = 3.02 (A1)
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b = 0.501 (A1)
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Part (c)
Uses the GDC for linear regression of v on d, giving v = 0.589d + 3.15 (M1)
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R² = 0.975 (A1)
✓ 1
Part (d)
The residuals from the linear model show a systematic pattern. At d = 1, 2, 4, 8 and 16 they are about −0.74, −0.03, +0.60, +0.54 and −0.37: negative, then positive, then negative. This indicates curvature that the power model captures, since it fits the transformed data almost perfectly (R² = 0.9996) (R1)
✓ 1
The linear model has intercept c ≈ 3.15. It predicts a wave speed of about 3.15 m s−1 at zero depth, which is unrealistic, whereas the power model gives v = 0 when d = 0 (R1)
✓ 1
[Accept other valid reasons, e.g. the power model's predictions are much closer to every observed value.]
—
Part (e)
Substitutes d = 12: v = 3.02 × 120.501 (M1)
✓ 1
v = 10.5 m s−1 (A1)
✓ 1
Part (f)
Unreliable: 100 m is far outside the data range of 1 m to 16 m (extrapolation), and the relationship may change in deeper water (R1)
✓ 1
Exam tip: For a power model \(y = ax^b\), the gradient of the ln y against ln x line is b and the intercept is ln a. R² for ln y against ln x is not on the same scale as R² for y against x, so support a choice of model with residuals or context as well.
221IBMI-033-03-r2-06Normal distribution and chi-squared testsPaper 2Hard16 marks
A seed company claims that the heights, H cm, of its tomato seedlings three weeks after planting are normally distributed, with 15% of seedlings shorter than 8 cm and 20% taller than 14 cm.
(a)
Find the mean μ and standard deviation σ of H implied by the company's claim.
(5)
(b)
Using the claim and the unrounded values of μ and σ from (a), find P(8 ≤ H < 11) and P(11 ≤ H < 14).
(3)
(c)
A gardener measures 200 of the company's seedlings. The results are: H < 8: 22; 8 ≤ H < 11: 70; 11 ≤ H < 14: 61; H ≥ 14: 47. She tests the company's claim at the 5% significance level. State the null hypothesis.
(1)
(d)
Find the expected frequencies for the four height groups.
(2)
(e)
Write down the number of degrees of freedom and find the χ2 statistic.
(3)
(f)
The critical value for this test is 7.815. State the gardener's conclusion, giving a reason.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Use of inverse normal for 0.15 and 0.80 (M1)
✓ 1
z = −1.0364 and z = 0.8416 (A1)
✓ 1
μ − 1.0364σ = 8 and μ + 0.8416σ = 14 (M1)
✓ 1
σ = 3.19 cm (3.1948…) (A1)
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μ = 11.3 cm (11.311…) (A1)
✓ 1
Part (b)
Use of normal cdf with the unrounded μ and σ from (a) (M1)
✓ 1
P(8 ≤ H < 11) = 0.311 (0.3112…) (A1)
✓ 1
P(11 ≤ H < 14) = 0.339 (0.3388…) (A1)
✓ 1
(Accept 0.312 and 0.338 from μ = 11.3, σ = 3.19.)
—
Part (c)
H0: the heights follow the distribution claimed by the company, N(11.3, 3.192) (A1)
✓ 1
Part (d)
H < 8: 30 and H ≥ 14: 40 (A1)
✓ 1
8 ≤ H < 11: 62.2 and 11 ≤ H < 14: 67.8 (A1)
✓ 1
(Accept 62.5 and 67.5 following through from 0.312 and 0.338.)
—
Part (e)
Degrees of freedom = 4 − 1 = 3, because the parameters come from the claim and are not estimated from the data (A1)
✓ 1
Correct substitution into Σ(O − E)2/E (M1)
✓ 1
χ2 = 5.00 (accept 4.88 following through from 62.5 and 67.5) (A1)
✓ 1
Part (f)
5.00 < 7.815, so do not reject H0 (R1)
✓ 1
There is insufficient evidence to doubt the company's claim about the seedling heights (A1)
✓ 1
Do not award A1 without R1.
—
Exam tip: Degrees of freedom are reduced only for parameters estimated from the sample data. Here μ and σ come from the company's claim, so df = 4 − 1 = 3. Keep unrounded values of μ and σ in your calculator for later parts.
222IBMI-033-08-r2-06Normal distribution and chi-squared testsPaper 2Hard13 marks
A biologist measures the lengths, in cm, of 200 trout caught in a lake. From the data, she estimates the mean length to be 32.0 cm and the standard deviation to be 4.0 cm. The lengths are grouped as follows.
Length L < 26: 18 26 ≤ L < 30: 41 30 ≤ L < 34: 72 34 ≤ L < 38: 52 L ≥ 38: 17
She wants to test, at the 5% significance level, whether the lengths can be modelled by a normal distribution.
(a)
State suitable hypotheses for this test.
(1)
(b)
Show that the expected frequency for the class 30 ≤ L < 34 is 76.6, correct to 3 significant figures.
(3)
(c)
Find the expected frequencies for the remaining four classes.
(2)
(d)
Determine the number of degrees of freedom for this test, giving a reason.
(2)
(e)
Find the χ2 test statistic and the p-value.
(3)
(f)
State the conclusion of the test in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
H0: trout lengths follow a normal distribution (accept N(32.0, 4.02)); H1: trout lengths do not follow a normal distribution
✓ 1
Part (b)
Uses L ~ N(32.0, 4.02) with P(30 ≤ L < 34)
✓ 1
Probability = 0.38292…
✓ 1
200 × 0.38292… = 76.58… ≈ 76.6
✓ 1
Part (c)
Probabilities 0.0668 and 0.2417 used (or symmetry noted)
✓ 1
L < 26: 13.4; 26–30: 48.3; 34–38: 48.3; L ≥ 38: 13.4
✓ 1
Part (d)
5 − 1 − 2 = 2 degrees of freedom
✓ 1
Reason: two parameters (mean and standard deviation) were estimated from the sample, in addition to the total being fixed
✓ 1
Part (e)
Correct use of \(\sum \frac{(O-E)^2}{E}\) with the five expected frequencies found above
✓ 1
χ2calc = 4.27 (4.268…)
✓ 1
p-value = 0.118 (with 2 degrees of freedom)
✓ 1
Part (f)
0.118 > 0.05, so do not reject H0
✓ 1
There is insufficient evidence to reject a normal model for the trout lengths; N(32.0, 4.02) is a reasonable model
✓ 1
Common error: Calculator goodness of fit programs often default to (number of classes − 1) degrees of freedom. Here that gives 4 instead of 2, so enter the degrees of freedom manually or the p-value will be wrong.
223IBMI3H-4-08-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A farmer records the crop yield, y tonnes per hectare, for different amounts of fertiliser, x (in units of 10 kg per hectare). For the whole data set the total sum of squares is \(\sum(y_i-\bar y)^2=482.5\). Three models are fitted. The sums of squares of the residuals are:
Linear model: 96.5 Exponential model: 42.4 Quadratic model: 14.5
(a)
Use the relationship \(R^2=1-\frac{SS_{res}}{SS_{tot}}\) to find \(R^2\) for the linear model.
(2)
(b)
Find \(R^2\) for the exponential model and for the quadratic model.
(2)
(c)
Identify the model that best fits the data and give one reason, other than R2, why R2 alone is not enough to decide a model is appropriate.
(2)
(d)
The quadratic model is \(y=-0.45x^2+5.1x+2.3\). Find the amount of fertiliser that gives the greatest predicted yield, and that yield.
(3)
(e)
Find the value of x at which the quadratic model predicts a yield of zero.
(2)
(f)
Comment on the validity of the quadratic model for values of x greater than your answer to part (e).
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(1-\frac{96.5}{482.5}\)
✓ 1
A1 0.800
✓ 1
Part (b)
A1 exponential: 0.912
✓ 1
A1 quadratic: 0.970
✓ 1
Part (c)
A1 quadratic (highest R2)
✓ 1
R1 a high R2 does not show the model is sensible outside the data or that residuals are random (e.g. a plot of residuals or context should also be checked)
✓ 1
Part (d)
M1 differentiate or use the vertex formula: \(-0.9x+5.1=0\)
✓ 1
A1 x = 5.67, i.e. 56.7 kg per hectare
✓ 1
A1 maximum yield 16.8 tonnes per hectare
✓ 1
Part (e)
M1 solve \(-0.45x^2+5.1x+2.3=0\)
✓ 1
A1 x = 11.8
✓ 1
Part (f)
R1 the model predicts negative yields, which is impossible, so it only applies within a limited range of fertiliser amounts
✓ 1
Exam tip: For a quadratic maximum, give both the input value and the output value, with units that match the context.
224IBMI3H-4-08-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
The resting heart rates of 15 randomly chosen patients have mean 62.4 beats per minute (bpm) and sample standard deviation 5.1 bpm. Heart rates are assumed to be normally distributed.
(a)
Find a 95% confidence interval for the population mean heart rate.
(3)
(b)
Find a 99% confidence interval for the population mean.
(2)
(c)
A doctor claims the population mean is 60 bpm. State whether the intervals support the claim.
(1)
(d)
Explain why the 99% interval is wider than the 95% interval.
(1)
(e)
Suppose that the population standard deviation is known to be exactly 5.1 bpm. Find the smallest sample size for which a 95% confidence interval for the mean has a margin of error of at most 1.5 bpm.
(3)
(f)
Find the width of the confidence interval for this sample size.
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of the t-distribution with 14 degrees of freedom, t = 2.145
✓ 1
A1 margin of error \(2.145\times\frac{5.1}{\sqrt{15}}=2.82\)
✓ 1
A1 (59.6, 65.2)
✓ 1
Part (b)
M1 t = 2.977 with 14 degrees of freedom
✓ 1
A1 (58.5, 66.3)
✓ 1
Part (c)
R1 60 lies inside both intervals, so the data are consistent with the claim
✓ 1
Part (d)
R1 greater confidence requires a larger critical value, so a wider interval is needed to capture the mean more often
Common error: When a population standard deviation is given, use the normal critical value; the t value is for an estimated standard deviation from a small sample. For sample size, always round up.
225IBMI3H-4-08-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard14 marks
A tyre company wishes to show that tyre A has a longer mean braking distance on a wet road than tyre B. Braking distances, in metres, from the same speed:
Tyre A: n = 12, mean 54.2, sample standard deviation 4.1 Tyre B: n = 10, mean 50.1, sample standard deviation 3.6
Assume both populations are normal with equal variances. A one-tailed pooled t-test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Calculate the pooled estimate of the common variance.
(2)
(c)
Calculate the t-statistic.
(2)
(d)
Write down the degrees of freedom and the critical value, or the p-value.
(2)
(e)
State the conclusion of the test in context.
(2)
(f)
State two assumptions required for the test to be valid.
(2)
(g)
Find the smallest difference between the sample means, for these sample sizes and standard deviations, that would lead to rejecting H0 at the 5% level.
Exam tip: Write the critical value and the t-statistic side by side before concluding; stating only 'significant' without the comparison does not earn the reasoning mark.
226IBMI3H-4-08b-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
At a bus interchange, buses of route A arrive at random at a mean rate of 3 per 10 minutes and buses of route B arrive at random at a mean rate of 2 per 10 minutes. The two routes are independent and the numbers of arrivals follow Poisson distributions.
(a)
State the distribution of the total number of buses, T, arriving in a 10-minute period, and state one assumption that is needed.
(3)
(b)
Find P(T = 7).
(2)
(c)
Find the probability that exactly 4 route A buses and 3 route B buses arrive in a 10-minute period.
(2)
(d)
Given that exactly 7 buses arrive in a 10-minute period, find the probability that exactly 4 of them are route A buses.
(3)
(e)
Show that your answer to part (d) equals the binomial probability P(Y = 4) where Y ~ B(7, 0.6), and explain why 0.6 is the appropriate probability.
(2)
Show mark scheme
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Notes
Part (a)
M1 sum of independent Poisson variables is Poisson
✓ 1
A1 T ~ Po(5)
✓ 1
R1 arrivals of the two routes are independent
✓ 1
Part (b)
M1 e−5 57 / 7!
✓ 1
A1 0.104
✓ 1
Part (c)
M1 P(A = 4) × P(B = 3)
✓ 1
A1 0.0303
✓ 1
Part (d)
M1 conditional probability P(A = 4 and B = 3) / P(T = 7)
✓ 1
A1 0.03032 / 0.1044
✓ 1
A1 0.29
✓ 1
Part (e)
A1 C(7, 4) × 0.64 × 0.43 = 0.290
✓ 1
R1 each of the 7 buses is independently from route A with probability 3/(3 + 2) = 0.6
✓ 1
Exam tip: Given a fixed total of Poisson events from independent sources, each source's count is binomial, with probability equal to its share of the total mean.
227IBMI3H-4-08b-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A bus operator records journey times, in minutes, which are normally distributed with a known standard deviation of 4.0 minutes. A 95% confidence interval for the mean journey time, based on a random sample of n journeys, is (49.02, 50.98).
(a)
Write down the sample mean.
(1)
(b)
Write down the half-width of the interval.
(1)
(c)
Find the value of n.
(3)
(d)
Use the same sample to find a 99% confidence interval for the mean journey time.
(3)
(e)
The operator advertises that the mean journey time is 52 minutes. State, with a reason, whether the 95% interval supports this.
(2)
(f)
State why a z value rather than a t value is used here.
(2)
Show mark scheme
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Notes
Part (a)
A1 midpoint = 50.0 minutes
✓ 1
Part (b)
A1 (50.98 − 49.02)/2 = 0.98
✓ 1
Part (c)
M1 1.96 × 4.0 / √n = 0.98
✓ 1
M1 √n = 8
✓ 1
A1 n = 64
✓ 1
Part (d)
M1 z = 2.576
✓ 1
A1 half-width 2.576 × 4/8 = 1.288
✓ 1
A1 (48.7, 51.3) minutes
✓ 1
Part (e)
M1 compares 52 with (49.02, 50.98)
✓ 1
R1 52 lies outside the interval, so the advertised mean is not plausible at this level
✓ 1
Part (f)
R1 the population standard deviation is known
✓ 1
R1 so the standardised sample mean is normal (not t-distributed)
✓ 1
Common error: A wider confidence level gives a wider interval from the same sample. Check that your 99% interval contains the 95% interval.
228IBMI3H-4-08b-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains and transition matrices; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
The condition of an e-bike battery is checked monthly and recorded as Good (G), Degraded (D) or Failed (F). Transitions follow a Markov chain with transition matrix (rows = this month, columns = next month, in the order G, D, F): G: 0.85, 0.12, 0.03 D: 0, 0.75, 0.25 F: 0, 0, 1 A new battery starts in state G.
(a)
Explain why F is called an absorbing state, and state the long-term probability that the battery is in state F.
(2)
(b)
Find the probability that the battery is still in state G after 3 months.
(2)
(c)
Find the smallest number of months n for which the probability that the battery is in state F exceeds 0.5.
(4)
(d)
Find the probability that the battery has failed within 4 months, by finding the full state distribution after 4 months.
(4)
Show mark scheme
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Mark
Notes
Part (a)
R1 once in F the battery stays in F with probability 1
✓ 1
A1 long-term probability of F is 1
✓ 1
Part (b)
M1 0.853 (the only way is to stay in G each month)
✓ 1
A1 0.614
✓ 1
Part (c)
M1 use of vTn with v = (1, 0, 0)
✓ 1
A1 P(F) after 7 months = 0.455
✓ 1
A1 P(F) after 8 months = 0.521
✓ 1
A1 n = 8
✓ 1
Part (d)
M1 repeated multiplication or T4
✓ 1
A1 P(G) = 0.854 = 0.522
✓ 1
A1 P(D) = 0.247
✓ 1
A1 P(F) = 1 − 0.522 − 0.247 = 0.231
✓ 1
Exam tip: With an absorbing state, the probability of having been absorbed by month n equals the probability of being in that state at month n.
229IBMI3H-4-04r7-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A Formula 1 team's cars suffer power-unit failures at random. Historically the number of failures in any 10-race period is modelled by a Poisson distribution with mean 8. After fitting a redesigned part, the team records \(X\) failures in the next 10 races and wishes to test, at the 5% significance level, whether the failure rate has decreased. They observe \(X=3\).
(a)
State suitable null and alternative hypotheses, defining the parameter used.
(2)
(b)
Calculate the \(p\)-value for this test.
(3)
(c)
State the conclusion of the test, in context.
(2)
(d)
Find the critical region for this test.
(2)
(e)
Suppose that the redesign has in fact reduced the true mean to 3 failures per 10 races. Find the probability of a Type II error for this test.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\lambda=8\), where \(\lambda\) is the mean number of failures per 10 races
✓ 1
A1 \(H_1:\lambda<8\)
✓ 1
Part (b)
M1 under \(H_0\), \(X\sim\mathrm{Po}(8)\)
✓ 1
M1 \(P(X\le3)\)
✓ 1
A1 0.0424
✓ 1
Part (c)
R1 \(0.0424<0.05\)
✓ 1
A1 reject \(H_0\): there is evidence at the 5% level that the failure rate has decreased
✓ 1
Part (d)
M1 evaluate \(P(X\le3)=0.0424\) and \(P(X\le4)=0.0996\)
✓ 1
A1 critical region is \(X\le3\)
✓ 1
Part (e)
M1 true distribution \(X\sim\mathrm{Po}(3)\)
✓ 1
M1 \(P(X\ge4)=1-P(X\le3)\) (ECF from (d))
✓ 1
A1 0.353
✓ 1
Common error: A Type II error means failing to reject \(H_0\) when it is false, so its probability is found with the true parameter value, not the value in \(H_0\).
230IBMI3H-4-04r7-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Each morning a patient in a hospital is in one of three states: Ward (W), Intensive care (I) or Home (H). The daily changes are modelled by a Markov chain with \(\mathbf{s}_{n+1}=\mathbf{T}\mathbf{s}_n\), where the entry in row \(i\), column \(j\) is the probability of moving from state \(j\) to state \(i\). In the order W, I, H: \(\mathbf{T}=\begin{pmatrix}0.6&0.35&0\\0.1&0.5&0\\0.3&k&1\end{pmatrix}\).
(a)
Find the value of \(k\).
(1)
(b)
On day 0, 80 patients are on the ward and 20 are in intensive care. Find the expected number of these patients at home on day 3, correct to the nearest whole number.
(3)
(c)
Explain why, in the long term, every patient is expected to be at home.
(2)
(d)
Find the smallest number of days after which at least 90% of the 100 patients are expected to be at home.
(3)
(e)
A patient is in intensive care today. Find the probability that the patient is at home in two days' time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(k=1-0.35-0.5=0.15\)
✓ 1
Part (b)
M1 attempt to find \(\mathbf{T}^3\begin{pmatrix}80\\20\\0\end{pmatrix}\)
✓ 1
A1 third entry 60.165
✓ 1
A1 60 patients
✓ 1
Part (c)
R1 \(H\to H\) has probability 1 (the column is \((0,0,1)^T\)), so a patient who reaches Home stays there
✓ 1
R1 from W and I there is always a positive chance of moving to H, so the proportion in W and I tends to 0
✓ 1
Part (d)
M1 attempt to find the expected number at home on successive days
✓ 1
A1 day 7: 87.9 (below 90) and day 8: 91.0 (at least 90)
✓ 1
A1 8 days
✓ 1
Part (e)
M1 attempt to use \(\mathbf{T}^2\) or add paths: \(I\to H\), \(I\to I\to H\) and \(I\to W\to H\)
✓ 1
A1 \(0.15+0.5\times0.15+0.35\times0.3\)
✓ 1
A1 0.33
✓ 1
Common error: A state with \(P(\text{stay})=1\) is absorbing; chains with an absorbing state do not have a steady state that depends on where you start, so check before you solve \(\mathbf{T}\mathbf{s}=\mathbf{s}\).
231IBMI3H-4-04r7-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
A Formula 1 team studies the extra time per lap, \(y\) seconds, lost as tyres age. The tyre age \(x\) is in laps. A quadratic model \(y=0.0065x^2+0.03x\) is proposed. The data and the model values are:
x (laps): 2, 6, 10, 14, 18, 22 y observed (s): 0.1, 0.4, 0.8, 1.5, 2.4, 3.6 y model (s): 0.086, 0.414, 0.950, 1.694, 2.646, 3.806
The coefficient of determination is \(R^2=1-\dfrac{SS_{res}}{SS_{tot}}\), where \(SS_{res}=\sum(y-\hat y)^2\) and \(SS_{tot}=\sum(y-\bar y)^2\).
(a)
Calculate \(SS_{res}\).
(3)
(b)
Show that \(\bar y=1.47\) (to 3 significant figures) and calculate \(SS_{tot}\).
(2)
(c)
Hence find \(R^2\) for the model, correct to 3 significant figures.
(2)
(d)
A linear model fitted to the same data has \(R^2=0.93\). Compare the two models, referring to the values of \(R^2\).
(2)
(e)
Use the quadratic model to predict the extra time lost when the tyre age is 26 laps. Comment briefly on the reliability of this prediction.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the residuals \(y-\hat y\)
✓ 1
M1 attempt to square and add
✓ 1
A1 0.163
✓ 1
Part (b)
A1 \(\bar y=\frac{8.8}{6}=1.4667\)
✓ 1
A1 \(SS_{tot}=8.87\)
✓ 1
Part (c)
M1 \(1-\frac{0.163}{8.87}\) (ECF from (a) and (b))
✓ 1
A1 0.982
✓ 1
Part (d)
R1 the quadratic model has the larger \(R^2\) (0.982 against 0.93)
✓ 1
R1 so it explains a greater proportion of the variation in extra time and fits the data better
✓ 1
Part (e)
A1 \(0.0065\times26^2+0.03\times26=5.17\) s
✓ 1
R1 26 is outside the data range (2 to 22 laps), so this is extrapolation and may be unreliable
✓ 1
Common error: A high \(R^2\) shows a good fit within the range of the data only; it does not make predictions beyond that range reliable.
232IBMI3H-4-03r9-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Seismic tremors beneath the Orlo volcano are recorded in 10-minute intervals. The number of tremors in one interval is modelled by a Poisson distribution with mean λ, and tremors occur independently. The probability that an interval contains no tremors is 0.2. Give final answers to 3 significant figures.
(a)
Show that λ = ln 5, and find its value to 3 significant figures.
(3)
(b)
Find the probability that an interval contains at least 3 tremors.
(2)
(c)
Find the probability that more than 8 tremors are recorded in a one-hour period (six intervals).
(3)
(d)
Given that an interval contains at least one tremor, find the probability that it contains exactly two tremors.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 e−λ = 0.2
✓ 1
A1 λ = −ln 0.2 = ln 5
✓ 1
A1 λ = 1.61
✓ 1
Part (b)
M1 X ~ Po(ln 5); P(X ≥ 3) = 1 − P(X ≤ 2)
✓ 1
A1 = 0.219
✓ 1
Part (c)
A1 One-hour mean = 6 ln 5 = 9.66
✓ 1
M1 P(Y > 8) = 1 − P(Y ≤ 8)
✓ 1
A1 = 0.627
✓ 1
Part (d)
A1 P(X = 2) = (ln 5)2 × 0.2 ÷ 2 = 0.259
✓ 1
A1 P(X ≥ 1) = 1 − 0.2 = 0.8
✓ 1
M1 Conditional probability = P(X = 2) ÷ P(X ≥ 1)
✓ 1
A1 = 0.324
✓ 1
Exam tip: For a conditional Poisson probability, divide by P(X ≥ 1), not by 1; the condition removes the zero outcome from the sample space.
233IBMI3H-4-03r9-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard13 marks
The mass, M tonnes, of a loaded shipping container is normally distributed with mean 24.0 and standard deviation 1.8. Container masses are independent. Three containers are stacked; their masses are M1, M2 and M3. Give probabilities to 3 significant figures.
(a)
Let S = M1 + M2 + M3. Find E(S) and Var(S).
(2)
(b)
A crane operator estimates the stack mass as 3M1. Find E(3M1) and Var(3M1), and explain why Var(3M1) is larger than Var(S).
(3)
(c)
The stack must not exceed 75.0 tonnes. Find P(S > 75.0).
(2)
(d)
Let W = M1 − (M2 + M3)/2. Find P(W > 2).
(4)
(e)
Find the value of c, to 3 significant figures, such that P(W > c) = 0.10.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(S) = 72.0 tonnes
✓ 1
A1 Var(S) = 3 × 1.82 = 9.72
✓ 1
Part (b)
A1 E(3M1) = 72.0
✓ 1
A1 Var(3M1) = 9 × 3.24 = 29.16
✓ 1
R1 Tripling one container makes its deviation count three times, whereas for three independent containers high and low masses partly cancel
✓ 1
Part (c)
M1 Standard deviation of S = √9.72 = 3.118
✓ 1
A1 P(S > 75.0) = 0.168
✓ 1
Part (d)
A1 E(W) = 0
✓ 1
M1 Var(W) = 1.82 + ¼ × 1.82 + ¼ × 1.82
✓ 1
A1 = 4.86
✓ 1
A1 P(W > 2) = 0.182
✓ 1
Part (e)
M1 Use z = 1.2816 for the upper 10%
✓ 1
A1 c = 1.2816 × √4.86 = 2.83
✓ 1
Common error: In Var(aX + bY) the coefficients are squared: halving a variable quarters its variance contribution.
234IBMI3H-4-03r9-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
A fleet of 240 container ships moves weekly between three ports A, B and C. A ship's weekly movement follows a Markov chain with transition matrix \(T=\begin{pmatrix}0.6&0.2&0.3\\0.3&0.7&0.1\\0.1&0.1&0.6\end{pmatrix}\) where the columns give the probabilities of moving from A, B and C respectively, and the rows give the probabilities of moving to A, B and C respectively. A ship starts in port A.
(a)
Find the probability that the ship is in port C after three weeks.
(2)
(b)
Given that the ship is in port C after two weeks, find the probability that it was in port B after one week.
(4)
(c)
In the long term, the fleet's ports settle to a fixed distribution. Find the long-term proportion of ships in each port, and hence the expected number of the 240 ships in port B.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Attempt to find T3 applied to (1, 0, 0)
✓ 1
A1 Probability = 0.175
✓ 1
Part (b)
A1 Probability of A→B→C = 0.3 × 0.1 = 0.03
✓ 1
A1 Probability of being in C after two weeks = 0.15
✓ 1
M1 Use of conditional probability, 0.03 ÷ 0.15
✓ 1
A1 = 0.2
✓ 1
Part (c)
M1 Set up Ts = s with the entries summing to 1
✓ 1
A1 Solve to obtain A = 11/30, B = 13/30, C = 1/5
✓ 1
M1 Expected number in B = 240 × 13/30
✓ 1
A1 = 104 ships
✓ 1
Exam tip: In a conditional probability on a Markov chain, the numerator is the probability of the specific path and the denominator is the total probability of the condition.
235IBMI3H-4-03r12-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
Flint fragments are scattered across an excavation site so that the number of fragments in a 1 m by 1 m square follows a Poisson distribution with mean 0.8. The numbers of fragments in different squares are independent.
(a)
Find the probability that a given square contains no fragments. Give your answer to 4 significant figures.
(2)
(b)
Five squares are chosen at random. Find the probability that exactly 2 of them contain no fragments. Use the unrounded value from (a) and give your answer to 3 significant figures.
(3)
(c)
The team excavates \(n\) squares and counts the total number of fragments, T. Find the least value of \(n\) for which \(P(T\geq 10)>0.95\).
(4)
(d)
For the value of \(n\) found in (c), find the probability that exactly 10 fragments are found in total. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(e^{-0.8}\)
✓ 1
A1 0.4493
✓ 1
Part (b)
M1 recognises binomial model with \(n=5\) and \(p=e^{-0.8}\) (the probability of an empty square)
✓ 1
M1 \(\binom{5}{2}p^2(1-p)^3\) (ECF from (a))
✓ 1
A1 0.337
✓ 1
Part (c)
M1 T follows a Poisson distribution with mean \(0.8n\)
✓ 1
M1 attempt to evaluate \(P(T\geq 10)=1-P(T\leq 9)\) for different values of \(n\)
✓ 1
A1 \(n=19\) gives 0.936, which is below 0.95 (or equivalent comparison)
✓ 1
A1 \(n=20\) gives 0.957, which exceeds 0.95, so least \(n=20\)
Common error: A Poisson model for counts and a binomial model for "how many squares are empty" are different layers: first find the probability for one square, then treat squares as repeated trials.
236IBMI3H-4-03r12-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
A server in a data centre is monitored once an hour and is in one of three states: Active (A), Idle (I) or Under maintenance (M). The hourly changes of state are modelled by a Markov chain. • From A: the server stays in A with probability 0.8, moves to I with probability 0.15 and moves to M with probability 0.05. • From I: the server moves to A with probability 0.4 and stays in I otherwise. • From M: the server moves to A with probability \(k\) and stays in M otherwise.
(a)
Write down the transition matrix T, using the order A, I, M for rows and columns, with each column giving the probabilities of moving from a state.
(2)
(b)
Show that the probability that a server that is under maintenance is again under maintenance two hours later is \((1-k)^2+0.05k\).
(2)
(c)
In the long run, 10% of the time is spent in state M. Find the value of \(k\), giving your answer to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 first column \(0.8,\,0.15,\,0.05\) and second column \(0.4,\,0.6,\,0\)
✓ 1
A1 third column \(k,\,0,\,1-k\)
✓ 1
Part (b)
M1 considers the routes M to M to M and M to A to M (M to I is impossible)
✓ 1
A1 \((1-k)(1-k)+k\times0.05\), which is the required expression (AG)
✓ 1
Part (c)
M1 steady state \(T\mathbf{s}=\mathbf{s}\) with \(\mathbf{s}=(a,i,m)\), \(a+i+m=1\)
✓ 1
A1 from the I row: \(0.4i=0.15a\), so \(i=0.375a\)
✓ 1
A1 from the M row: \(km=0.05a\)
✓ 1
M1 with \(m=0.1\): \(a+0.375a=0.9\), so \(a=0.6545\ldots\) (that is, \(\frac{36}{55}\))
✓ 1
A1 \(k=\frac{0.05a}{0.1}=\frac{18}{55}=0.327\)
✓ 1
Common error: A steady-state equation set is redundant: one equation from \(T\mathbf{s}=\mathbf{s}\) is always implied by the others, so you must also use that the proportions sum to 1.
237IBMI3H-4-03r12-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A cycling shop compares two brands of tyre. The distance (in km) ridden before the first puncture was recorded for a random sample of tyres of each brand. Brand A (7 tyres): 412, 438, 395, 451, 420, 433, 407 Brand B (8 tyres): 389, 402, 371, 415, 380, 396, 408, 384 The shop owner wishes to test, at the 5% significance level, whether Brand A tyres last longer on average than Brand B tyres. Assume the distances are normally distributed with the same population variance for both brands.
(a)
State the null and alternative hypotheses.
(2)
(b)
Calculate the pooled estimate of the common variance, giving your answer to 4 significant figures.
(3)
(c)
Carry out a pooled two-sample t-test, stating the p-value to 3 significant figures and your conclusion in context.
(4)
(d)
Find the smallest difference \(\bar x_A-\bar x_B\) in sample means, to 3 significant figures, that would lead to \(H_0\) being rejected at the 5% level with these sample sizes and the pooled variance from (b). The critical t-value for 13 degrees of freedom (one-tailed, 5%) is 1.771.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_A=\mu_B\)
✓ 1
A1 \(H_1:\mu_A>\mu_B\)
✓ 1
Part (b)
M1 sample variances \(s_A^2=379.24\) and \(s_B^2=221.27\)
✓ 1
M1 \(s_p^2=\frac{6s_A^2+7s_B^2}{13}\)
✓ 1
A1 294.2 km2
✓ 1
Part (c)
M1 pooled t-test with 13 degrees of freedom
✓ 1
A1 p-value = 0.00296
✓ 1
R1 p < 0.05
✓ 1
A1 reject \(H_0\): there is sufficient evidence that Brand A tyres last longer on average
✓ 1
Part (d)
M1 standard error of the difference \(\sqrt{s_p^2\left(\frac17+\frac18\right)}\) (ECF from (b))
✓ 1
M1 difference $=t_{crit}\times$ standard error with $t_{crit}=1.771$
✓ 1
A1 15.7 km
✓ 1
Common error: A pooled test requires the equal-variance assumption: state it, and use the weighted combination \(\frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}\), not a simple average of the two variances.
238IBMI3H-4-03r18-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard13 marks
A food laboratory counts mould spores on agar plates exposed to air in a bakery. Historically the number of spores on a plate follows a Poisson distribution with mean 2.5. After a new air-filtration system is installed, the manager wants to test at the 5% significance level whether the mean number of spores per plate has decreased. Four plates are exposed under identical conditions and the total number of spores counted is 4.
(a)
State suitable null and alternative hypotheses.
(2)
(b)
Write down the distribution of the total number of spores on four plates if \(H_0\) is true. Hence find the p-value for this test and state the conclusion in context.
(5)
(c)
Find the critical region for this test and state the probability of a Type I error.
(4)
(d)
The system in fact reduces the mean to 1.5 spores per plate. Find the probability of a Type II error.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\) mean number of spores per plate is 2.5
✓ 1
A1 \(H_1:\) mean number of spores per plate is less than 2.5
✓ 1
Part (b)
M1 sum of independent Poisson variables is Poisson
✓ 1
A1 \(S\sim\mathrm{Po}(10)\)
✓ 1
M1 \(p=P(S\le4)\)
✓ 1
A1 \(p=0.0293\) (3 s.f.)
✓ 1
R1 \(0.0293<0.05\), so reject \(H_0\): there is evidence at the 5% level that the mean has decreased
✓ 1
Part (c)
M1 compares cumulative probabilities with 0.05
✓ 1
A1 \(P(S\le4)=0.0293<0.05\) and \(P(S\le5)=0.0671>0.05\)
✓ 1
A1 critical region \(S\le4\)
✓ 1
A1 probability of Type I error \(=0.0293\) (ECF from critical region)
✓ 1
Part (d)
M1 under the true mean the total is \(\mathrm{Po}(6)\) and a Type II error means \(S\ge5\) (ECF from critical region), so \(1-P(S\le4)\)
✓ 1
A1 \(1-e^{-6}(1+6+18+36+54)=0.715\) (3 s.f.)
✓ 1
Exam tip: For a one-tailed Poisson test, define the distribution of the total count over all plates before looking up probabilities. A critical region is the set of outcomes whose cumulative probability stays below the significance level.
239IBMI3H-4-03r18-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
A telecoms regulator monitors the market share of three mobile networks X, Y and Z. Each month, customers may switch network. The proportions of customers on (X, Y, Z) change according to \(\mathbf{s}_{n+1}=T\mathbf{s}_n\), where \(T=\begin{pmatrix}0.8&0.1&0.2\\k&0.7&0.2\\0.2-k&0.2&0.6\end{pmatrix}\) and \(0Initially 50% of customers are on X, 30% on Y and 20% on Z. After one month, 31% of customers are on network Y.
(a)
Find the value of k.
(3)
(b)
Find the proportion of customers on each network after one month.
(2)
(c)
Find the long-term market share of each network.
(4)
(d)
The market has 24 000 customers in total. Find the long-term number of customers on network Z, and state the eigenvalue of T to which the steady state corresponds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(0.5k+0.7(0.3)+0.2(0.2)=0.31\) (Y row of \(T\mathbf{s}_0\))
✓ 1
M1 \(0.5k+0.25=0.31\)
✓ 1
A1 \(k=0.12\)
✓ 1
Part (b)
M1 evaluates \(T\mathbf{s}_0\) using \(k=0.12\)
✓ 1
A1 \((0.47,\,0.31,\,0.22)\)
✓ 1
Part (c)
M1 \(T\mathbf{s}=\mathbf{s}\) with \(x+y+z=1\)
✓ 1
M1 forms and solves a correct system, e.g. from \(0.2x=0.1y+0.2z\) and \(0.12x+0.7y+0.2z=y\)
✓ 1
A1 \(x=\frac{5}{12}\)
✓ 1
A1 \(y=\frac13\), \(z=\frac14\)
✓ 1
Part (d)
A1 \(24\,000\times\frac14=6000\) customers on Z
✓ 1
A1 the steady-state vector is an eigenvector for eigenvalue 1
✓ 1
Exam tip: Check your matrix first: every column must sum to 1. A quick column check also catches an incorrect value of k before you start the steady-state algebra.
240IBMI3H-4-03r18-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
On a mountaineering route, rockfalls occur independently at two sections. The number of rockfalls per hour on section A follows a Poisson distribution with mean 1.8, and on section B a Poisson distribution with mean 3.2, independently of A. Let \(T\) be the total number of rockfalls on the two sections in one hour.
(a)
Find the probability that exactly 6 rockfalls occur in total in a given hour.
(2)
(b)
Let \(A\) and \(B\) be the numbers of rockfalls on sections A and B. Show that \(P(A=2\mid T=6)=\binom{6}{2}(0.36)^2(0.64)^4\).
(4)
(c)
Evaluate the probability in part (b).
(1)
(d)
Given that exactly 6 rockfalls occurred in total, find the probability that at least 5 of them were on section B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(T\sim\mathrm{Po}(5)\)
✓ 1
A1 \(P(T=6)=0.146\) (3 s.f.)
✓ 1
Part (b)
M1 \(P(A=2\mid T=6)=\dfrac{P(A=2)P(B=4)}{P(T=6)}\) using independence
M1 uses the binomial result with \(n=6\), \(p=0.64\) for section B
✓ 1
A1 \(P(5)=6(0.64)^5(0.36)=0.2319\) and \(P(6)=0.64^6=0.0687\)
✓ 1
A1 \(0.301\) (3 s.f.)
✓ 1
Exam tip: When a "show that" result is given, write every step: the cancellation of the exponential factors is the key mark and must be visible.
241IBMI3H-4-05r18-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
The number of violin strings that break during a 4-hour orchestral rehearsal follows a Poisson distribution. The probability that no strings break during a 4-hour rehearsal is 0.30. Breaks occur independently and at a constant average rate.
(a)
Show that the mean number of breaks in a 4-hour rehearsal is 1.20, correct to 3 significant figures.
(2)
(b)
Using the exact value of \(\lambda\) from (a), find the probability that at least one string breaks during a 90-minute portion of the rehearsal.
(3)
(c)
In the same hall a cello section has its own strings, which break independently of the violins at a mean rate of 0.85 per 4-hour rehearsal. For the violins use a mean of 1.20 per 4-hour rehearsal. Find the probability that exactly 3 strings (violin and cello together) break in one 4-hour rehearsal.
(2)
(d)
The orchestra manager will bring spare strings if the probability of at least \(n\) breaks in a 4-hour rehearsal is less than 0.05. Using \(\mathrm{Po}(2.05)\), find the smallest value of \(n\) for which this is true.
(2)
(e)
Given that exactly 3 strings break in a 4-hour rehearsal, find the probability that exactly 1 of them is a violin string.
M1 mean for 1.5 hours \(= \frac{1.5}{4}\times 1.2039\ldots = 0.4515\ldots\)
✓ 1
M1 \(1 - e^{-0.4515\ldots}\)
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A1 \(0.363\)
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Part (c)
M1 total \(\sim \mathrm{Po}(1.20 + 0.85) = \mathrm{Po}(2.05)\)
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A1 \(0.185\)
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Part (d)
M1 evidence of comparing \(P(T \ge n)\) with 0.05, e.g. \(P(T \ge 5) = 0.0573\) and \(P(T \ge 6) = 0.0184\)
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A1 \(n = 6\)
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Part (e)
M1 \(\dfrac{P(V=1)\,P(C=2)}{P(T=3)}\) with \(V \sim \mathrm{Po}(1.20)\), \(C \sim \mathrm{Po}(0.85)\)
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M1 correct substitution, e.g. \(\dfrac{0.3614\ldots \times 0.1301\ldots}{0.1848\ldots}\) or \(3\times\frac{1.20}{2.05}\times\left(\frac{0.85}{2.05}\right)^2\)
✓ 1
A1 \(0.302\)
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Common error: The sum of independent Poisson variables is Poisson with the means added. For a conditional probability such as part (e), divide the probability of the specific combination by the probability of the total.
242IBMI3H-4-05r18-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard13 marks
A fruit grower packs apples into cartons. The mass of one apple is \(A \sim N(150,\ 12^2)\) grams and the mass of an empty carton is \(C \sim N(85,\ 4^2)\) grams. Each carton holds 6 apples. All masses are independent. Let \(W\) be the total mass of a full carton (carton and 6 apples).
(a)
Find \(\mathrm{E}(W)\) and the standard deviation of \(W\).
(3)
(b)
A carton labelled '1 kg' is rejected if \(W < 1000\). Find the probability that a full carton is rejected.
(2)
(c)
Two apples have masses \(A_1\) and \(A_2\). State, with a reason, whether \(\mathrm{Var}(2A) = \mathrm{Var}(A_1 + A_2)\), and give both values.
(2)
(d)
Find the probability that one randomly chosen apple is more than 20 g heavier than a second randomly chosen apple.
(3)
(e)
The grower changes variety so that the mean apple mass becomes \(\mu\) grams, while the standard deviation of one apple and the carton distribution stay the same. Find the value of \(\mu\), to the nearest 0.1 g, such that only 2% of full cartons are rejected.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\mathrm{E}(W) = 6\times150 + 85 = 985\) g
✓ 1
M1 \(\mathrm{Var}(W) = 6\times 12^2 + 4^2 = 880\)
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A1 sd \(= \sqrt{880} = 29.7\) g
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Part (b)
M1 \(P(W<1000)\) with \(W\sim N(985,\ 880)\)
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A1 \(0.693\) (ECF from (a))
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Part (c)
A1 \(\mathrm{Var}(2A) = 4\times144 = 576\) and \(\mathrm{Var}(A_1+A_2) = 288\)
✓ 1
R1 not equal: \(2A\) is one apple doubled (perfectly dependent), whereas \(A_1+A_2\) are two independent apples
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Part (d)
M1 \(D = A_1 - A_2\) with mean 0
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M1 \(\mathrm{Var}(D) = 288\) and attempt \(P(D>20)\)
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A1 \(0.119\)
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Part (e)
M1 \(z = 2.0537\ldots\) from \(P(W<1000) = 0.02\) (inverse normal)
Exam tip: A multiple of one random variable (\(2A\)) and a sum of independent copies (\(A_1+A_2\)) have the same mean but different variances. Always write down the mean and variance of the combination before using the normal distribution.
243IBMI3H-4-05r18-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard13 marks
Each week an aircraft in an airline's fleet is in one of three states: in service (S), in maintenance (M) or grounded awaiting parts (G). The transition matrix is \(\mathbf{T} = \begin{pmatrix} 0.90 & 0.70 & 0.30 \\ 0.08 & 0.20 & 0.30 \\ 0.02 & 0.10 & 0.40 \end{pmatrix}\), with the states in the order S, M, G. Each column gives the probabilities of moving from that state, so the state vector after one week is \(\mathbf{T}\) multiplied by the previous state vector.
(a)
An aircraft is in service this week. Find the probability that it is in maintenance in two weeks' time.
(2)
(b)
This week the fleet of 60 aircraft has 50 in service, 8 in maintenance and 2 grounded. Find the expected number in service next week.
(2)
(c)
Find the long-run proportion of the fleet in each state.
(3)
(d)
Each grounded aircraft costs the airline €1800 per week. For a fleet of 60 aircraft in the long run, find the expected weekly cost of grounded aircraft, to the nearest euro.
(2)
(e)
The airline changes its parts policy. From G the probability of moving to S becomes \(k\), the probability of moving to M stays 0.30, and the probability of staying in G becomes \(0.70 - k\). All other probabilities are unchanged. Find the value of \(k\) for which the long-run proportion of the fleet in service is exactly 0.86. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\mathbf{T}^2\) (or sum of the three two-step routes)
✓ 1
A1 \(0.094\)
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Part (b)
M1 \(0.90\times50 + 0.70\times8 + 0.30\times2\)
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A1 \(51.2\)
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Part (c)
M1 solve \(\mathbf{T}\mathbf{s} = \mathbf{s}\) for \(\mathbf{s}\)
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M1 use \(s_S + s_M + s_G = 1\)
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A1 \(S = 0.852,\ M = 0.102,\ G = 0.0455\) (exact: \(\frac{75}{88},\ \frac{9}{88},\ \frac{1}{22}\))
M1 matrix with third column \((k,\ 0.30,\ 0.70-k)\) and \(\mathbf{T}\mathbf{s}=\mathbf{s}\)
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M1 use \(\sum s = 1\) to express \(s_S\) in terms of \(k\), e.g. \(s_S = \dfrac{5(80k+21)}{44(10k+3)}\)
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M1 set \(s_S = 0.86\) and solve
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A1 \(k = \frac{71}{180} = 0.394\)
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Exam tip: For long-run proportions use the steady-state equation together with the condition that the proportions sum to 1. For an unknown matrix entry, keep it as a symbol, express the steady state in terms of it, then solve.
244IBMI3H-4-03r25-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
Two separate vents, A and B, on an island volcano emit ash plumes independently of each other. Vent A produces plumes at a mean rate of 1.4 per month and vent B at a mean rate of 2.2 per month. The numbers of plumes from each vent in any period are Poisson distributed.
(a)
Let T be the total number of plumes from the two vents in one month. Write down the distribution of T and find P(T = 5). Give your answer to 3 significant figures.
(3)
(b)
Find the probability that at least 12 plumes occur in total during a 3-month period. Give your answer to 3 significant figures.
(3)
(c)
Given that exactly 5 plumes occur in a particular month, find the probability that exactly 2 of them came from vent A. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 T ~ Po(3.6) (sum of independent Poisson variables)
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M1 P(T = 5) = e−3.6 × 3.65/5! (or GDC)
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A1 0.138
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Part (b)
A1 total over 3 months ~ Po(10.8)
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M1 1 − P(X ≤ 11)
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A1 0.397
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Part (c)
M1 conditional probability: P(A = 2 and B = 3) ÷ P(T = 5)
A1 denominator P(T = 5) from part (a) (ECF from (a))
✓ 1
A1 0.345
✓ 1
Exam tip: The sum of independent Poisson variables is Poisson with the means added; for 'given that' questions write the conditional probability as a fraction before using the GDC.
245IBMI3H-4-03r25-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A cargo ship carries two sizes of container. The mass of a small container is S tonnes, where S ~ N(14.2, 1.12). The mass of a large container is L tonnes, where L ~ N(27.5, 1.62). All container masses are independent of each other.
(a)
A large container and two different small containers are chosen at random. Find the probability that the large container is heavier than the two small containers combined. Give your answer to 3 significant figures.
(5)
(b)
A student models the combined mass of the two small containers as 2S instead of S1 + S2. Explain why this is not appropriate.
(1)
(c)
A crane lifts one large and three different small containers together. Find the load, in tonnes, that is exceeded with probability 0.01. Give your answer to 3 significant figures.
(6)
Show mark scheme
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Mark
Notes
Part (a)
M1 define D = L − S1 − S2 with S1, S2 independent copies of S
✓ 1
A1 E(D) = 27.5 − 28.4 = -0.9
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M1 Var(D) = 1.62 + 1.12 + 1.12 (variances add)
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A1 Var(D) = 4.98
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A1 P(D > 0) = 0.343
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Part (b)
R1 2S assumes both containers have exactly the same mass, whereas two different containers vary independently (the variance of 2S would be 4 × 1.21, which is too large)
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Part (c)
M1 load Q = L + S1 + S2 + S3, mean 70.1
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A1 E(Q) = 70.1
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M1 Var(Q) = 1.62 + 3 × 1.12
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A1 Var(Q) = 6.19, standard deviation 2.488
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M1 inverse normal with upper-tail probability 0.01, z = 2.3263
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A1 75.9 tonnes (accept 75.9)
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Common error: Writing 3S for the total of three different containers. Use S1 + S2 + S3, which has variance 3σ², not 9σ².
246IBMI3H-4-03r25-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
A container ship regularly calls at three ports, A, B and C. The port visited next depends only on the port currently visited. The transition matrix, where columns give the current port (A, B, C in order) and rows give the next port, is
State the range of values of q for which T is a valid transition matrix.
(2)
(b)
In the long run, the ship calls at port C for exactly one quarter of its stops. Find the value of q.
(5)
(c)
Using q = 0.2, the ship starts at port A. Find the probability that its third subsequent stop is its first call at port C (that is, it does not visit C at the first or second subsequent stop).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 q ≥ 0
✓ 1
A1 0.6 − q ≥ 0, so 0 ≤ q ≤ 0.6
✓ 1
Part (b)
M1 steady-state vector (a, b, c) with Ts = s and a + b + c = 1
✓ 1
A1 c = 0.25, so a + b = 0.75
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M1 use the second equation b = 0.3a + 0.6b + 0.4c (independent of q) to solve for a and b
✓ 1
A1 a = 2/7, b = 13/28
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A1 from the first equation a = 0.5a + 0.2b + 0.25q, giving q = 0.2
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Part (c)
M1 identify routes avoiding C on stops 1 and 2: A→A→A→C, A→A→B→C, A→B→A→C, A→B→B→C
Exam tip: In steady-state problems, use equations that do not contain the unknown parameter first; then substitute back into the remaining equation to find it.
247IBMI3H-4-07r27-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
During a performance of a short piece, the number of wrong notes played by the violin section, V, has a Poisson distribution with mean 1.4. The number of wrong notes played by the brass section, B, has a Poisson distribution with mean 0.8. The two sections play independently. Let T = V + B.
(a)
State the distribution of T and find P(T ≥ 4).
(3)
(b)
Given that exactly 3 wrong notes are played in total, find the probability that exactly 2 of them were played by the violin section.
(4)
(c)
The piece is performed 4 times in a concert, independently. Find the probability that the total number of wrong notes over the 4 performances is at least 6 and at most 10.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 T ~ Po(2.2)
✓ 1
M1 1 − P(T ≤ 3)
✓ 1
A1 0.181
✓ 1
Part (b)
M1 conditional probability: P(V = 2 and B = 1) ÷ P(T = 3)
✓ 1
M1 P(V = 2) × P(B = 1) = 0.2417 × 0.3595
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A1 P(T = 3) = 0.1966
✓ 1
A1 0.442
✓ 1
Part (c)
A1 total ~ Po(8.8)
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M1 P(total ≤ 10) − P(total ≤ 5)
✓ 1
A1 0.601
✓ 1
Exam tip: For a range "at least a and at most b", use P(X ≤ b) − P(X ≤ a − 1) so that both end values are included.
248IBMI3H-4-07r27-09Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
A streaming service plays songs from two playlists, Classical (C) and Jazz (J). After a Classical song, the next song is Classical with probability 0.7. After a Jazz song, the next song is Classical with probability p. The transition matrix, whose columns give the probabilities of moving from C and J (in that order), is T = [[0.7, p], [0.3, 1 − p]] (rows listed in turn).
(a)
In the long run, 60% of songs played are Classical. Show that p = 0.45.
(3)
(b)
The first song played is Classical. Using p = 0.45, find the probability that the fourth song is Classical.
(3)
(c)
The first song played is Jazz. Find the probability that both the third song and the fourth song are Classical.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 steady state: Ts = s with s = (0.6, 0.4)
✓ 1
A1 0.7 × 0.6 + 0.4p = 0.6 (or 0.42 + 0.4p = 0.6)
✓ 1
A1 p = 0.18 ÷ 0.4 = 0.45 (AG)
✓ 1
Part (b)
M1 initial state (1, 0) and T3 (three transitions)
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A1 correct matrix power or successive state vectors (0.7, 0.3), (0.625, 0.375), …
✓ 1
A1 0.606
✓ 1
Part (c)
M1 state after two transitions from (0, 1)
✓ 1
A1 P(third song Classical) = 0.5625
✓ 1
A1 0.394 (multiplying by 0.7 for Classical followed by Classical)
✓ 1
Exam tip: "Both the third and fourth" needs the probability of the earlier event multiplied by the next transition probability; do not use the matrix power for the fourth song alone.
249IBMI3H-4-07r27-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
An agronomist records the yield of a wheat crop, y tonnes per hectare, for different amounts of nitrogen fertiliser, x kg per hectare: x: 0, 30, 60, 90, 120, 150, 180 y: 3.1, 4.9, 6.2, 6.9, 7.0, 6.4, 5.6 She suggests a model of the form y = ax2 + bx + c.
(a)
Use quadratic regression to find the values of a, b and c, each correct to 3 significant figures.
(3)
(b)
The coefficient of determination for this model is R2 = 0.998. A linear model fitted to the same data has R2 = 0.404. State, with a reason, which model is more suitable.
(2)
(c)
Use the 3 significant figure values from part (a) to find the amount of fertiliser that maximises the yield according to the model, and the maximum yield predicted.
(4)
(d)
Use the model to find the predicted yield when x = 400, and comment on the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 quadratic regression on the data
✓ 1
A1 a = −0.000313
✓ 1
A1 b = 0.0699 and c = 3.11
✓ 1
Part (b)
A1 The quadratic model
✓ 1
R1 Its R2 is much closer to 1, so it explains far more of the variation in yield (the data are curved, rising then falling)
✓ 1
Part (c)
M1 x = −b ÷ (2a) (or use of the GDC maximum)
✓ 1
A1 x ≈ 112 kg per hectare (111.7)
✓ 1
M1 substitute this value into the model (or use c − b2 ÷ 4a)
✓ 1
A1 maximum yield ≈ 7.01 tonnes per hectare
✓ 1
Part (d)
M1 substitute x = 400 into the 3 s.f. model
✓ 1
A1 y ≈ −19.0, which is impossible (negative yield), so the model should not be extrapolated this far
✓ 1
Exam tip: A high coefficient of determination only measures fit within the data range; always comment on whether predictions outside that range are sensible.
250IBMI3H-4-04r28-06Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A fish processor packs eight fish into each crate. The mass of one fish is normally distributed with mean 1.8 kg and standard deviation 0.3 kg. The mass of an empty crate is normally distributed with mean 1.2 kg and standard deviation 0.1 kg. Let T be the total mass of a packed crate (crate plus eight fish).
(a)
Find E(T) and Var(T).
(3)
(b)
State one assumption needed for your calculation of Var(T).
(1)
(c)
Find the probability that a packed crate has a mass greater than 16.5 kg.
(2)
(d)
A lorry carries five packed crates, whose masses are independent. Find the probability that their total mass exceeds 80 kg.
(3)
(e)
The processor wants a mass limit k kg such that 95% of packed crates have mass below k. Find k to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(T) = 1.2 + 8(1.8) = 15.6 kg
✓ 1
M1 Var(T) = 0.12 + 8(0.32)
✓ 1
A1 Var(T) = 0.73
✓ 1
Part (b)
R1 The masses of the eight fish and of the crate are independent
✓ 1
Part (c)
M1 use of T ~ N(15.6, 0.73)
✓ 1
A1 0.146
✓ 1
Part (d)
M1 mean 5(15.6) = 78 and variance 5(0.73) = 3.65
✓ 1
A1 normal model with these parameters used
✓ 1
A1 0.148
✓ 1
Part (e)
M1 P(T < k) = 0.95
✓ 1
A1 z = 1.645 (1.6449)
✓ 1
A1 k = 17.0 kg
✓ 1
Common error: Eight fish form a sum of eight independent masses (variance 8σ²), which is different from one fish multiplied by 8 (variance 64σ²).
251IBMI3H-4-04r28-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Customers in a town buy their clothes from one of three brands, A, B or C, each month. The monthly switching between brands is modelled by a Markov chain with transition matrix
from A
from B
from C
to A
0.8
0.2
p
to B
0.1
0.7
0.2
to C
0.1
0.1
0.8 − p
where 0 ≤ p ≤ 0.8 and each column sums to 1. In parts (a) to (c), p = 0.1.
(a)
Find the probability that a customer who buys from brand C this month buys from brand A two months from now.
(2)
(b)
Initially 50% of customers use brand A, 30% use B and 20% use C. Find the proportion of customers using each brand after three months. Give your answers to 3 significant figures.
(3)
(c)
Find the long-run proportion of customers using each brand when p = 0.1.
(3)
(d)
A marketing campaign changes the value of p. Find the value of p for which the long-run proportion of customers using brand A is exactly 0.5.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the (1,3) entry of T2, or sums the routes via A, B and C
✓ 1
A1 0.8 × 0.1 + 0.2 × 0.2 + 0.1 × 0.7 = 0.19
✓ 1
Part (b)
M1 uses T3 applied to (0.5, 0.3, 0.2), or applies T three times
✓ 1
A1 A: 0.456
✓ 1
A1 B: 0.304 and C: 0.239
✓ 1
Part (c)
M1 solves Ts = s with the entries summing to 1
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A1 A: 7/16 = 0.4375
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A1 B: 5/16 = 0.3125 and C: 1/4 = 0.25
✓ 1
Part (d)
M1 forms equations from Ts = s with (a, b, c) and a + b + c = 1
✓ 1
M1 eliminates b and c to express a in terms of p
✓ 1
A1 a = (15p + 2)/(20p + 6)
✓ 1
A1 15p + 2 = 0.5(20p + 6), so p = 0.2
✓ 1
Exam tip: For a steady-state calculation with a parameter, use any two rows of Ts = s together with the normalising equation; the third row is redundant.
252IBMI3H-4-05r29-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
In practice sessions, the lap time of driver A, in seconds, is \(A\sim N(88.4,\,0.6^2)\) and the lap time of driver B is \(B\sim N(88.9,\,0.5^2)\). All lap times are independent.
(a)
Find the probability that, on a randomly chosen lap each, driver A is quicker than driver B.
(4)
(b)
Each driver completes ten laps. Find the probability that driver A's total time for the ten laps is less than driver B's total time.
(5)
(c)
A student models driver A's ten-lap total as \(10A\). State why this is not appropriate, and calculate \(\mathrm{Var}(10A)\).
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 consider \(D=B-A\) and require \(D>0\)
✓ 1
A1 \(E(D)=0.5\)
✓ 1
A1 \(\mathrm{Var}(D)=0.61\) (variances add)
✓ 1
A1 0.739
✓ 1
Part (b)
M1 totals \(S_A\sim N(884,\,10\times0.36)\) and \(S_B\sim N(889,\,10\times0.25)\), i.e. variances 3.6 and 2.5
✓ 1
A1 both variances correct
✓ 1
M1 consider \(S_B-S_A\)
✓ 1
A1 mean 5 and variance 6.1
✓ 1
A1 0.979
✓ 1
Part (c)
R1 the ten laps are independent, so the total is a sum of ten independent variables, not ten copies of the same lap time
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A1 \(\mathrm{Var}(10A)=100\times0.36=36\)
✓ 1
A1 the correct variance is \(10\times0.36=3.6\), so \(10A\) greatly overestimates the spread
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Exam tip: For a total of \(n\) independent observations use \(\mathrm{Var}=n\sigma^2\); for \(nX\) use \(n^2\sigma^2\).
253IBMI3H-4-05r29-11Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Patients on a hospital ward are in one of three states each day: low dependency (L), medium dependency (M) or high dependency (H). The daily transition matrix is \(T=\begin{pmatrix}0.8&0.1&0\\0.2&0.7&p\\0&0.2&1-p\end{pmatrix}\), with states in the order L, M, H and each column giving the probabilities from the current state.
(a)
Find the probability that a patient in M on one day is in L on the next day, and write down the probability that a patient in H on one day is in M on the next day.
(2)
(b)
In the long run, the proportion of patients in L is 0.2. Find the value of \(p\).
(5)
(c)
Using \(p=0.2\), on day 0 the proportions of patients in L, M, H are 0.6, 0.3, 0.1. Find the proportion in H on day 2.
(3)
(d)
A ward has 90 patients. Using \(p=0.2\), find the expected number in H in the long run.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 0.1
✓ 1
A1 \(p\)
✓ 1
Part (b)
M1 \(T\pi=\pi\) with \(\pi_L+\pi_M+\pi_H=1\)
✓ 1
A1 from row 1, \(\pi_M=2\pi_L=0.4\)
✓ 1
A1 from row 3, \(0.2\pi_M=p\,\pi_H\)
✓ 1
M1 \(\pi_H=1-0.2-0.4=0.4\)
✓ 1
A1 \(p=0.2\)
✓ 1
Part (c)
M1 \(T^2\begin{pmatrix}0.6\\0.3\\0.1\end{pmatrix}\) or two steps
✓ 1
A1 day 1 vector \((0.51,\,0.35,\,0.14)\)
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A1 0.182
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Part (d)
M1 \(90\times0.4\)
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A1 36
✓ 1
Exam tip: When a matrix has an unknown, use the steady-state equations from the rows that contain it, then use the sum of the probabilities to finish.
254IBMI3H-4-05r29-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
At a hospital, the number of ambulance arrivals between 02:00 and 03:00 has historically followed a Poisson distribution with mean 6. After a public-health campaign, the hospital believes the mean has decreased. It will count arrivals in one randomly chosen night hour and test at the 5% significance level.
(a)
State suitable hypotheses.
(2)
(b)
Find the critical region for the test.
(3)
(c)
Write down the actual significance level of the test.
(1)
(d)
Four arrivals are observed. State the conclusion.
(2)
(e)
Suppose the mean has in fact fallen to 3. Find the probability of a Type II error.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\lambda=6\)
✓ 1
A1 \(H_1:\lambda<6\)
✓ 1
Part (b)
M1 evaluate \(P(X\le c)\) for \(\lambda=6\)
✓ 1
A1 \(P(X\le1)=0.0174\) and \(P(X\le2)=0.0620\)
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A1 critical region \(X\le1\)
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Part (c)
A1 0.0174
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Part (d)
R1 4 is not in the critical region
✓ 1
A1 insufficient evidence at the 5% level that the mean has decreased
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Part (e)
M1 \(P(X\ge2)\) with \(\lambda=3\)
✓ 1
M1 \(1-P(X\le1)\)
✓ 1
A1 0.801
✓ 1
Common error: A Type II error occurs when \(H_0\) is not rejected although it is false, so use the true mean and the complement of the critical region.
255IBMI3H-4-07r33-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
Waiting times, in minutes, at a vaccination centre are normally distributed with a known standard deviation of 12 minutes. A random sample of 64 patients has a mean waiting time of 38.4 minutes.
(a)
Find a 95% confidence interval for the population mean waiting time.
(3)
(b)
The centre manager says, "95% of patients wait between these two times." Explain why this interpretation is incorrect.
(2)
(c)
Find the smallest sample size for which the width of a 95% confidence interval for the mean is at most 4.0 minutes.
(3)
(d)
Find the width of the 95% confidence interval when the sample size is 139.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of the normal distribution with critical value 1.960 (σ known)
✓ 1
M1 38.4 ± 1.96 × 12 ÷ √64
✓ 1
A1 (35.5, 41.3)
✓ 1
Part (b)
R1 the interval is for the population mean, not for individual waiting times
✓ 1
R1 individual waits vary much more (standard deviation 12) than the sample mean does (12 ÷ 8 = 1.5), so most individual waits fall well outside this narrow interval; the 95% refers to the long-run proportion of such intervals that contain the true mean
✓ 1
Part (c)
M1 width = 2 × 1.96 × 12 ÷ √n ≤ 4
✓ 1
M1 n ≥ 138.3... (√n ≥ 11.76...)
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A1 n = 139
✓ 1
Part (d)
M1 2 × 1.96 × 12 ÷ √139
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A1 3.99 minutes (accept 3.98 to 3.99)
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Common error: Round a minimum sample size up, never down, because rounding down would leave the interval wider than the limit.
256IBMI3H-4-07r33-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
The wind at a site is classified each day as high (H) or low (L). Transitions are modelled by T with columns giving the next-day probabilities: from H the next day is H with probability 0.9 and L with probability 0.1; from L the next day is H with probability q and L with probability 1 − q. In the long term, the proportion of high-wind days is 0.75.
(a)
Find the value of q.
(3)
(b)
Show that the eigenvalues of T are 1 and 0.6.
(3)
(c)
Today the wind is low. Find the probability that it is high in 3 days' time.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 steady state: 0.1sH = qsL, with sH + sL = 1
✓ 1
M1 sH = q/(0.1 + q) = 0.75 (or 0.075 = 0.25q)
✓ 1
A1 q = 0.3
✓ 1
Part (b)
M1 det(T − λI) = 0 with T = (0.9, 0.3; 0.1, 0.7) in rows
M1 use of (0, 1)T with T3, or successive days: day 1 (0.3, 0.7)
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M1 day 2 (0.48, 0.52), then day 3 high = 0.9 × 0.48 + 0.3 × 0.52
✓ 1
A1 0.588
✓ 1
Exam tip: In a two-state chain the second eigenvalue is 1 minus the sum of the two switching probabilities (trace minus 1); it controls how fast the chain settles.
257IBMI3H-4-05r34-02Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
A telescope has two independent cameras that count meteor streaks. Camera A records streaks at a rate of 3.5 per hour and camera B at a rate of 2.1 per hour. Let X and Y be the numbers of streaks recorded by A and B in one hour; assume X ~ Po(3.5) and Y ~ Po(2.1).
(a)
Write down the distribution of the total number of streaks, T = X + Y, and find P(T ≥ 8).
(3)
(b)
Given that exactly 6 streaks are recorded in an hour, find the probability that exactly 2 of them were recorded by camera A.
(4)
(c)
Hence, or otherwise, find the expected number of the 6 streaks that were recorded by camera A, given that the total is 6.
Given T = 6, the number from A is binomial with n = 6 and p = 3.5 ÷ 5.6 = 0.625 (or sum xP(X = x | T = 6) over x = 0 to 6)
✓ 1
Expected number = 6 × 0.625 = 3.75
✓ 1
Exam tip: For a conditional probability involving a sum of independent variables, write the joint event as a product before dividing by the probability of the condition.
258IBMI3H-4-05r34-04Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
A rowing crew consists of eight rowers and one cox. The mass of each rower, in kg, is independently normally distributed with mean 82 and standard deviation 6. The mass of the cox is independently normally distributed with mean 52 and standard deviation 3. Let W be the total mass of the nine people.
(a)
Find E(W) and Var(W).
(2)
(b)
Find the probability that the total mass of the crew exceeds 700 kg.
(3)
(c)
A coach suggests modelling the total mass of the eight rowers as 8X, where X is the mass of one rower. Explain why this gives a different variance from the sum of eight independent rower masses, and state both variances.
(2)
(d)
The boat is safe for a total mass of up to L kg, where L is chosen so that P(W > L) = 0.01. Find L to the nearest kilogram.
(3)
Show mark scheme
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Mark
Notes
Part (a)
E(W) = 8 × 82 + 52 = 708 kg
✓ 1
Var(W) = 8 × 36 + 9 = 297
✓ 1
Part (b)
W ~ N(708, 297); standard deviation = 17.2
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P(W > 700) = P(Z > −0.464)
✓ 1
= 0.679 (3 s.f.)
✓ 1
Part (c)
Var(8X) = 64 × 36 = 2304, whereas the sum of eight independent masses has variance 8 × 36 = 288
✓ 1
8X assumes all rowers have identical mass (perfect correlation); independent masses partly cancel each other's variation
✓ 1
Part (d)
z = 2.326 (inverse normal for 0.99)
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L = 708 + 2.326 × 17.2…
✓ 1
L = 748 kg (748.09 before rounding)
✓ 1
Exam tip: Use the unrounded standard deviation from your calculator in later parts so that final answers are not affected by intermediate rounding.
259IBMI3H-4-05r34-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
On any night, the sky above an observatory is either clear (C) or cloudy (Y). The transition matrix is \(\mathbf{T}=\begin{pmatrix}0.8 & p\\ 0.2 & 1-p\end{pmatrix}\), with columns giving tonight's sky (C, Y) and rows giving tomorrow's sky (C, Y). In the long run, 75% of nights are clear.
(a)
Find the value of p.
(3)
(b)
Tonight is cloudy. Find the smallest number of nights from now, n, for which the probability that the night is clear exceeds 0.74.
(4)
(c)
In the long run, find the expected number of clear nights in a 28-night observing period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Steady state s = (0.75, 0.25)T satisfies Ts = s
✓ 1
0.8 × 0.75 + p × 0.25 = 0.75
✓ 1
p = 0.6
✓ 1
Part (b)
s0 = (0, 1)T and use of Tns0
✓ 1
n = 2 gives P(clear) = 0.72, which is not enough
✓ 1
n = 3 gives P(clear) = 0.744 > 0.74
✓ 1
Smallest n = 3
✓ 1
Part (c)
28 × 0.75
✓ 1
21
✓ 1
Exam tip: When asked for the smallest n satisfying a condition, show the values for n and for n − 1 so that the boundary is clear.
260IBMI3H-4-05r37-03Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A university library has found that late returns at its desk occur at random at a mean rate of 3.2 per day. After introducing text-message reminders, the library counts the late returns over 5 consecutive days and records a total of 8. The librarian tests, at the 5% significance level, whether the mean daily rate has decreased. Assume late returns still occur independently and at random.
(a)
State suitable null and alternative hypotheses, and the distribution of the total number of late returns in 5 days under the null hypothesis.
(3)
(b)
Find the p-value for the observed total.
(2)
(c)
State the conclusion of the test, in context.
(2)
(d)
Find the critical region for the total number of late returns in 5 days.
(3)
(e)
Write down the actual probability of a Type I error for this test, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: λ = 3.2 per day (16 in 5 days)
✓ 1
A1 H1: λ < 3.2 per day (λ < 16 in 5 days)
✓ 1
A1 under H0 the total ~ Po(16)
✓ 1
Part (b)
M1 P(X ≤ 8) with X ~ Po(16)
✓ 1
A1 0.0220
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Part (c)
M1 compares 0.0220 with 0.05 (ECF from (b))
✓ 1
A1 reject H0; there is evidence that the mean daily rate of late returns has decreased
Common error: Comparing a single-day count with the daily mean is misleading when the data are totals; scale λ to the same period as the count.
261IBMI3H-4-05r37-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard11 marks
A swimming club compares two training programmes. The 100 m times (seconds) after eight weeks of programme A are: 58.4, 59.1, 57.8, 58.9, 57.5, 58.2 (6 swimmers). Times after programme B are: 59.6, 60.2, 58.8, 60.5, 59.3, 61.0, 59.9 (7 different swimmers). The coach wants to know whether programme A gives a lower mean time than programme B, at the 5% significance level.
(a)
Find the mean and sample standard deviation (sn−1) for each programme, to 3 significant figures.
(2)
(b)
State two assumptions needed for a pooled two-sample t-test.
(2)
(c)
State the hypotheses for the test.
(2)
(d)
Calculate the pooled t statistic, to 3 significant figures.
(3)
(e)
The p-value (11 degrees of freedom) is 0.000839. State your conclusion in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 A: mean 58.3 s, standard deviation 0.618 s
✓ 1
A1 B: mean 59.9 s, standard deviation 0.744 s
✓ 1
Part (b)
A1 both populations are normally distributed
✓ 1
A1 the populations have equal variances (and the samples are independent random samples)
R1 reject H0: there is strong evidence that programme A gives a lower mean 100 m time
✓ 1
Common error: In a one-tailed test the p-value is not doubled; check the direction in H₁ matches the sign of t.
262IBMI3H-4-04r38-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Calculator
A depot packs boxed equipment for delivery. The mass of the equipment placed in a box, \(A\) kg, is normally distributed with mean 18.4 and standard deviation 1.2. The mass of its packaging, \(P\) kg, is normally distributed with mean 2.1 and standard deviation 0.3. \(A\) and \(P\) are independent. The mass of a complete package is \(A+P\) kg.
(a)
Find the mean and standard deviation of the mass of a complete package. Give the standard deviation to three significant figures.
(2)
(b)
Find the probability that a randomly chosen package has a mass greater than 22 kg.
(2)
(c)
A pallet carries 12 independently chosen packages with total mass \(T\) kg. Find the probability that \(T\) exceeds 250 kg.
(3)
(d)
Two pallets, each of 12 packages, have total masses \(T_1\) and \(T_2\). Find the probability that \(T_1\) exceeds \(T_2\) by more than 8 kg.
(3)
(e)
Show that the standard deviation of the mean mass of the packages on one pallet, \(\frac{T}{12}\), is 0.357 kg, correct to three significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean \(=20.5\) kg
✓ 1
A1 variance \(=1.2^2+0.3^2=1.53\), so standard deviation \(=1.24\) kg
✓ 1
Part (b)
M1 \(A+P\sim N(20.5,\,1.53)\) and \(P(A+P>22)\)
✓ 1
A1 \(0.113\)
✓ 1
Part (c)
M1 \(\mathrm{E}(T)=246\) and \(\mathrm{Var}(T)=12\times1.53=18.36\)
✓ 1
M1 \(P(T>250)\) using \(N(246,\,18.36)\)
✓ 1
A1 \(0.175\)
✓ 1
Part (d)
M1 \(T_1-T_2\sim N(0,\,2\times18.36)\), variances are added (variance \(36.72\))
✓ 1
M1 \(P(T_1-T_2>8)\)
✓ 1
A1 \(0.0934\)
✓ 1
Part (e)
M1 \(\mathrm{Var}\left(\frac{T}{12}\right)=\frac{18.36}{144}\) \((=0.1275)\) or \(\frac{1.53}{12}\)
✓ 1
A1 \(\sqrt{0.1275}=0.357\)
✓ 1
Common error: When subtracting independent variables the variances are added, not subtracted; \(\mathrm{Var}(T_1-T_2)=\mathrm{Var}(T_1)+\mathrm{Var}(T_2)\).
263IBMI3H-4-04r38-08Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Calculator
A species of fish moves between three zones X, Y and Z of a reef once each day. The transition matrix is \(\mathbf{T}=\begin{pmatrix}0.6&0.1&p\\0.3&0.7&r\\0.1&0.2&q\end{pmatrix}\), where each column gives the probabilities for a fish at the zone it heads. The long-term distribution of the fish across X, Y and Z is \(\begin{pmatrix}0.25\\0.5\\0.25\end{pmatrix}\).
(a)
Write down an equation connecting \(p\), \(r\) and \(q\).
(1)
(b)
Find the values of \(p\), \(r\) and \(q\).
(4)
(c)
A fish starts in zone X. Find the probability that it is in zone Y after 3 days.
(3)
(d)
A fish starts in zone X. Find the probability that it is in zone Y after 2 days, by considering the possible zones on day 1.
(2)
(e)
Explain why the long-term distribution does not depend on the zone in which a fish starts.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(p+r+q=1\) (columns of a transition matrix sum to 1)
✓ 1
Part (b)
M1 use of \(\mathbf{T}\mathbf{s}=\mathbf{s}\) with \(\mathbf{s}=(0.25,0.5,0.25)^T\)
✓ 1
M1 forming equations from two rows, e.g. \(0.15+0.05+0.25p=0.25\) and \(0.075+0.35+0.25r=0.5\)
✓ 1
A1 \(p=0.2\) and \(r=0.3\)
✓ 1
A1 \(q=0.5\)
✓ 1
Part (c)
M1 use of \(\mathbf{T}^3\begin{pmatrix}1\\0\\0\end{pmatrix}\) (ECF from (b))
✓ 1
M1 selecting the Y entry (second row, first column of \(\mathbf{T}^3\))
✓ 1
A1 \(0.468\)
✓ 1
Part (d)
M1 \(0.3\times0.6+0.7\times0.3+0.3\times0.1\) (via X, Y, Z respectively)
✓ 1
A1 \(0.42\)
✓ 1
Part (e)
R1 every entry of \(\mathbf{T}\) is positive, so any zone can be reached from any other zone
✓ 1
R1 for such a chain the distribution converges to the same steady state whatever the initial state
✓ 1
Exam tip: Check each column of a transition matrix sums to 1 before using it: this single condition often provides the missing probability immediately.
264IBMI3H-4-04r45-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard14 marks
Daily activity at a volcano is classified as Quiet (Q), Minor (M) or Eruptive (E). The transition matrix is T, where the columns give today's state (Q, M, E from left to right) and the rows give tomorrow's state (Q, M, E from top to bottom): T = [0.7, 0.3, 0.1; 0.25, 0.5, 0.9 − p; 0.05, 0.2, p] (rows separated by semicolons), where 0 < p < 0.9. In the long run the volcano is Eruptive on 25 % of days.
(a)
Find the value of p.
(4)
(b)
Use p = 0.65 for the rest of the question. The volcano is Quiet on Monday. Find the probability that it is Eruptive on Wednesday.
(3)
(c)
The volcano is Quiet on day 0. Find the smallest number of days n after which the probability of an Eruptive day is within 0.01 of the long-run value 0.25.
(3)
(d)
Explain why the chain has a long-run distribution that does not depend on the starting state.
(2)
(e)
Estimate the number of Eruptive days in a 30-day period in the long run.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 steady state π = (q, m, 0.25) with q + m = 0.75
✓ 1
M1 uses Tπ = π, e.g. 0.7q + 0.3m + 0.025 = q
✓ 1
A1 q = 5/12 and m = 1/3
✓ 1
A1 from the third row 0.05q + 0.2m + 0.25p = 0.25, p = 0.65
✓ 1
Part (b)
M1 identifies the three routes Q → Q → E, Q → M → E, Q → E → E, or calculates T2
✓ 1
M1 0.7 × 0.05 + 0.25 × 0.2 + 0.05 × 0.65
✓ 1
A1 0.1175
✓ 1
Part (c)
M1 calculates Tn multiplied by (1, 0, 0) for successive n
✓ 1
A1 n = 7 gives 0.239 (difference 0.011, too large) and n = 8 gives 0.243
✓ 1
A1 n = 8
✓ 1
Part (d)
A1 every entry of T is positive, so the chain is regular
✓ 1
R1 a regular chain has a unique steady-state vector, which any starting state approaches
✓ 1
Part (e)
M1 30 × 0.25
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A1 7.5 days
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Exam tip: If a steady-state proportion is given, put the known value into the vector and use as few of the three equations as needed to solve for the unknowns.
265IBMI3H-4-04r45-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard15 marks
The number of tickets, y, sold in week x after a show opens is recorded for seven weeks: x: 1, 2, 3, 4, 5, 6, 7 y: 420, 515, 540, 505, 430, 300, 140
(a)
Find the quadratic regression model y = ax2 + bx + c, giving a, b and c to 3 significant figures.
(3)
(b)
Use the unrounded model to find the predicted value and residual (y minus predicted value) for each week, to 1 decimal place.
(3)
(c)
Calculate the sum of squares of the residuals, SSres, to the nearest whole number.
(2)
(d)
Calculate SStot = Σ(y − ȳ)2, to the nearest whole number.
M1 sums the squares of the residuals (using unrounded values)
✓ 1
A1 368
✓ 1
Part (d)
M1 mean ȳ = 407.1 and sums the squared differences
✓ 1
A1 122 393
✓ 1
Part (e)
M1 1 − 367.86/122 392.86
✓ 1
A1 0.997
✓ 1
Part (f)
R1 about 99.7 % of the variation in weekly ticket sales is explained by the quadratic model in the week number
✓ 1
Part (g)
A1 −94.3
✓ 1
R1 negative ticket sales are impossible, so the model is not valid beyond week 7
✓ 1
Common error: Using rounded residuals or rounded regression coefficients in sums of squares: keep unrounded values until the final answer.
266IBMI3H-4-06r45-07Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
The number of small earthquakes recorded per day beneath a volcano is modelled by a Poisson distribution with mean 3.2. Counts on different days are independent.
(a)
Find the probability that at least 6 earthquakes are recorded on a given day. Give your answer to 4 decimal places.
(2)
(b)
Find the probability that at most 6 earthquakes are recorded over a period of 3 days. Give your answer to 3 significant figures.
(2)
(c)
A day is called ‘unusually active’ if the number of earthquakes is so high that the chance of getting at least that many is below 1%. Find the least number of earthquakes that makes a day unusually active.
(3)
(d)
Over a period of 10 days, find the probability that exactly 2 days have at least 6 earthquakes. Use the unrounded probability from part (a) and give your answer to 3 significant figures.
M1 defines Y as the number of such days, Y ~ B(10, 0.10540…)
✓ 1
M1 attempts P(Y = 2)
✓ 1
A1 0.205
✓ 1
Common error: A different time period needs a different Poisson mean: the mean for 3 days is 3 × 3.2 = 9.6, not 3.2.
267IBMI3H-4-06r45-13Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
A volcano has three daily alert levels: Green (G), Amber (A) and Red (R). The probabilities of moving from today’s level (columns) to tomorrow’s level (rows), in the order G, A, R, are given by T = [[0.9, 0.2, 0], [0.1, 0.7, k], [0, 0.1, 1 − k]] (written row by row), where k is a constant. In parts (a) to (c), k = 0.8.
(a)
Given that today is Amber, find the probability that the day after tomorrow is Red.
(3)
(b)
Given that today is Green, find the probability that the level is Amber in 3 days’ time.
(3)
(c)
Show that (0.64, 0.32, 0.04) is the long-term (steady-state) distribution.
(3)
(d)
Now k is unknown. Find the value of k for which, in the long term, the proportion of Red days is 0.0625.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 considers routes A→A→R, A→R→R and A→G→R (or T2)
✓ 1
A1 0.7 × 0.1 + 0.1 × 0.2 + 0.2 × 0
✓ 1
A1 0.09
✓ 1
Part (b)
M1 uses T3 applied to (1, 0, 0) (or equivalent step-by-step method)
✓ 1
A1 distribution after 3 days (0.779, 0.203, 0.018)
A1 0.1 × 0.32 + 0.2 × 0.04 = 0.04 and 0.64 + 0.32 + 0.04 = 1, so the vector is unchanged and valid
✓ 1
Part (d)
M1 steady-state equations give g = 2a (from the G row) and 0.1a = kr (from the R row), where g, a, r are the long-term proportions
✓ 1
M1 uses g + a + r = 1 with r = 0.0625 to get a = 0.3125
✓ 1
A1 k = 0.1 × 0.3125/0.0625 = 0.5
✓ 1
Exam tip: For a steady state, any two of the three row equations together with ‘proportions add to 1’ are enough; the third row equation is then automatically satisfied.
268IBMI3H-4-07r47-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Calculator
On a freight line the mass of a loaded container, X tonnes, is normally distributed with mean 24 and standard deviation 1.8. A wagon carries two containers chosen at random, with masses X1 and X2, which are independent. Let Y = X1 + X2.
(a)
Find the probability that the load on the wagon exceeds 52 tonnes.
(4)
(b)
A planner wrongly models the wagon load as W = 2X. Find P(W > 52).
(3)
(c)
Explain why the two probabilities differ.
(2)
(d)
The railway wants the load limit k such that P(Y > k) = 0.01. Find k to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(Y) = 48
✓ 1
M1 Var(Y) = 2 × 1.82 = 6.48
✓ 1
A1 Y ~ N(48, 6.48)
✓ 1
A1 0.0581
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Part (b)
M1 E(W) = 48 and Var(W) = 22 × 1.82 = 12.96
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A1 standard deviation 3.6
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A1 0.133
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Part (c)
R1 Var(2X) = 4Var(X) but Var(X1 + X2) = 2Var(X);
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R1 two independent containers are unlikely to both be heavy, so their variations partly cancel, but 2X assumes the same extreme value twice
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Part (d)
M1 z = 2.326 from inverse normal for 0.99
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M1 k = 48 + 2.326 × √6.48
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A1 53.9 (53.92 unrounded)
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Common error: 2X (one value doubled) and X₁ + X₂ (two independent values added) have the same mean but different variances.
269IBMI3H-4-07r47-14Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard10 marks
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At a railway station, arriving trains use platform A, B or C. The timetable planner models the proportions of arrivals as 35% for A, 45% for B and 20% for C. In a sample of 120 arrivals, 38 used platform A, x used platform B and 29 used platform C.
(a)
Find x.
(1)
(b)
State the null hypothesis for a χ2 goodness-of-fit test and find the expected frequencies.
(2)
(c)
Carry out the test at the 5% significance level, stating the χ2 value, the degrees of freedom and the p-value, and write a conclusion in context.
(4)
(d)
In another sample of 120 arrivals, 38 again used platform A, c used platform B and the rest used platform C. Find the set of integer values of c for which the planner's model is not rejected at the 5% significance level.
(3)
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Marking point
Mark
Notes
Part (a)
A1 x = 120 − 38 − 29 = 53
✓ 1
Part (b)
A1 H0: the planner's proportions (35%, 45%, 20%) fit the arrivals
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A1 expected frequencies 42, 54, 24
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Part (c)
M1 use of Σ(O − E)2/E
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A1 χ2 = 1.44
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A1 3 − 1 = 2 degrees of freedom and p = 0.486
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R1 p > 0.05: no evidence against the planner's model
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Part (d)
M1 statistic = 16/42 + (c − 54)2/54 + (58 − c)2/24 compared with critical value 5.99 (or p > 0.05)
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M1 solving or trial of integer values of c (e.g. c = 47 gives 6.33 and c = 48 gives 5.21; c = 66 gives 5.71 and c = 67 gives 6.89)
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A1 48 ≤ c ≤ 66
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Exam tip: Expected frequencies come from the proportions in H₀ multiplied by the total; the degrees of freedom are the number of categories minus 1 when no parameters are estimated.
270IBMI3H-4-07r48-05Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
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At a container terminal, the mass of a single loaded container, X tonnes, is normally distributed with mean 18.4 and standard deviation 1.2. Masses of different containers are independent.
(a)
A crane lifts three containers chosen at random in a single lift. Let S be their total mass. Find \(E(S)\) and \(\text{Var}(S)\).
(3)
(b)
The crane's safe working limit is 60 tonnes. Find the probability that a lift of three random containers exceeds the limit.
(2)
(c)
A different lift assumes all three containers have the same mass as one container chosen at random, giving a total \(3X\). Find \(\text{Var}(3X)\) and the probability that \(3X\) exceeds 60 tonnes.
(3)
(d)
Explain why \(S\) and \(3X\) have the same mean but \(P(S>60)\) and \(P(3X>60)\) are different.
(1)
(e)
A ship carries 40 such containers on deck, with independent masses. Find the smallest load limit \(L\), in tonnes to 1 decimal place, such that the probability that the total mass exceeds \(L\) is no more than 0.01.
(3)
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Marking point
Mark
Notes
Part (a)
A1 \(E(S)=3\times18.4=55.2\)
✓ 1
M1 \(\text{Var}(S)=3\times1.2^2\)
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A1 \(\text{Var}(S)=4.32\)
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Part (b)
M1 \(S\sim N(55.2,4.32)\), \(P(S>60)\)
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A1 0.0105 (0.010461…)
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Part (c)
A1 \(\text{Var}(3X)=3^2\times1.2^2=12.96\)
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M1 \(3X\sim N(55.2,12.96)\)
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A1 \(P(3X>60)=0.0912\) (0.09121…)
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Part (d)
R1 the sum of independent containers has smaller variance because high and low masses partly cancel, whereas \(3X\) multiplies a single deviation by 3
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Part (e)
A1 total \(\sim N(736,57.6)\)
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M1 \(L=736+z\sqrt{57.6}\) with \(z=2.3263\) (inverse normal for 0.99)
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A1 \(L=753.7\) (753.656…)
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Exam tip: Sums of independent copies add variances (\(n\sigma^2\)), whereas scaling one variable multiplies variance by \(n^2\); keep the two situations apart.
271IBMI3H-4-07r48-10Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard9 marks
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A container barge visits one of three ports, A, B or C, each week. The transition matrix T has each column giving the probabilities of moving from a port, with states in the order A, B, C: $\(\mathbf T=\begin{pmatrix}0.6&0.2&k\\0.3&0.5&0.4\\0.1&0.3&0.6-k\end{pmatrix}\)$ where $k$ is a constant.
(a)
Write down the range of possible values of \(k\).
(1)
(b)
In the long term the barge is at port C for a proportion 0.28 of weeks. Find the value of \(k\).
(5)
(c)
Using \(k=0.15\), the barge starts at port A. Find the probability that it is at port C two weeks later.
(2)
(d)
A company owns 500 such barges, each following this model independently. Estimate the number at port A in a typical week in the long term.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(0\le k\le0.6\)
✓ 1
Part (b)
M1 uses \(\mathbf T\mathbf s=\mathbf s\) with \(\mathbf s=(a,b,0.28)^T\) and \(a+b=0.72\)
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M1 second row: \(b=0.3a+0.5b+0.4(0.28)\), eliminates \(a\)
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A1 \(b=0.41\) and \(a=0.31\)
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M1 first row: \(a=0.6a+0.2b+0.28k\), so \(0.124=0.082+0.28k\)
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A1 \(k=0.15\)
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Part (c)
M1 \(\mathbf T^2(1,0,0)^T\) or paths A→A→C, A→B→C, A→C→C: \(0.6\times0.1+0.3\times0.3+0.1\times0.45\)
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A1 0.195
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Part (d)
A1 \(500\times0.31=155\)
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Exam tip: Unknown parameters in a transition matrix can often be found from the steady-state equations: use the rows that avoid \(k\) first, then substitute into the row that contains it.
272IBMI3H-4-07r48-12Poisson distribution and linear combinations of random variables; t-tests and confidence intervals; Markov chains; non-linear regression and the coefficient of determinationPaper 2Hard12 marks
Calculator
The concentration of volcanic ash, \(y\) μg m−3, was measured at distances \(x\) km downwind of a vent:
\(x\)
0
1
2
3
4
5
\(y\)
82
61
47
34
25
19
A volcanologist proposes the model \(y=84\times0.74^x\).
(a)
Find the values of \(y\) predicted by this model at \(x=1\) and \(x=3\), to 1 decimal place.
(2)
(b)
Calculate the sum of the squares of the residuals, \(SS_{res}=\sum(y_i-\hat y_i)^2\), for the model, to 3 significant figures.
(3)
(c)
The mean of the observed values of \(y\) is 44.67 (44.666…). Find \(SS_{tot}=\sum(y_i-\bar y)^2\) and hence \(R^2=1-\frac{SS_{res}}{SS_{tot}}\). Use unrounded values from earlier parts.
(3)
(d)
Linear regression gives \(R^2=0.961\) for these data. Compare the two models.
(2)
(e)
Use the model to estimate the distance at which the concentration falls to 10 μg m−3. Give your answer to 3 significant figures.
Exam tip: A high \(R^2\) shows how well a model fits the data you have; it does not guarantee that predictions far outside the data range are reliable.
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