Edexcel International A Level Maths · S2 · Statistics 2
Continuous uniform distribution and Normal approximations
28 free, high-quality past-paper-style questions on Continuous uniform distribution and Normal approximations for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 240 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-S2-3-B-01S2 · A2Easy6 marks
A straight line segment AB has length 12 cm. A point P is chosen at random on AB, so that every position is equally likely. The random variable X is the length AP in cm.
Figure: P is a random point on AB; not to scale
(a)
Write down the probability density function of X.
(2)
(b)
Find the probability that AP is more than three times as long as PB.
(2)
(c)
Find E(X) and the standard deviation of X.
(2)
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Answer
Mark
Notes
Part (a)
f(x) = 1/12
B1
for 0 ≤ x ≤ 12, and f(x) = 0 otherwise
B1
Part (b)
PB = 12 − X, so X > 3(12 − X) ⟹ X > 9
M1
P(X > 9) = 3/12 = 0.25
A1
Part (c)
E(X) = (0 + 12)/2 = 6
B1
Var(X) = 122/12 = 12, so σ = √12 = 3.46 cm (awrt)
B1
Exam tip: Turn a geometric condition into an inequality in X first. Once it reads X > 9, the probability is just the length of the interval divided by 12.
At a museum, 40% of visitors choose to hire an audio guide, independently of each other. On one afternoon 90 visitors arrive. The random variable X is the number of these visitors who hire an audio guide.
(a)
Explain why a normal distribution can be used to approximate the distribution of X, and state its parameters.
(2)
(b)
Using this approximation, find P(X ≥ 40).
(4)
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Answer
Mark
Notes
Part (a)
X ~ B(90, 0.4) with n large and p close to 0.5 (np = 36 and n(1 − p) = 54 both large)
B1
Y ~ N(36, 21.6)
B1
Part (b)
continuity correction: P(Y > 39.5)
M1
standardises: (39.5 − 36)/√21.6 (= 0.753…)
M1
1 − Φ(0.75) = 1 − 0.7734
M1
= 0.2266 (awrt 0.227; calculator 0.2257 accepted)
A1
Model sketch
Figure: N(36, 21.6) with the continuity-corrected region y > 39.5 shaded
Exam tip: “At least 40” includes 40, so its bar starts at 39.5. List the integers included, then move half a unit outwards.
A ferry's actual departure time is compared with its timetabled time each day. The random variable D minutes is the actual departure time minus the timetabled time, so a negative value of D means that the ferry leaves early. Figure 1 shows the probability density function of D.
(a)
Find the value of k.
(1)
(b)
Find the probability that, on a randomly chosen day, the ferry leaves early.
(1)
(c)
Write down E(D) and find Var(D).
(2)
(d)
Find the probability that D is more than one standard deviation above its mean.
(2)
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Answer
Mark
Notes
Part (a)
the rectangle runs from −4 to 8, width 12, and has area 1, so k = 1/12
Exam tip: Read the two ends of the rectangle from the axis first. The height then follows from total area 1, and every probability is a width times that height.
The time T seconds that a laser printer takes to warm up has the cumulative distribution function F(t) shown in Figure 1. The graph consists of straight-line sections.
(a)
Explain how Figure 1 shows that T has a continuous uniform distribution, and state the interval on which it is defined.
(2)
(b)
Specify fully the probability density function of T.
(1)
(c)
Find P(2 < T < 3.2).
(2)
(d)
Find the value of t such that P(T > t) = 0.1.
(1)
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Answer
Mark
Notes
Part (a)
F rises at a constant rate (a straight line) between its two corners, so the pdf, its gradient, is constant there
B1
T ~ U[1.5, 4]
B1
Part (b)
f(t) = 0.4 for 1.5 ≤ t ≤ 4, and f(t) = 0 otherwise
B1
Part (c)
(3.2 − 2)/2.5 or F(3.2) − F(2) = 0.68 − 0.2
M1
= 0.48
A1
Part (d)
(t − 1.5)/2.5 = 0.9 ⟹ t = 3.75
B1
Common error: Giving the gradient of the cdf as the probability. The gradient, 1/2.5 = 0.4, is the height of the pdf; probabilities are differences of F values.
A harbour keeper reads the water level on the vertical staff shown in Figure 1, which is marked every 5 cm. She always records the level as the value of the highest mark that is below the water surface, so the recorded value is never greater than the true level. The random variable R cm is the true level minus the recorded level, and is modelled by a continuous uniform distribution.
(a)
Write down the level she records for the situation shown in Figure 1.
(1)
(b)
Write down the distribution of R.
(1)
(c)
Find E(R) and Var(R).
(2)
(d)
Three independent readings are taken. Find the probability that for at least one of them R is more than 4.
(2)
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Answer
Mark
Notes
Part (a)
130 cm
B1
Part (b)
R ~ U[0, 5]
B1
Part (c)
E(R) = 2.5
B1
Var(R) = 52/12 = 25/12 (≈ 2.08)
B1
Part (d)
P(R > 4) = 1/5, so 1 − (4/5)3
M1
= 0.488
A1
Exam tip: Ask which way the recording rule moves the value. Always reading down gives an error between 0 and 5, not between −2.5 and 2.5, so the mean error is 2.5 cm, not 0.
In a large batch of tulip bulbs, 30% produce red flowers. A bag contains 80 bulbs, which may be regarded as a random sample from the batch. The random variable X is the number of bulbs in the bag that produce red flowers. Figure 1 shows the distribution of X; the shaded bars represent the event A.
(a)
Write P(A) in the form P(r ≤ X ≤ s), stating the values of r and s.
(1)
(b)
Give a reason why a normal distribution can be used to approximate the distribution of X, and state its parameters.
(2)
(c)
Use this approximation to find P(A).
(3)
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Answer
Mark
Notes
Part (a)
P(20 ≤ X ≤ 27)
B1
Part (b)
n = 80 is large and p = 0.3 is not far from 0.5 (np = 24 and n(1 − p) = 56 both large; the chart is roughly symmetrical)
Φ(0.85) − (1 − Φ(1.10)) = 0.8023 − 0.1357 = 0.6666 (awrt 0.667 from tables; a calculator gives 0.6673, also accepted)
A1
Exam tip: Each bar has width 1 and is centred on its integer, so the shaded block starts at the left edge of the bar for 20 (19.5) and ends at the right edge of the bar for 27 (27.5).
The continuous random variable X is uniformly distributed on the interval [a, b]. Given that E(X) = 9 and P(X < 6) = 0.3,
(a)
find the values of a and b,
(4)
(b)
find Var(X),
(1)
(c)
find P(X > 12 | X > 6).
(2)
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Answer
Mark
Notes
Part (a)
(a + b)/2 = 9, so a + b = 18
B1
(6 − a)/(b − a) = 0.3
M1
substitutes b = 18 − a: 6 − a = 0.3(18 − 2a) ⟹ 0.4a = 0.6
M1
a = 1.5, b = 16.5
A1
Part (b)
152/12 = 18.75 ft their a, b
B1
Part (c)
P(X > 12) ÷ P(X > 6) = (4.5/15) ÷ (10.5/15)
M1
= 3/7 (≈ 0.429)
A1
Model sketch
Figure: X uniform on [1.5, 16.5], with P(X < 6) = 0.3 shaded
Exam tip: Two facts about a uniform distribution give two equations in a and b. The mean always gives a + b; a probability gives a length as a fraction of b − a.
A computer program generates a random number R from the continuous uniform distribution on [−3, 5]. Successive numbers are independent.
(a)
Specify fully the cumulative distribution function of R.
(3)
(b)
Find the 20th percentile of R.
(1)
(c)
Find E(4R − 3) and Var(4R − 3).
(3)
(d)
The program generates three numbers. Find the probability that all three are positive.
(2)
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Answer
Mark
Notes
Part (a)
F(r) = (r + 3)/8 for −3 ≤ r ≤ 5
B1
F(r) = 0 for r < −3
B1
F(r) = 1 for r > 5
B1
Part (b)
(r + 3)/8 = 0.2 ⟹ r = −1.4
B1
Part (c)
E(R) = 1, so E(4R − 3) = 1
B1
Var(R) = 82/12 = 16/3
B1
Var(4R − 3) = 16 × 16/3 = 256/3 (≈ 85.3)
B1
Part (d)
P(R > 0) = 5/8, cubed for three independent numbers
M1
(5/8)3 = 125/512 = 0.244 (awrt)
A1
Model sketch
Figure: graph of the cumulative distribution function F(r) = (r + 3)/8 for −3 ≤ r ≤ 5
Common error: Writing Var(4R − 3) = 4 Var(R) − 3. Adding a constant does not change the spread, and multiplying by 4 multiplies the variance by 4² = 16.
On a bottling line, 1.5% of bottles are underfilled and 45% of bottles carry a promotional label. A crate contains 300 bottles, which may be treated as a random sample.
(a)
Using a suitable approximation, find the probability that a crate contains at most 3 underfilled bottles. Give a reason for your choice of approximation.
(3)
(b)
Using a suitable approximation, find the probability that a crate contains more than 145 bottles with a promotional label. Give a reason for your choice of approximation.
(5)
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Answer
Mark
Notes
Part (a)
n = 300 is large and p = 0.015 is small, so use a Poisson approximation
B1
B(300, 0.015) ≈ Po(4.5)
B1
P(U ≤ 3) = 0.3423 (tables)
B1
Part (b)
n large and p = 0.45 close to 0.5, so use a normal approximation
B1
L ≈ Y ~ N(135, 74.25)
B1
continuity correction: P(Y > 145.5)
M1
standardises: (145.5 − 135)/√74.25 = 1.22
M1
1 − Φ(1.22) = 1 − 0.8888 = 0.1112 (awrt 0.111)
A1
Exam tip: The size of p decides between the two approximations: a small p gives a skewed distribution with mean close to variance (Poisson); a p near ½ gives a symmetrical, bell-shaped one (normal).
At a railway station, 55% of passengers pay by contactless card. A random sample of 200 passengers is taken, and X is the number who pay by contactless card.
(a)
State, with a reason, an approximate distribution for X.
Figure 1 shows a spinner used in a game. When it is spun, the pointer comes to rest at an angle Θ degrees, measured clockwise from the fixed line OA, where Θ is modelled by the continuous uniform distribution on [0, 360]. The player wins if the pointer comes to rest in the shaded sector.
(a)
Find the probability that the player wins on one spin.
(1)
(b)
Show, by integration, that Var(Θ) = 10 800.
(3)
(c)
Find the probability that the pointer comes to rest within 30° of the line OA.
(2)
(d)
The spinner is spun 300 times. Using a suitable approximation, find the probability that the player wins at least 80 times.
(4)
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Answer
Mark
Notes
Part (a)
(140 − 50)/360 = 0.25
B1
Part (b)
E(Θ) = 180 (by symmetry or ∫ θ/360 dθ)
B1
E(Θ2) = ∫ θ2/360 dθ from 0 to 360 = [θ3/1080] = 43 200
Common error: Treating “within 30° of OA” as 0 < Θ < 30 only. The angle is measured all the way round, so the region next to OA on the other side is 330 < Θ < 360.
Figure 1 shows a square ABCD of side 10 cm. A point P is chosen at random on the side AB, so that the distance AP = X cm, where X ~ U[0, 10]. The segment PC is drawn.
(a)
Find the probability that PC is shorter than 12 cm.
(3)
(b)
Show, by integration, that E(PC2) = 400/3.
(3)
(c)
Find the mean and the standard deviation of the area of triangle PBC.
(2)
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Answer
Mark
Notes
Part (a)
PC2 = (10 − X)2 + 102
M1
(10 − X)2 < 44 ⟹ X > 10 − √44 = 3.366…
M1
P = √44/10 = 0.663 (awrt)
A1
Part (b)
E(PC2) = ∫ (x2 − 20x + 200)/10 dx from 0 to 10, using f(x) = 1/10
M1
= [x3/30 − x2 + 20x] from 0 to 10
M1
= 100/3 − 100 + 200 = 400/3 (working seen)
A1
Part (c)
area = ½ × 10 × (10 − X) = 50 − 5X, mean 50 − 5 × 5 = 25 cm2
B1
standard deviation 5 × 10/√12 = 14.4 cm2 (awrt)
B1
Exam tip: Write the length or area as a function of X from the figure first, then turn the condition on the length into an interval of X. The probability is the length of that interval over 10.
The number X of birds that visit a garden feeder in a day is modelled by a Poisson distribution with mean 32. Figure 1 shows the curve of the normal distribution Y ~ N(32, 32) that is used to approximate X. The shaded area represents P(a ≤ X ≤ b) for two integers a and b.
(a)
Give a reason why a normal approximation is suitable here.
(1)
(b)
Write down the values of a and b.
(2)
(c)
Find the shaded area.
(3)
(d)
Use the same approximation to find the probability that more than 40 birds visit the feeder in a day.
(2)
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Answer
Mark
Notes
Part (a)
the mean λ = 32 is large (so the Poisson distribution is roughly symmetrical)
B1
Part (b)
27.5 is the lower edge of the bar for 28: a = 28
B1
36.5 is the upper edge of the bar for 36: b = 36
B1
Part (c)
standardises: ±4.5/√32 = ±0.80
M1
2Φ(0.80) − 1
M1
= 2 × 0.7881 − 1 = 0.5762 (awrt 0.576 from tables; a calculator gives 0.5737, awrt 0.574, also accepted)
A1
Part (d)
P(Y > 40.5): (40.5 − 32)/√32 = 1.50
M1
1 − Φ(1.50) = 1 − 0.9332 = 0.0668 (awrt 0.067 from tables; a calculator gives 0.0665, also accepted)
A1
Common error: Reading 27.5 as “27 or more”. Under the continuity correction the boundary 27.5 separates 27 from 28, and the shaded area lies to its right, so 28 is the smallest value included.
A courier firm finds that 60% of its parcels are delivered before noon and that 2% of its parcels are damaged, independently of each other and of other parcels. For a random sample of 50 parcels, X is the number delivered before noon, so X ~ B(50, 0.6). For a random sample of 200 parcels, Y is the number damaged, so Y ~ B(200, 0.02). Figure 1 shows the distributions of X and Y, labelled A and B in some order.
(a)
State which of the charts, A or B, shows the distribution of Y. Give a reason.
(1)
(b)
With reference to Figure 1, explain which of X and Y is suited to a normal approximation and which to a Poisson approximation.
(2)
(c)
Using a Poisson approximation, find P(Y ≥ 6).
(2)
(d)
Using a normal approximation, find P(X ≥ 32).
(4)
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Answer
Mark
Notes
Part (a)
A, since E(Y) = 200 × 0.02 = 4 and chart A is centred near 4, with its peak at 3 and 4 (the two tallest bars are almost equal; accept a peak at 3, at 4 or at 3–4), while chart B is centred near 50 × 0.6 = 30 (or: A is skewed, as expected for small p)
B1
Part (b)
X: chart B is roughly symmetrical and bell-shaped (n large, p close to 0.5), so a normal approximation
B1
Y: chart A is positively skewed (n large, p small), so a Poisson approximation
B1
Part (c)
Y ≈ Po(4), so P(Y ≥ 6) = 1 − P(Y ≤ 5)
M1
= 1 − 0.7851 = 0.2149
A1
Part (d)
X ≈ W ~ N(30, 12)
B1
continuity correction: P(W > 31.5)
M1
(31.5 − 30)/√12 = 0.43
M1
1 − Φ(0.43) = 1 − 0.6664 = 0.3336 (awrt 0.334 from tables; a calculator gives 0.3325, awrt 0.333, also accepted)
A1
Exam tip: The shape tells you the approximation: a skewed chart squeezed against zero needs a Poisson model with λ = np; a symmetrical bell needs a normal model with mean np and variance np(1 − p).
A machine produces strips of ribbon 40 cm long. Each strip contains exactly one defect, and X cm is the distance of the defect from the left-hand end of the strip. The histogram in Figure 1 shows the values of X for a random sample of 200 strips.
(a)
Explain why Figure 1 supports modelling X by a continuous uniform distribution, and state the model.
(2)
(b)
Use Figure 1 to find how many of the 200 strips had the defect between 10 cm and 20 cm from the left-hand end.
(2)
(c)
Using your model, find the expected number of the 200 strips with the defect between 10 cm and 20 cm, and comment on your answer to part (b).
(2)
(d)
Each strip is cut at the defect into two pieces. Using your model, find the probability that the shorter piece is less than 8 cm long.
(2)
(e)
Find Var(X) under your model.
(1)
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Answer
Mark
Notes
Part (a)
the bars all have roughly the same height (frequency density close to 5 everywhere)
B1
X ~ U[0, 40]
B1
Part (b)
5 × 4.8 + 5 × 5.4 (frequency = class width × frequency density)
M1
= 24 + 27 = 51
A1
Part (c)
200 × 10/40 = 50
B1
51 is close to 50, so the observed data agree well with the model
B1
Part (d)
X < 8 or X > 32
M1
16/40 = 0.4
A1
Part (e)
402/12 = 400/3 (≈ 133)
B1
Common error: Reading the bar heights as frequencies. In a histogram the height is frequency density, so each frequency is the height multiplied by the class width of 5 cm.
Calls to a council helpline arrive at random. In each hour of the working day, the number of calls has a Poisson distribution with the mean shown for that hour in Figure 1. The numbers of calls in different hours are independent.
(a)
Find the probability that at most 2 calls arrive between 13:00 and 14:00.
(2)
(b)
Write down the distribution of the number of calls, C, that arrive between 10:00 and 13:00.
(1)
(c)
Using a suitable approximation, find the probability that at least 30 calls arrive between 10:00 and 13:00.
(4)
(d)
Explain why a normal approximation would not be suitable for the number of calls between 16:00 and 17:00.
(1)
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Answer
Mark
Notes
Part (a)
Po(5)
B1
P(≤ 2) = 0.1247 (tables)
B1
Part (b)
C ~ Po(9 + 11 + 7) = Po(27)
B1
Part (c)
λ = 27 is large, so Y ~ N(27, 27)
B1
continuity correction: P(Y > 29.5)
M1
(29.5 − 27)/√27 = 0.48
M1
1 − Φ(0.48) = 1 − 0.6844 = 0.3156 (awrt 0.316 from tables; a calculator gives 0.3152, awrt 0.315, also accepted)
A1
Part (d)
the mean is only 4, too small: the distribution is noticeably skewed
B1
Exam tip: Independent Poisson counts over neighbouring hours add: read each hour's mean from the chart and add them before you approximate.
A surveyor records lengths to the nearest centimetre. The rounding error E cm in a recorded length is the true length minus the recorded length, and is modelled by the continuous uniform distribution on [−0.5, 0.5]. Errors in different lengths are independent.
(a)
Find P(|E| < 0.1).
(2)
(b)
Find the standard deviation of E.
(1)
(c)
The surveyor records 200 lengths. The random variable N is the number of these lengths for which |E| < 0.1. Using a suitable approximation, find P(N > 45).
(5)
(d)
Using the same approximation, find the smallest integer k such that P(N > k) < 0.05.
(4)
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Answer
Mark
Notes
Part (a)
P(−0.1 < E < 0.1) = 0.2 ÷ 1
M1
= 0.2
A1
Part (b)
√(1/12) = 0.289 cm (awrt)
B1
Part (c)
N ~ B(200, 0.2)
B1
n large, np = 40 and n(1 − p) = 160 both large: Y ~ N(40, 32)
B1
continuity correction: P(Y > 45.5)
M1
(45.5 − 40)/√32 = 0.97
M1
1 − Φ(0.97) = 1 − 0.8340 = 0.1660 (awrt 0.166)
A1
Part (d)
P(N > k) ≈ P(Y > k + 0.5)
M1
(k + 0.5 − 40)/√32 > 1.6449
B1
k > 39.5 + 1.6449√32 = 48.80…
M1
k = 49
A1
Model sketch
Figure: E uniform on [−0.5, 0.5], with |e| < 0.1 shaded
Exam tip: In a “smallest k” part, set up the inequality with the continuity correction, solve it, then round up. A quick check of k = 48 and k = 49 with the tables confirms the answer.
Each seed of a certain variety germinates with probability 0.85, independently of all other seeds.
(a)
A gardener sows 50 seeds. Find the probability that at least 45 of them germinate.
(3)
(b)
A nursery sows 240 seeds. Using a suitable approximation, find the probability that at least 200 of them germinate.
(4)
(c)
The nursery wants the probability that at least 200 seeds germinate to be at least 0.95. Using a normal approximation, find the smallest number of seeds, n, that should be sown.
(5)
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Answer
Mark
Notes
Part (a)
counts failures: F ~ B(50, 0.15)
M1
at least 45 germinate ⟺ F ≤ 5, P(F ≤ 5)
M1
= 0.2194 (tables)
A1
Part (b)
X ~ B(240, 0.85) ≈ Y ~ N(204, 30.6)
B1
continuity correction: P(Y > 199.5)
M1
(199.5 − 204)/√30.6 = −0.81
M1
Φ(0.81) = 0.7910 (awrt 0.791)
A1
Part (c)
P(Y > 199.5) ≥ 0.95 with Y ~ N(0.85n, 0.1275n)
M1
(199.5 − 0.85n)/√(0.1275n) ≤ −1.6449
B1
0.85n − 1.6449√(0.1275n) − 199.5 ≥ 0, a quadratic in √n: 0.85(√n)2 − 0.5873√n − 199.5 = 0
M1
√n = 15.67…, so n ≥ 245.5…
M1
n = 246
A1
Exam tip: When n is unknown, both the mean and the standard deviation contain n. Substitute u = √n to turn the inequality into a quadratic, take the positive root, and round n up.
Flaws occur at random in a roll of fabric, at a mean rate of 0.4 per square metre.
(a)
Find the probability that a 5 m2 piece of the fabric has at most 1 flaw.
(2)
(b)
A full roll has an area of 60 m2. Using a suitable approximation, find the probability that a roll has at most 20 flaws.
(4)
(c)
A roll is rejected if it has more than k flaws. Using the same approximation, find the smallest value of k for which the probability that a roll is rejected is less than 0.01.
(4)
(d)
Five rolls are chosen at random. Using your answer to part (b), find the probability that at least one of them has at most 20 flaws.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
Po(2)
M1
P(≤ 1) = 0.4060 (tables)
A1
Part (b)
X ~ Po(24)
B1
λ large, so Y ~ N(24, 24)
B1
P(Y < 20.5): (20.5 − 24)/√24 = −0.71
M1
1 − Φ(0.71) = 1 − 0.7611 = 0.2389 (awrt 0.239)
A1
Part (c)
P(X > k) ≈ P(Y > k + 0.5) < 0.01
M1
(k + 0.5 − 24)/√24 > 2.3263
B1
k > 23.5 + 2.3263√24 = 34.89…
M1
k = 35
A1
Part (d)
1 − (1 − 0.2389)5
M1
= 0.745 (awrt) ft their (b)
A1
Common error: Using √24 as the variance in N(24, 24). For a Poisson approximation both the mean and the variance equal λ; only the standardising step uses √λ.
A commuter arrives at a shuttle stop at a random time. The time W minutes that she waits for the shuttle is modelled by the continuous uniform distribution on [0, 12]. Waiting times on different days are independent.
(a)
Find P(W > 9).
(1)
(b)
The commuter rates her annoyance on a day by A = W2/4. Find E(A).
(3)
(c)
Over 160 days, L is the number of days on which she waits more than 9 minutes. Using a suitable approximation, find P(35 ≤ L ≤ 45).
(5)
(d)
On another route, the probability that the wait on any day is more than 20 minutes is 0.0075. Using a suitable approximation, find the probability that in 200 days there are at least 3 days with a wait of more than 20 minutes.
Trains leave a station at the times shown in Figure 1. A passenger arrives at the station at A minutes past 08:00, where A ~ U[0, 40], and takes the next train to leave. The random variable W minutes is the time she waits.
(a)
Find P(W > 8).
(3)
(b)
Find the probability that she catches the 08:22 train.
(1)
(c)
Show that E(W) = 6.5.
(3)
(d)
She makes this journey on 150 days, arriving independently each day. Using a suitable approximation, find the probability that she catches the 08:22 train on at most 18 of these days.
(4)
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Answer
Mark
Notes
Part (a)
waits of more than 8 minutes are only possible before the 08:16 and 08:40 trains (the other gaps are 4 and 6 minutes)
M1
4 < A < 8 or 22 < A < 32, total length 4 + 10 = 14
M1
14/40 = 0.35
A1
Part (b)
16 < A < 22: 6/40 = 0.15
B1
Part (c)
on a gap of length L the wait falls from L to 0, contributing ∫ (L − u)/40 du = L2/80
M1
gaps 4, 12, 6 and 18 minutes
M1
(16 + 144 + 36 + 324)/80 = 520/80 = 6.5
A1
Part (d)
N ~ B(150, 0.15) ≈ Y ~ N(22.5, 19.125)
B1
continuity correction: P(Y < 18.5)
M1
(18.5 − 22.5)/√19.125 = −0.91
M1
1 − Φ(0.91) = 1 − 0.8186 = 0.1814 (awrt 0.181 from tables; a calculator gives 0.1802, awrt 0.180, also accepted)
A1
Exam tip: Split the time line at the departures. Within each gap the wait is a simple straight-line function of the arrival time, so each probability is a sum of interval lengths over 40.
Figure 1 shows a straight road OE of length 25 km with two fuel stations, S1 and S2. Vehicles that break down on the road do so at a distance X km from O, where X ~ U[0, 25]. The random variable D km is the distance from a breakdown to the nearer fuel station.
(a)
Show that P(D < 3) = 0.44.
(3)
(b)
Find P(D > 5).
(3)
(c)
Given that a breakdown occurs less than 3 km from a fuel station, find the probability that the nearer station is S1.
(2)
(d)
There are 150 breakdowns on the road in a year, independently of each other. Using a suitable approximation, find the probability that at most 60 of them occur less than 3 km from a fuel station.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
D < 3 when 0 ≤ X < 5 (near S1, cut off at O) or 13 < X < 19 (near S2)
M1
total length 5 + 6 = 11
M1
11/25 = 0.44
A1
Part (b)
the stations are equidistant at X = 9; D > 5 when 7 < X < 11 or 21 < X ≤ 25 (no values near O)
M1
total length 4 + 4 = 8
M1
8/25 = 0.32
A1
Part (c)
P(0 ≤ X < 5) ÷ P(D < 3) = (5/25) ÷ (11/25)
M1
= 5/11 (≈ 0.455)
A1
Part (d)
N ~ B(150, 0.44) ≈ Y ~ N(66, 36.96)
B1
continuity correction: P(Y < 60.5)
M1
(60.5 − 66)/√36.96 = −0.90
M1
1 − Φ(0.90) = 1 − 0.8159 = 0.1841 (awrt 0.184 from tables; a calculator gives 0.1828, awrt 0.183, also accepted)
A1
Common error: Counting a full 6 km around S1. The station is only 2 km from O, so the interval within 3 km of it is cut off at the end of the road and has length 5 km.
In a large town, a proportion p of households have a heat pump. For a random sample of n households, X is the number with a heat pump, so X ~ B(n, p). Figure 1 shows the curve of the normal distribution Y used to approximate X, with its maximum point and its two points of inflexion marked. (The points of inflexion of a normal curve are at μ ± σ.)
(a)
Use Figure 1 to write down the mean and the variance of Y.
(2)
(b)
Find the values of n and p.
(3)
(c)
Use the approximation to find P(X < 45).
(3)
(d)
Use the approximation to find P(X = 48).
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
mean = 48
B1
σ = 54 − 48 = 6, variance = 36
B1
Part (b)
np = 48 and np(1 − p) = 36
M1
1 − p = 36/48, so p = 0.25
A1
n = 48/0.25 = 192
A1
Part (c)
continuity correction: P(Y < 44.5)
M1
(44.5 − 48)/6 = −0.58
M1
1 − Φ(0.58) = 1 − 0.7190 = 0.2810 (awrt 0.281 from tables; a calculator gives 0.2798, awrt 0.280, also accepted)
A1
Part (d)
P(47.5 < Y < 48.5)
M1
±0.5/6 = ±0.08
M1
2Φ(0.08) − 1 = 2 × 0.5319 − 1 = 0.0638 (awrt 0.064 from tables; a calculator gives 0.0664, awrt 0.066, also accepted)
A1
Exam tip: Dividing the two equations np = 48 and np(1 − p) = 36 removes n at once and leaves 1 − p = 0.75.
When cable is cut to length, offcuts are left over. The length X mm of an offcut is modelled by the continuous uniform distribution whose probability density function is shown in Figure 1. Lengths of different offcuts are independent.
(a)
Find the value of b.
(1)
(b)
Find E(X) and Var(X).
(2)
(c)
Offcuts longer than 40 mm are recycled. On one day there are 400 offcuts, and R is the number that are recycled. Using a suitable approximation, find P(70 ≤ R ≤ 90).
(5)
(d)
Offcuts shorter than 20.5 mm are too short to use. Using a suitable approximation, find the probability that more than 12 of the 400 offcuts are too short to use. Give a reason for your choice of approximation.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.04(b − 20) = 1 ⟹ b = 45
B1
Part (b)
E(X) = (20 + 45)/2 = 32.5
B1
Var(X) = 252/12 = 625/12 (≈ 52.1)
B1
Part (c)
P(X > 40) = 5/25 = 0.2
B1
R ~ B(400, 0.2) ≈ Y ~ N(80, 64)
B1
continuity correction: P(69.5 < Y < 90.5)
M1
±10.5/8 = ±1.31
M1
2Φ(1.31) − 1 = 2 × 0.9049 − 1 = 0.8098 (awrt 0.810 from tables; a calculator gives 0.8106, awrt 0.811, also accepted)
A1
Part (d)
P(X < 20.5) = 0.5/25 = 0.02
B1
n = 400 large and p = 0.02 small, so B(400, 0.02) ≈ Po(8)
B1
1 − P(≤ 12)
M1
= 1 − 0.9362 = 0.0638
A1
Exam tip: The same random variable can lead to two different approximations. Work out p first: p = 0.2 gives a symmetrical binomial (normal), p = 0.02 a skewed one (Poisson).
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