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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 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1AASL4-q00014.1 Sampling, data and outliersPaper 1Easy9 marks
A school has 600 students: 240 in Grade 10, 200 in Grade 11 and 160 in Grade 12. The principal wants to survey a sample of 60 students about how they travel to school.
(a)(i)
State whether each of the following variables is discrete or continuous.
The number of siblings a student has.
(1)
(a)(ii)
The time a student takes to travel to school.
(1)
(b)
A stratified sample by grade is chosen. Find the number of students from each grade in the sample.
(2)
(c)
A teacher suggests surveying the first 60 students who arrive at school on Monday. Name this sampling technique and give one reason why it may produce a biased sample.
(2)
(d)
Describe how a systematic sample of 60 students could be chosen from an alphabetical list of all 600 students.
(1)
(e)
State one difference between a quota sample and a stratified sample.
(1)
(f)
One student in the sample leaves the question about travel time blank. Suggest how this missing value should be dealt with when the data are analysed.
(1)
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Notes
(a)(i)
discrete
A1
(a)(ii)
continuous
A1
(b)
\(\dfrac{240}{600}\times60\) (or using the fraction \(\tfrac{60}{600}=\tfrac{1}{10}\) of each grade)
(M1)
Grade 10: 24, Grade 11: 20, Grade 12: 16
A1
All three required.
(c)
convenience sampling
A1
e.g. the earliest arrivals are likely to share a way of travelling (for example they come by car or live nearby), so they are not representative of all students
R1
Accept any reason that refers to the sample not being representative of the population.
(d)
choose a random starting student among the first 10 on the list, then select every 10th student after that
A1
Must see the interval \(600\div60=10\) and a random start.
(e)
in a quota sample the members of each group are not chosen at random (they are chosen until each quota is filled), whereas in a stratified sample they are chosen randomly from each stratum
A1
Accept any valid distinction about random selection within groups.
(f)
leave that student out of the analysis of travel time (use the other 59 responses), rather than recording the value as 0 or guessing it
R1
Accept replacing the student with another randomly chosen student from the same grade.
2AASL4-q00024.1 Sampling, data and outliersPaper 1Easy7 marks
A city council wants to know the opinion of the 24 000 adults who live in the city about a proposed new cycle lane. The council places a poll on the website of a local cycling club and receives 850 responses. A local newspaper later reports that “70% of residents support the cycle lane”, based on these responses.
(a)
State the population the council is interested in.
(1)
(b)
Write down the size of the sample.
(1)
(c)
Explain why this sample is likely to be biased.
(2)
(d)
Describe how the council could select a simple random sample of 400 adults from the city.
(2)
(e)
Suggest one reason why the newspaper’s figure of 70% may be unreliable, other than the method of sampling.
(1)
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Notes
(a)
all (24 000) adults who live in the city
A1
Do not accept “people who use the website” or “cyclists”.
(b)
\(850\)
A1
(c)
the respondents chose themselves from visitors to a cycling-club website (self-selected / convenience sample)
R1
so cyclists, who are more likely to support a cycle lane, are over-represented; adults who do not cycle or do not visit the site are unlikely to be included
R1
Must refer to the context for this mark.
(d)
use a complete list of the adults in the city (e.g. the electoral register) and number them 1 to 24 000
M1
use a random number generator to choose 400 different numbers (ignoring repeats) and survey those adults
A1
Award A1 only if each adult has an equal chance of being chosen.
(e)
e.g. the 70% applies only to those who responded, not to all residents; the newspaper may have its own interest in the result; people could respond more than once online
R1
Accept any reasonable reason relating to the reliability of the source or of the data.
Answers: (a) all adults living in the city · (b) 850
3AASL4-q00034.1 Sampling, data and outliersPaper 1Easy6 marks
A factory makes 2000 light bulbs each day on three machines. Every bulb is given a serial number as it is made. A quality inspector needs to test 40 of the bulbs made on one day.
(a)(i)
Name the sampling technique described in each of the following.
The inspector tests the first 40 bulbs made that morning.
(1)
(a)(ii)
The inspector uses a random number generator to choose 40 different serial numbers from 1 to 2000.
(1)
(a)(iii)
The inspector tests every 50th bulb made, starting with a bulb chosen at random from the first 50.
(1)
(a)(iv)
The inspector chooses bulbs from each machine in proportion to the number of bulbs it makes, selecting them at random.
(1)
(b)
The bulbs made in the afternoon are more likely to be faulty. State, with a reason, which method in part (a) is most likely to underestimate the proportion of faulty bulbs.
(1)
(c)
Machine 1 has 50 moulds that are used in turn, so every 50th bulb comes from the same mould. Explain why the method in part (a)(iii) may give a misleading result.
(1)
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(a)(i)
convenience sampling
A1
(a)(ii)
simple random sampling
A1
(a)(iii)
systematic sampling
A1
(a)(iv)
stratified sampling
A1
Do not accept quota: the bulbs are chosen at random within each machine.
(b)
method (i) (convenience): all the bulbs tested were made in the morning, so the faulty afternoon bulbs are not included
R1
(c)
the sample may contain bulbs from only one mould (of machine 1), so a faulty (or perfect) mould would be over-represented and the sample would not represent all the moulds
R1
Answers: (a)(i) convenience (ii) simple random (iii) systematic (iv) stratified · (b) (i)
4AASL4-q00104.2 Presentation of dataPaper 1Easy8 marks
The times, in minutes, taken by 15 students to complete a quiz are
Time (min)
23
17
29
12
26
21
33
18
52
24
20
30
15
27
21
(a)
Find the median, the lower quartile and the upper quartile of the times.
(3)
(b)
Show that the time of 52 minutes is an outlier.
(2)
(c)
There are no other outliers. Draw a box-and-whisker diagram to represent the data on a scale from 10 to 55 minutes.
Answers: (a) median 23, \(Q_{1}=18\), \(Q_{3}=29\) · (c) box 18–29, median 23, whiskers 12 and 33, cross at 52
5AASL4-q00124.2 Presentation of dataPaper 1Easy7 marks
The box-and-whisker diagram shows the departure delays, in minutes, of 120 flights from an airport on one day.
Diagram drawn to scale
(a)
Write down the median delay.
(1)
(b)
Find the interquartile range.
(1)
(c)
Write down the approximate number of flights delayed by between 14 and 27 minutes.
(1)
(d)
The airport aims for at least 75% of its flights to be delayed by less than 30 minutes. Determine, with a reason, whether this aim was met on this day.
(2)
(e)
Explain why the upper whisker ends at 44 rather than at 52.
(1)
(f)
Explain why the delays are unlikely to be normally distributed.
(1)
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Notes
(a)
\(19\) minutes
A1
(b)
\(27-14=13\) minutes
A1
(c)
\(60\) flights
A1
Half of 120.
(d)
the upper quartile is 27, so (about) 75% of the flights were delayed by at most 27 minutes
R1
\(27\lt 30\), so the aim was met
A1
Do not award A1R0.
(e)
the whisker extends to the largest value that is not an outlier; 52 is more than \(27+1.5\times13=46.5\), so it is an outlier and is shown separately with a cross
R1
(f)
the box and whiskers are not symmetrical: the median is much closer to \(Q_{1}\) than to \(Q_{3}\) and the upper whisker is much longer than the lower whisker
6AASL4-q00294.4 Correlation and regression of y on xPaper 1Easy5 marks
The four scatter diagrams A, B, C and D are drawn on the same scales.
Diagram drawn to scale
(a)
The values of Pearson’s product-moment correlation coefficient, \(r\), for the four data sets are \(0.98\), \(-0.74\), \(-0.04\) and \(-0.99\). Match each diagram with its value of \(r\).
(3)
(b)
Using diagram C, explain why a value of \(r\) close to zero does not mean that there is no relationship between \(x\) and \(y\).
(1)
(c)
Diagram A shows, for 12 summer days, the number of ice creams sold at a beach, \(x\) (hundreds), and the number of people treated for sunburn, \(y\) (tens). A journalist says that eating ice cream causes sunburn. Comment on this claim.
(1)
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Notes
(a)
A: \(0.98\)
A1
B: \(-0.74\) and D: \(-0.99\)
A1
Both required for this mark.
C: \(-0.04\)
A1
(b)
the points in C follow a clear curve (U-shape); \(r\) only measures the strength of a linear relationship
R1
(c)
correlation does not imply causation; both variables are probably caused by a third factor, such as hot sunny weather
8AASL4-q00404.5 Probability and expected occurrencesPaper 1Easy7 marks
A fair four-sided die, numbered 1 to 4, and a fair six-sided die, numbered 1 to 6, are rolled. The score, \(D\), is the positive difference between the two numbers obtained (so \(D=0\) when the numbers are equal).
(a)
Find \(\mathrm{P}(D=0)\).
(2)
(b)
Find \(\mathrm{P}(D\ge 3)\).
(2)
(c)
The two dice are rolled 120 times. Find the expected number of times that \(D=1\).
(2)
(d)
In an experiment the two dice were rolled 120 times and \(D=1\) occurred 41 times. Write down the relative frequency of \(D=1\) in this experiment.
(1)
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Mark
Notes
(a)
sample space of 24 equally likely outcomes, e.g. a table of differences
(M1)
\(\dfrac{4}{24}=\dfrac{1}{6}\)
A1
(b)
\(4+2+1=7\) outcomes with \(D=3,\,4,\,5\)
(A1)
\(\dfrac{7}{24}\)
A1
(c)
\(\mathrm{P}(D=1)=\dfrac{7}{24}\), and \(120\times\dfrac{7}{24}\)
9AASL4-q00414.5 Probability and expected occurrencesPaper 1Easy5 marks
A spinner is divided into four sectors coloured red, blue, yellow and green. The angles of the sectors are shown in the diagram, which is drawn to scale. When the spinner is spun, the probability that it stops on a colour is proportional to the angle of that sector.
(a)
Find the probability that the spinner stops on red.
(1)
(b)
Find the probability that the spinner does not stop on yellow.
(2)
(c)
The spinner is spun 240 times. Find the expected number of times it stops on green or blue.
(2)
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Mark
Notes
(a)
\(\dfrac{150}{360}=\dfrac{5}{12}\)
A1
(b)
\(1-\dfrac{75}{360}\)
(M1)
Or \(\dfrac{150+90+45}{360}\).
\(\dfrac{285}{360}=\dfrac{19}{24}\)
A1
(c)
\(\mathrm{P}(\text{green or blue})=\dfrac{135}{360}=\dfrac{3}{8}\), and \(240\times\dfrac{3}{8}\)
10AASL4-q00464.6 Combined events and diagramsPaper 1Easy7 marks
In a group of 50 students, 20 play a musical instrument (\(M\)), 26 play a sport (\(S\)) and 12 do neither. The Venn diagram shows the events \(M\) and \(S\). The values \(p\), \(q\), \(r\) and \(s\) are the numbers of students in each region.
Diagram not to scale
(a)
Find the values of \(p\), \(q\) and \(r\).
(3)
(b)
A student is chosen at random from the group. Find the probability that the student does exactly one of the two activities.
(2)
(c)
A student is chosen at random from those who play a sport. Find the probability that this student also plays a musical instrument.
13AASL4-q00514.6 Combined events and diagramsPaper 1Easy6 marks
Spinner P has three equal sectors numbered 1, 2 and 3. Spinner Q has four equal sectors numbered 1, 2, 3 and 4. Both spinners are spun and the score is the product of the two numbers obtained.
(a)
Draw a sample space diagram showing all the possible scores.
(2)
(b)
Find the probability that the score is an even number.
(2)
(c)
Given that at least one of the spinners shows a 3, find the probability that the score is greater than 6.
14AASL4-q00604.7 Discrete random variablesPaper 1Easy7 marks
At a school fair, a player pays $5 to draw one card at random from 20 cards. One card is gold, four cards are silver and the rest are white. A gold card wins $40, a silver card wins $10 and a white card wins nothing. Let \(X\) be the player’s gain, in dollars, from one game (the amount won minus the $5 paid).
(a)
Find the probability distribution of \(X\), giving your answer as a table.
(3)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Explain whether the game is fair.
(1)
(d)
The game is played 200 times during the fair. Find the profit that the fair organisers can expect to make.
(1)
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Notes
(a)
values of \(X\): \(35,\ 5,\ -5\)
A1
\(\mathrm{P}(X=35)=\dfrac{1}{20}\) and \(\mathrm{P}(X=5)=\dfrac{4}{20}\)
15AASL4-q00644.7 Discrete random variablesPaper 1Easy8 marks
The probability that Kai passes his driving test at any attempt is 0.6, independently of previous attempts. He will take the test until he passes, but he is allowed at most three attempts. Let \(X\) be the number of attempts that Kai takes.
(a)
Show that \(\mathrm{P}(X=3)=0.16\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(3)
(c)
Each attempt costs €45. Find the expected total amount that Kai pays for his attempts.
(1)
(d)
Find the probability that Kai passes his driving test.
(2)
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Mark
Notes
(a)
he fails the first two attempts: \(0.4\times0.4\)
M1
Award M0 for \(0.4\times0.4\times0.6\).
third attempt is taken whatever its result, so \(\mathrm{P}(X=3)=0.4^{2}=0.16\)
16AASL4-q00814.9 The normal distributionPaper 1Easy6 marks
The random variables \(X\), \(Y\) and \(W\) are normally distributed, with \(X\sim\mathrm{N}(50,\,5^{2})\), \(Y\sim\mathrm{N}(50,\,10^{2})\) and \(W\sim\mathrm{N}(65,\,5^{2})\). Their probability density curves are labelled \(P\), \(Q\) and \(R\) on the diagram.
Graph drawn to scale
(a)
Write down which curve represents each of \(X\), \(Y\) and \(W\).
(2)
(b)(i)
Using the fact that approximately 68% of a normal distribution lies within one standard deviation of the mean and 95% lies within two standard deviations of the mean, estimate
\(\mathrm{P}(X\gt 60)\);
(1)
(b)(ii)
\(\mathrm{P}(Y\gt 60)\);
(2)
(b)(iii)
\(\mathrm{P}(W\gt 60)\).
(1)
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Notes
(a)
\(X\): curve \(R\) and \(W\): curve \(Q\)
A1
Same spread, different centres.
\(Y\): curve \(P\)
A1
Award A1A1 for all three correct; A1A0 for one correct.
(b)(i)
\(60=\mu+2\sigma\), so \(\mathrm{P}(X\gt 60)\approx\dfrac{1-0.95}{2}=0.025\)
A1
Accept 2.5%.
(b)(ii)
\(60=\mu+\sigma\) for \(Y\), so \(\mathrm{P}(Y\gt 60)=\dfrac{1-0.68}{2}\)
(M1)
\(0.16\)
A1
Accept 16%.
(b)(iii)
\(60=\mu-\sigma\) for \(W\), so \(\mathrm{P}(W\gt 60)\approx0.5+0.34=0.84\)
19AASL4-q01074.11 Conditional probability and independencePaper 1Easy7 marks
An integer \(n\) is chosen at random from the integers 1 to 30 inclusive. Let \(A\) be the event “\(n\) is a multiple of 3”, \(B\) the event “\(n\) is a multiple of 5” and \(C\) the event “\(n\) is a prime number”.
(a)
Write down \(\mathrm{P}(A)\).
(1)
(b)
Find \(\mathrm{P}(A\mid B)\).
(2)
(c)
State, with a reason, whether \(A\) and \(B\) are independent.
(1)
(d)
Find \(\mathrm{P}(A\mid C)\) and hence determine whether \(A\) and \(C\) are independent.
(3)
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Mark
Notes
(a)
\(\dfrac{10}{30}=\dfrac{1}{3}\)
A1
(b)
\(B=\{5,10,15,20,25,30\}\), of which 15 and 30 are multiples of 3
(M1)
Or \(\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}=\dfrac{2/30}{6/30}\).
\(\dfrac{2}{6}=\dfrac{1}{3}\)
A1
(c)
independent, because \(\mathrm{P}(A\mid B)=\dfrac{1}{3}=\mathrm{P}(A)\)
R1
Accept \(\mathrm{P}(A\cap B)=\dfrac{1}{15}=\dfrac{1}{3}\times\dfrac{1}{5}\). FT from (a) and (b).
(d)
10 primes \(\{2,3,5,7,11,13,17,19,23,29\}\), of which only 3 is a multiple of 3
(M1)
\(\mathrm{P}(A\mid C)=\dfrac{1}{10}\)
A1
\(\dfrac{1}{10}\ne\dfrac{1}{3}=\mathrm{P}(A)\), so \(A\) and \(C\) are not independent
20AASL4-q00094.2 Presentation of dataPaper 1Medium8 marks
The box-and-whisker diagrams show the scores of the students in two classes, A and B, in the same test. An outlier is indicated with a cross.
Diagram drawn to scale
(a)
Write down the median score for Class B.
(1)
(b)
Find the interquartile range for each class.
(2)
(c)
Show that the score of 18 in Class B is an outlier.
(2)
(d)
Compare the distributions of the scores in the two classes.
(2)
(e)
State, with a reason, which class has scores that are more likely to be normally distributed.
(1)
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Mark
Notes
(a)
\(58\)
A1
(b)
Class A: \(63-41=22\)
A1
Class B: \(62-48=14\)
A1
(c)
\(Q_{1}-1.5\times\text{IQR}=48-1.5\times14\)
M1
FT from their IQR for Class B.
\(=27\) and \(18\lt 27\)
A1
Must see 27 and a comparison.
(d)
Class B has the higher median (58 compared with 52), so Class B scored higher on average
R1
Must refer to the median (or a measure of central tendency) with values or a clear comparison.
Class B has the smaller IQR (14 compared with 22), so its scores are less spread out (more consistent)
R1
Accept range (A: 62; B: 52) as the measure of spread.
(e)
Class A: its box and whiskers are approximately symmetrical about the median
R1
Award R1 only with a reason referring to symmetry. Class B is not symmetrical: the median is much closer to \(Q_{3}\).
Answers: (a) 58 · (b) A: 22, B: 14 · (e) Class A
21AASL4-q00114.2 Presentation of dataPaper 1Medium6 marks
Each of the frequency histograms P, Q and R shows a set of whole-number data with values from 1 to 9. The box-and-whisker diagrams 1, 2 and 3 represent the same three data sets, in a different order.
Diagram drawn to scale
(a)
Write down which box-and-whisker diagram represents each histogram.
(2)
(b)
Using histogram P, show that the median of this data set is 5.
(2)
(c)
State, with a reason, which data set could be modelled by a normal distribution.
(1)
(d)
All three box-and-whisker diagrams have the same whiskers. Explain what this shows about using only the range to compare the data sets.
(1)
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Notes
(a)
P and 2
A1
Q and 3; R and 1
A1
(b)
\(n=1+2+4+7+9+7+4+2+1=37\), so the median is the 19th value
M1
cumulative frequencies \(1,3,7,14,23\): the 15th to 23rd values are 5, so the 19th value is 5
A1
median \(=5\)
AG
(c)
P (box plot 2), because its histogram and box plot are symmetrical about the median, with frequencies highest in the middle
R1
Award R1 only for P with a reason referring to symmetry.
(d)
the range is the same (8) for all three sets, although the sets have very different centres and shapes; the range alone does not distinguish them
R1
Answers: (a) P–2, Q–3, R–1 · (c) P
22AASL4-q00194.3 Measures of central tendency and dispersionPaper 1Medium8 marks
The table shows the number of books, \(x\), borrowed from a library last month by each student in a group.
Number of books, \(x\)
0
1
2
3
4
Frequency
2
5
\(k\)
3
1
The mean number of books borrowed is 1.8.
(a)
Find the value of \(k\).
(3)
(b)
Find the median number of books borrowed.
(2)
(c)
In a reading scheme, each student receives 3 points for every book borrowed plus 1 bonus point. The variance of the number of books borrowed is 0.96. Find the mean and the variance of the number of points received.
(3)
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Mark
Notes
(a)
\(\dfrac{0(2)+1(5)+2k+3(3)+4(1)}{11+k}=1.8\)
M1
\(18+2k=1.8(11+k)\), i.e. \(18+2k=19.8+1.8k\)
A1
\(k=9\)
A1
(b)
20 students, so the median is the mean of the 10th and 11th values; cumulative frequencies 2, 7, 16
(M1)
FT from their \(k\).
median \(=2\)
A1
(c)
mean \(=3(1.8)+1=6.4\)
A1
variance multiplied by \(3^{2}\); adding 1 does not change it
(M1)
Award M0 for \(3\times0.96\) or \(3\times0.96+1\).
23AASL4-q00204.3 Measures of central tendency and dispersionPaper 1Medium6 marks
A list of five positive integers has mode 6, median 8, mean 11 and range 15.
(a)
Find the five integers.
(4)
(b)
A sixth positive integer, \(k\), is added to the list. The median of the six integers is 9. Find the value of \(k\).
(2)
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Mark
Notes
(a)
sum of the integers \(=5\times11=55\)
M1
the mode 6 must occur at least twice, and the 3rd (middle) value is 8, so the two smallest values are 6 and 6
R1
largest value \(=6+15=21\)
M1
fourth value \(=55-6-6-8-21=14\); the integers are \(6,\,6,\,8,\,14,\,21\)
A1
(b)
the median is the mean of the 3rd and 4th values, which must be 8 and \(k\) (\(8\le k\le14\)): \(\dfrac{8+k}{2}=9\)
M1
If \(k\ge14\) the median is 11; if \(k\le8\) the median is at most 8.
\(k=10\)
A1
Answers: (a) 6, 6, 8, 14, 21 · (b) \(k=10\)
24AASL4-q00214.3 Measures of central tendency and dispersionPaper 1Medium7 marks
A test is marked out of 120. In class A, 12 students took the test and their mean mark was 64. In class B, 18 students took the test and their mean mark was 74.
(a)
Find the total of the marks for class A.
(1)
(b)
Find the mean mark of all 30 students.
(2)
(c)
One more student takes the test later. The mean mark of all 31 students is 71. Find this student’s mark.
(2)
(d)
It is later found that one mark in class B was recorded as 47 instead of 74. Find the correct mean mark for class B.
every time at the lower class boundary: \(\dfrac{0(4)+10(7)+20(10)+30(6)+40(3)}{30}=\dfrac{570}{30}\)
M1
\(19\) minutes
A1
(d)
every time is less than the upper boundary of its class, so the total is less than \(10(4)+20(7)+30(10)+40(6)+50(3)=870\) and the mean is less than \(\dfrac{870}{30}=29\)
R1
Answers: (a) 11 · (b) 24 min · (c) 19 min
26AASL4-q00314.4 Correlation and regression of y on xPaper 1Medium8 marks
For 40 apartments in a city, the floor area, \(x\) m2, and the monthly rent, \(y\) euros, were recorded. The floor areas ranged from 35 m2 to 110 m2. The regression line of \(y\) on \(x\) is \(y=9.5x+120\), and \(r=0.83\).
(a)
Interpret the value 9.5 in the context of the question.
(1)
(b)
Estimate the monthly rent of an apartment with a floor area of 80 m2.
(2)
(c)
Explain why the value 120 should not be interpreted as the rent of an apartment with a floor area of 0 m2.
(1)
(d)
Explain why the equation should not be used to estimate the floor area of an apartment with a monthly rent of €1000.
(1)
(e)(i)
The mean floor area of the 40 apartments is 68 m2.
Find the mean monthly rent of the 40 apartments.
(2)
(e)(ii)
Describe the correlation between floor area and monthly rent.
(1)
Show mark scheme
Working
Mark
Notes
(a)
for each extra square metre of floor area, the monthly rent increases by about €9.50
R1
(b)
\(9.5(80)+120\)
(M1)
€880
A1
(c)
\(x=0\) is far outside the range of the data (35 to 110), and an apartment with no floor area is not meaningful
R1
(d)
this is a regression line of \(y\) on \(x\); it should not be used to predict \(x\) from \(y\) (the regression line of \(x\) on \(y\) is needed)
R1
(e)(i)
the regression line passes through the mean point \((\bar{x},\,\bar{y})\): \(\bar{y}=9.5(68)+120\)
M1
€766
A1
(e)(ii)
(fairly) strong positive (linear) correlation
A1
Answers: (b) €880 · (e)(i) €766 (ii) strong positive
27AASL4-q00424.5 Probability and expected occurrencesPaper 1Medium8 marks
Two fair six-sided dice, each numbered 1 to 6, are rolled. The score on the first die is \(a\) and the score on the second die is \(b\). These values are used to form the equation \(x^{2}+ax+b=0\).
(a)
Show that the equation has real roots if and only if \(a^{2}\ge 4b\).
(1)
(b)
Find the probability that the equation has equal roots.
(2)
(c)
Find the probability that the equation has real roots.
(3)
(d)
The two dice are rolled 72 times and a new equation is formed each time. Find the expected number of equations that have no real roots.
(2)
Show mark scheme
Working
Mark
Notes
(a)
real roots \(\Leftrightarrow\) discriminant \(a^{2}-4b\ge 0\)
R1
\(a^{2}\ge 4b\)
AG
(b)
\(a^{2}=4b\): \((a,b)=(2,1)\) or \((4,4)\)
(M1)
\(\dfrac{2}{36}=\dfrac{1}{18}\)
A1
(c)
systematic listing of outcomes by value of \(a\), or a table of 36 outcomes
28AASL4-q00434.5 Probability and expected occurrencesPaper 1Medium8 marks
A bag contains \(r\) red counters and \(b\) blue counters only. A counter is chosen at random and the probability that it is red is \(\dfrac{2}{5}\).
(a)
Show that \(3r=2b\).
(2)
(b)
Six more red counters are added to the bag. The probability that a counter chosen at random is red is now \(\dfrac{1}{2}\). Find the value of \(r\) and the value of \(b\).
(4)
(c)
After the extra red counters have been added, a counter is chosen at random, its colour noted and it is replaced. This is done 50 times. Find the expected number of times a blue counter is chosen.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{r}{r+b}=\dfrac{2}{5}\)
M1
\(5r=2r+2b\)
A1
\(3r=2b\)
AG
(b)
\(\dfrac{r+6}{r+b+6}=\dfrac{1}{2}\)
M1
\(2r+12=r+b+6\), so \(b=r+6\)
A1
substituting into \(3r=2b\): \(3r=2(r+6)\)
M1
\(r=12,\ b=18\)
A1
Both values required.
(c)
\(\mathrm{P}(\text{blue})=\dfrac{18}{36}\), and \(50\times\dfrac{1}{2}\)
(M1)
FT from (b).
\(25\)
A1
Answers: (b) \(r=12,\ b=18\) · (c) 25
29AASL4-q00474.6 Combined events and diagramsPaper 1Medium7 marks
Events \(A\) and \(B\) are such that \(\mathrm{P}(A)=0.3\) and \(\mathrm{P}(A\cup B)=0.65\).
(a)
Find \(\mathrm{P}(B)\) if \(A\) and \(B\) are mutually exclusive.
mutually exclusive means \(\mathrm{P}(C\cap D)=0\)
R1
independent means \(\mathrm{P}(C\cap D)=\mathrm{P}(C)\,\mathrm{P}(D)\), which is greater than 0, so both cannot hold
R1
Answers: (a) 0.35 · (b) \(\tfrac{7}{15}\)
30AASL4-q00504.6 Combined events and diagramsPaper 1Medium9 marks
In a club of 60 members, 30 play football (\(F\)), 24 play basketball (\(B\)) and 18 play tennis (\(T\)). Of these, 12 play both football and basketball, 9 play both football and tennis, 8 play both basketball and tennis, and 5 play all three sports. The Venn diagram shows the events, with the number of members in each region labelled \(a\) to \(h\).
Diagram not to scale
(a)
Find the values of \(g\), \(d\), \(e\) and \(f\).
(2)
(b)
Find the values of \(a\), \(b\), \(c\) and \(h\).
(3)
(c)
A member is chosen at random. Find the probability that the member plays exactly two of the three sports.
(2)
(d)
A member who plays at least two of the sports is chosen at random. Find the probability that this member plays tennis.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(g=5\) and subtracting \(g\) from a pairwise total, e.g. \(d=12-5\)
35AASL4-q00624.7 Discrete random variablesPaper 1Medium7 marks
The probability distribution of a discrete random variable \(X\) is shown in the table, where \(k\) is a constant.
\(x\)
1
2
3
4
\(\mathrm{P}(X=x)\)
0.15
\(k\)
\(k^{2}\)
0.1
(a)
Show that \(k^{2}+k-0.75=0\).
(1)
(b)
Find the value of \(k\), justifying your answer.
(3)
(c)
Find \(\mathrm{E}(X)\).
(2)
(d)
Find the probability that \(X\) is an even number.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.15+k+k^{2}+0.1=1\)
M1
\(k^{2}+k-0.75=0\)
AG
(b)
\((k+1.5)(k-0.5)=0\), or \(4k^{2}+4k-3=0\Rightarrow(2k+3)(2k-1)=0\)
(M1)
\(k=0.5\) or \(k=-1.5\)
A1
\(k=0.5\) only, since a probability cannot be negative
R1
Award R1 only if \(k=-1.5\) is explicitly rejected with a reason.
(c)
\(1(0.15)+2(0.5)+3(0.25)+4(0.1)\)
(M1)
FT from their positive k.
\(2.3\)
A1
(d)
\(0.5+0.1=0.6\)
A1
FT from their k.
Answers: (b) \(k=0.5\) · (c) 2.3 · (d) 0.6
36AASL4-q00634.7 Discrete random variablesPaper 1Medium7 marks
A company sells one-year insurance for mobile phones. In a year, the probability that a policy holder makes a claim for a repair, costing the company €300, is 0.08. The probability that a policy holder makes a claim for a replacement, costing the company €900, is 0.03. Otherwise no claim is made. Assume that each policy holder makes at most one claim in a year.
(a)
Let \(C\) be the amount, in euros, paid out by the company on one policy in a year. Write down the probability distribution of \(C\).
(2)
(b)
Find \(\mathrm{E}(C)\).
(2)
(c)
The company wants an expected profit of €15 per policy. Find the price that it should charge for a policy.
(2)
(d)
The company sells 2000 policies at this price. Find the expected total profit.
(1)
Show mark scheme
Working
Mark
Notes
(a)
values \(0,\ 300,\ 900\)
A1
probabilities \(0.89,\ 0.08,\ 0.03\)
A1
Must see 0.89.
(b)
\(0(0.89)+300(0.08)+900(0.03)\)
(M1)
€51
A1
(c)
\(\text{price}-51=15\)
(M1)
FT from (b).
€66
A1
(d)
\(2000\times15=\)€30 000
A1
Answers: (b) €51 · (c) €66 · (d) €30 000
37AASL4-q00684.8 The binomial distributionPaper 1Medium7 marks
A random variable \(X\) follows a binomial distribution with mean 6 and variance 4.2.
(a)
Find the probability of success, \(p\), and the number of trials, \(n\).
(4)
(b)
Write down an expression for \(\mathrm{P}(X=0)\).
(1)
(c)
Explain why no binomial random variable can have a mean of 5 and a variance of 6.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(np=6\) and \(np(1-p)=4.2\)
M1
dividing: \(1-p=\dfrac{4.2}{6}\)
M1
\(p=0.3\)
A1
\(n=20\)
A1
(b)
\(0.7^{20}\)
A1
FT from (a).
(c)
\(\dfrac{\text{variance}}{\text{mean}}=1-p\)
M1
\(1-p\lt 1\) since \(p\gt 0\), so the variance must be less than the mean, but \(6\gt 5\)
R1
Or solving gives \(1-p=1.2\Rightarrow p=-0.2\), which is impossible.
Answers: (a) \(p=0.3,\ n=20\) · (b) \(0.7^{20}\)
38AASL4-q00694.8 The binomial distributionPaper 1Medium9 marks
A fair six-sided die is rolled four times. Let \(X\) be the number of rolls that show a 5 or a 6.
(a)
Write down the distribution of \(X\), giving its parameters.
(1)
(b)
Find \(\mathrm{P}(X=0)\) and \(\mathrm{P}(X=4)\).
(2)
(c)
Write down \(\mathrm{E}(X)\).
(1)
(d)
By considering the possible orders of the outcomes, show that \(\mathrm{P}(X=1)=\dfrac{32}{81}\).
39AASL4-q00804.9 The normal distributionPaper 1Medium5 marks
A random variable \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that \(\mathrm{P}(X\lt 42)=0.2\) and \(\mathrm{P}(X\gt 58)=0.2\). This information is shown on the diagram, where each shaded region has area 0.2.
Diagram not to scale
(a)
Write down the value of \(\mu\).
(1)
(b)
Write down \(\mathrm{P}(42\lt X\lt 58)\).
(1)
(c)
By considering the percentage of the distribution that lies within one standard deviation of the mean, deduce whether \(\sigma\) is greater than or less than 8.
(2)
(d)
It is now given that \(\sigma=9.5\). Write down an estimate of \(\mathrm{P}(31\lt X\lt 69)\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\mu=50\)
A1
By symmetry, the mean is halfway between 42 and 58.
(b)
\(0.6\)
A1
(c)
about 68% of the distribution lies between \(\mu-\sigma\) and \(\mu+\sigma\), but only 60% lies between \(50-8\) and \(50+8\)
R1
so \(50\pm8\) lies within one standard deviation of the mean, hence \(\sigma\gt 8\)
A1
Award A1 only with the R1.
(d)
\(0.95\)
A1
\(31\) and \(69\) are \(\mu\pm2\sigma\). Accept 95%.
40AASL4-q00824.9 The normal distributionPaper 1Medium6 marks
A random variable \(X\) is normally distributed with mean 30. The diagram shows the probability density curve of \(X\). The shaded region, to the left of 24, has area 0.15.
Diagram not to scale
(a)
Write down \(\mathrm{P}(X\gt 36)\).
(1)
(b)
Find \(\mathrm{P}(24\lt X\lt 36)\).
(2)
(c)
Find \(\mathrm{P}(X\lt 36\mid X\gt 24)\).
(2)
(d)
In a sample of 200 independent observations of \(X\), find the expected number that lie between 24 and 36.
42AASL4-q01034.11 Conditional probability and independencePaper 1Medium8 marks
A fair spinner with five equal sectors numbered 1 to 5 is spun and a fair six-sided die, numbered 1 to 6, is rolled. Let \(A\) be the event “the product of the two scores is even”, \(B\) the event “the sum of the two scores is 7” and \(C\) the event “the sum of the two scores is at least 9”.
(a)
Find \(\mathrm{P}(A)\).
(2)
(b)
Write down \(\mathrm{P}(A\mid B)\).
(1)
(c)
Hence determine whether \(A\) and \(B\) are independent.
(1)
(d)
Find \(\mathrm{P}(A\mid C)\).
(2)
(e)
Find \(\mathrm{P}(C\mid A)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
sample space of 30 equally likely outcomes, or \(\mathrm{P}(\text{product odd})=\dfrac{3}{5}\times\dfrac{3}{6}=\dfrac{3}{10}\)
(M1)
\(\dfrac{21}{30}=\dfrac{7}{10}\)
A1
(b)
\(1\)
A1
Every pair with sum 7 has one odd and one even score, so the product is even.
(c)
\(\mathrm{P}(A\mid B)=1\ne\dfrac{7}{10}=\mathrm{P}(A)\), so not independent
R1
FT from (a) and (b).
(d)
\(C=\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6)\}\), 6 outcomes, of which 5 have an even product
43AASL4-q01044.11 Conditional probability and independencePaper 1Medium8 marks
In a medical study of a large population, 30% of adults smoke. The probability that a smoker develops a certain disease is three times the probability that a non-smoker develops it. Overall, 4.8% of adults develop the disease.
(a)
Find the probability that a non-smoker develops the disease and the probability that a smoker develops the disease.
(3)
(b)
Given that an adult develops the disease, find the probability that the adult smokes.
(2)
(c)
Given that an adult does not develop the disease, find the probability that the adult smokes.
(2)
(d)
State, with a reason, whether smoking and developing the disease are independent events.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.3(3k)+0.7k=0.048\), where \(k=\mathrm{P}(D\mid S')\)
45AASL4-q01124.12 Standardization and inverse normalPaper 1Medium7 marks
The marks in a history test are normally distributed with mean 61 and standard deviation 9. The marks in a geography test are normally distributed with mean 55 and standard deviation 7.5.
(a)
Find the standardized value, \(z\), of a mark of 70 in the history test.
(1)
(b)
Hence estimate the percentage of students who scored more than 70 in history.
(1)
(c)
Find the history mark that has a standardized value of \(-1.5\).
(2)
(d)
Lena’s history mark and her geography mark have the same standardized value. Her geography mark is 5 marks lower than her history mark. Find this standardized value and Lena’s two marks.
46AASL4-q00534.6 Combined events and diagramsPaper 1Hard9 marks
In a large college, a proportion \(p\) of the students take a course online and the rest take it in a classroom. The probability that a student passes the final test is 0.6 for a student who takes the course online and 0.8 for a student who takes it in a classroom.
(a)
Show that the probability that a student chosen at random passes the final test is \(0.8-0.2p\).
(2)
(b)
Given that a student passes the final test, the probability that this student took the course online is \(\dfrac{1}{3}\). Find the value of \(p\).
(3)
(c)
A student chosen at random fails the final test. Find the probability that this student took the course online.
(2)
(d)
Two students are chosen at random, independently. Find the probability that exactly one of them passes the final test.
(2)
Show mark scheme
Working
Mark
Notes
(a)
tree diagram or \(0.6p+0.8(1-p)\)
M1
\(0.6p+0.8-0.8p\)
A1
\(=0.8-0.2p\)
AG
(b)
\(\dfrac{0.6p}{0.8-0.2p}=\dfrac{1}{3}\)
M1
\(1.8p=0.8-0.2p\), or \(2p=0.8\)
A1
\(p=0.4\)
A1
(c)
\(\mathrm{P}(\text{fail})=1-0.72=0.28\) and \(\mathrm{P}(\text{online}\cap\text{fail})=0.4\times0.4=0.16\)
(M1)
FT from (b).
\(\dfrac{0.16}{0.28}=\dfrac{4}{7}\)
A1
(d)
\(2\times0.72\times0.28\)
(M1)
Award M0 for \(0.72\times0.28\) without the factor 2. FT from (b).
48AASL4-q00564.6 Combined events and diagramsPaper 1Hard15 marks
A bag contains 5 red counters and 3 blue counters. Two counters are taken at random from the bag, one after the other, without replacement. The tree diagram shows the possible outcomes.
R = red, B = blue
(a)
Write down the value of \(a\) and the value of \(b\).
(2)
(b)
Find the probability that both counters are the same colour.
(2)
(c)
Given that both counters are the same colour, find the probability that they are both red.
(3)
(d)(i)
Let \(X\) be the number of blue counters taken.
Find the probability distribution of \(X\).
(3)
(d)(ii)
Show that \(\mathrm{E}(X)=\dfrac{3}{4}\).
(2)
(e)
In a game, a player pays $\(k\) to take two counters in this way and then receives $3 for each blue counter taken. Find the value of \(k\) for which the game is fair.
49AASL4-q00654.7 Discrete random variablesPaper 1Hard8 marks
In a game, a player rolls a fair six-sided die. The player may either keep this score or roll the die once more, in which case the score of the second roll is the final score. A player decides to use the strategy “roll again if the first score is \(m\) or less”, where \(m\in\{0,1,2,3,4,5\}\).
(a)
Write down the expected final score when \(m=0\) (the player never rolls again).
(1)
(b)
Show that the expected final score when \(m=2\) is \(\dfrac{25}{6}\).
(3)
(c)
Determine the value of \(m\) that gives the greatest expected final score, and state this expected score.
(3)
(d)
Explain, without further calculation, why a player should never roll again after scoring 4.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(3.5\)
A1
(b)
first score 3, 4, 5 or 6 is kept: \(\dfrac{3+4+5+6}{6}\)
M1
first score 1 or 2 (probability \(\tfrac{2}{6}\)) followed by an expected 3.5: \(\dfrac{2}{6}\times3.5\)
M1
\(\dfrac{18}{6}+\dfrac{7}{6}\)
A1
\(=\dfrac{25}{6}\)
AG
(c)
general expression or at least two further values: \(m=1:\dfrac{23.5}{6}\), \(m=3:\dfrac{15+10.5}{6}\), \(m=4:\dfrac{11+14}{6}\)
M1
\(m=3\)
A1
\(\dfrac{25.5}{6}=\dfrac{17}{4}\)
A1
(d)
rolling again gives an expected score of 3.5, which is less than the 4 already obtained
Since \(A\subset A\cup B\), \(A\cap(A\cup B)=A\). FT from their \(\mathrm{P}(A)\).
Answers: (b)(i) \(\mathrm{P}(A)=0.3\) · (ii) \(\tfrac{5}{12}\)
52AASL4-q01104.11 Conditional probability and independencePaper 1Hard13 marks
A supermarket buys bags of rice from two suppliers, A and B. Supplier A provides 60% of the bags and supplier B provides the rest. Of the bags from supplier A, 20% are underweight; of the bags from supplier B, 5% are underweight. The tree diagram shows this information.
U = underweight, N = not underweight
(a)
Write down the value of \(p\), of \(q\) and of \(r\).
(2)
(b)
A bag is chosen at random. Find the probability that it is underweight.
(2)
(c)
Given that a bag is underweight, find the probability that it came from supplier A.
(2)
(d)
Determine whether the events “from supplier A” and “underweight” are independent.
(2)
(e)
Two bags are chosen at random. Find the probability that exactly one of them is underweight.
(2)
(f)(i)
The masses, \(X\) grams, of bags from supplier B are normally distributed with mean \(\mu\). A bag is underweight if its mass is less than 490 g, and it is known that \(\mathrm{P}(X\gt 510)=0.05\).
Write down the value of \(\mu\).
(1)
(f)(ii)
A bag from supplier B is not underweight. Find the probability that its mass is more than 510 g.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(p=0.4\)
A1
\(q=0.8\) and \(r=0.95\)
A1
Both required.
(b)
\(0.6(0.2)+0.4(0.05)=0.12+0.02\)
(M1)
\(0.14\)
A1
(c)
\(\dfrac{0.12}{0.14}\)
(M1)
\(\dfrac{6}{7}\)
A1
(d)
\(\mathrm{P}(\text{underweight}\mid A)=0.2\) and \(\mathrm{P}(\text{underweight})=0.14\), or \(\mathrm{P}(A)\,\mathrm{P}(U)=0.084\ne0.12=\mathrm{P}(A\cap U)\)
M1
not independent
A1
Must follow from a correct comparison.
(e)
\(2\times0.14\times0.86\)
(M1)
Award for 0.14 × 0.86 seen.
\(0.2408\)
A1
Accept \(\dfrac{301}{1250}\).
(f)(i)
\(\mu=500\)
A1
\(\mathrm{P}(X\lt 490)=0.05=\mathrm{P}(X\gt 510)\), so by symmetry \(\mu\) is midway between 490 and 510.
53AASL4-q00134.2 Presentation of dataPaper 2Easy7 marks
The frequency histogram shows the time, \(h\) hours per week, that a group of students spend on homework. The class intervals are \(0\le h\lt 5\), \(5\le h\lt 10\), …, \(30\le h\lt 35\).
Graph drawn to scale
(a)
Write down the number of students in the group.
(1)
(b)
Write down the modal class.
(1)
(c)
Find an estimate for the mean time spent on homework.
(2)
(d)
Find an estimate for the standard deviation of the time spent on homework.
(1)
(e)
Find the percentage of the students who spend at least 20 hours per week on homework.
(1)
(f)
Explain why the answer to part (c) is an estimate.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(120\)
A1
(b)
\(10\le h\lt 15\)
A1
(c)
use of mid-interval values \(2.5,\,7.5,\,\ldots,\,32.5\) with frequencies \(10,\,20,\,34,\,26,\,18,\,8,\,4\)
(M1)
e.g. \(\dfrac{2.5(10)+7.5(20)+\cdots+32.5(4)}{120}\)
\(15.1\) hours
A1
\(15.083\ldots\)
(d)
\(7.39\) hours
A1
\(7.3875\ldots\) (GDC \(\sigma_{x}\)).
(e)
\(\dfrac{18+8+4}{120}=25\%\)
A1
(f)
the exact times are not known; every value in a class is assumed to equal its mid-interval value
55AASL4-q00234.3 Measures of central tendency and dispersionPaper 2Easy8 marks
The maximum daily temperatures, in °C, in a town on ten days in May were
Day
1
2
3
4
5
6
7
8
9
10
Temperature (°C)
18.4
21.0
19.6
22.3
20.8
17.9
23.5
21.7
19.2
20.6
(a)
Find the mean temperature.
(1)
(b)(i)
Find
the standard deviation;
(1)
(b)(ii)
the variance.
(1)
(c)
Use your calculator to find the interquartile range of the temperatures.
(2)
(d)
The temperatures are converted to degrees Fahrenheit, °F, using \(F=1.8C+32\). Find the mean and the standard deviation of the temperatures in °F.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(20.5\) °C
A1
(b)(i)
\(1.66\) °C
A1
\(1.66433\ldots\) (\(\sigma_{x}\)).
(b)(ii)
\(2.77\) °C2
A1
Accept 2.76 from squaring 1.66. FT from (b)(i).
(c)
\(Q_{1}=19.2\) and \(Q_{3}=21.7\)
(A1)
GDC values.
IQR \(=2.5\) °C
A1
Other methods of finding quartiles can give different values (e.g. 19.3 and 21.525, giving 2.225); accept a correct IQR from a clearly stated alternative method.
(d)
mean \(=1.8(20.5)+32\)
(M1)
\(68.9\) °F
A1
standard deviation \(=1.8\times1.66433\ldots=3.00\) °F
A1
Award A0 for adding 32 to the standard deviation. Accept 2.99 or 3.
Answers: (a) 20.5 °C · (b)(i) 1.66 °C, (ii) 2.77 · (c) 2.5 °C · (d) 68.9 °F, 3.00 °F
56AASL4-q00244.3 Measures of central tendency and dispersionPaper 2Easy9 marks
The table shows the number of goals, \(g\), scored in each of the 46 matches played in a football league in one month.
Goals, \(g\)
0
1
2
3
4
5
6
Number of matches
8
14
11
7
3
2
1
(a)
Write down the mode.
(1)
(b)
Find the median.
(1)
(c)
Find the mean number of goals per match.
(2)
(d)
Find the interquartile range.
(2)
(e)
Find the standard deviation.
(1)
(f)
Determine whether the match with 6 goals is an outlier.
Find the equation of the regression line of \(y\) on \(x\), in the form \(y=ax+b\).
(2)
(c)(i)
In the context of the question, interpret
the value of \(a\);
(1)
(c)(ii)
the value of \(b\).
(1)
(d)
Use the regression line to estimate the volume of water in the reservoir halfway through week 6 \((x=5.5)\).
(2)
(e)
Use the regression line to find the volume it predicts at the end of week 15, and comment on your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.997\)
A1
\(-0.997281\ldots\)
(b)
\(a=-2.05\)
A1
\(-2.05238\ldots\)
\(b=26.0\)
A1
\(26.0357\ldots\) Award A1A0 if not written as an equation.
(c)(i)
the volume of water decreases by about 2.05 million m3 each week
R1
Must refer to a decrease in volume per week.
(c)(ii)
the estimated volume at the start of the summer (week 0) is about 26.0 million m3
R1
(d)
\(-2.05238(5.5)+26.0357\)
(M1)
\(14.7\) million m3
A1
\(14.7476\ldots\)
(e)
\(-4.75\) million m3
A1
Accept \(-4.74\) to \(-4.76\).
a negative volume is impossible; week 15 is far outside the data range (weeks 1 to 8), so the prediction is an unreliable extrapolation (e.g. rain may fall or the reservoir empties)
59AASL4-q00444.5 Probability and expected occurrencesPaper 2Easy7 marks
A drawing pin is dropped onto a table and lands either point up or point down. Selin drops the pin repeatedly and records the total number of times it has landed point up.
Total number of drops
50
100
200
500
1000
Total number landing point up
21
38
74
182
371
(a)
Find the relative frequency of the pin landing point up after 100 drops.
(1)
(b)
Write down the best estimate of the probability that the pin lands point up. Give a reason for your choice.
(2)
(c)
The pin is dropped another 1200 times. Use your answer to part (b) to find the expected number of times the pin lands point up.
(2)
(d)
Two of these pins are dropped together, and they land independently. Estimate the probability that both land point down.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{38}{100}=0.38\)
A1
(b)
\(\dfrac{371}{1000}=0.371\)
A1
it is based on the greatest number of trials (relative frequencies settle down as the number of trials increases)
R1
Award R1 only with 0.371.
(c)
\(1200\times0.371\)
(M1)
FT from (b).
\(445\)
A1
\(445.2\) Accept 445.2.
(d)
\(\mathrm{P}(\text{point down})=1-0.371=0.629\), and \(0.629^{2}\)
61AASL4-q00704.8 The binomial distributionPaper 2Easy6 marks
The probability that a seed of a certain type fails to germinate is 0.12, independently of other seeds. A packet contains 20 of these seeds. Let \(X\) be the number of seeds in the packet that fail to germinate.
(a)
Find \(\mathrm{P}(X=3)\).
(2)
(b)
Find the probability that at most two seeds fail to germinate.
(2)
(c)
Find the mean and the variance of \(X\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(20,\,0.12)\) and attempt to find \(\mathrm{P}(X=3)\)
63AASL4-q00834.9 The normal distributionPaper 2Easy7 marks
The lengths, \(L\) cm, of fish in a lake are normally distributed with mean 18.0 cm and standard deviation 1.5 cm. The shaded region on the diagram represents \(\mathrm{P}(17\lt L\lt 19)\).
Graph drawn to scale
(a)
Find \(\mathrm{P}(L\gt 20)\).
(2)
(b)
Find \(\mathrm{P}(17\lt L\lt 19)\).
(2)
(c)
A sample of 400 fish is caught. Find the expected number of these fish that are longer than 20 cm.
(2)
(d)
Write down the approximate percentage of fish with lengths between 15.0 cm and 21.0 cm.
(1)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf with lower 20, upper large, \(\mu=18\), \(\sigma=1.5\)
(M1)
\(0.0912\)
A1
\(0.0912112\ldots\)
(b)
normalcdf(17, 19, 18, 1.5)
(M1)
\(0.495\)
A1
\(0.495014\ldots\)
(c)
\(400\times0.0912112\ldots\)
(M1)
\(36.5\)
A1
\(36.484\ldots\) An expected number need not be a whole number: award A0 for 36 or 37.
64AASL4-q00854.9 The normal distributionPaper 2Easy6 marks
The time, \(T\) minutes, that adults take to complete a particular logic puzzle is normally distributed with mean 42 minutes and standard deviation 3.2 minutes.
(a)
Find the probability that a randomly chosen adult takes more than 45 minutes.
(2)
(b)
Hence, or otherwise, find \(\mathrm{P}(39\lt T\lt 45)\).
(2)
(c)
Find the time \(t\) such that 95% of adults complete the puzzle in less than \(t\) minutes.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(T\gt 45)\) with \(\mu=42,\ \sigma=3.2\), e.g. normalcdf(45, large, 42, 3.2)
(M1)
\(0.174\)
A1
\(0.174250\ldots\)
(b)
\(1-2\times0.174250\ldots\) (symmetry about 42), or normalcdf(39, 45, 42, 3.2)
(M1)
\(0.651\)
A1
\(0.651498\ldots\) FT from (a) for the symmetry method.
(c)
\(\mathrm{P}(T\lt t)=0.95\), inverse normal
(M1)
\(t=47.3\) minutes
A1
\(47.2635\ldots\)
Answers: (a) 0.174 · (b) 0.651 · (c) 47.3 min
65AASL4-q00874.9 The normal distributionPaper 2Easy8 marks
The lengths, \(L\) cm, of leaves on a type of shrub are modelled by a normal distribution with mean 16.5 cm and standard deviation 2.2 cm. A biologist collects 200 leaves and uses the model to find the expected number of leaves in each length class.
Length, \(L\) cm
\(L\lt 13\)
\(13\le L\lt 15\)
\(15\le L\lt 17\)
\(17\le L\lt 19\)
\(19\le L\lt 21\)
\(L\ge 21\)
Expected frequency
\(a\)
38.4
\(b\)
56.4
21.5
\(c\)
(a)
Find \(\mathrm{P}(L\lt 13)\).
(2)
(b)
Hence find the value of \(a\).
(1)
(c)
Find the value of \(b\) and the value of \(c\).
(3)
(d)
Find the length exceeded by 5% of the leaves, according to the model.
(2)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(−large, 13, 16.5, 2.2)
(M1)
\(0.0558\)
A1
\(0.0558150\ldots\)
(b)
\(a=200\times0.0558150\ldots=11.2\)
A1
\(11.1630\ldots\) FT from (a).
(c)
\(200\times\mathrm{P}(15\le L\lt 17)\) or \(200\times\mathrm{P}(L\ge 21)\)
(M1)
\(b=68.4\)
A1
\(68.4434\ldots\)
\(c=4.08\)
A1
\(4.08100\ldots\) Accept 4.1. Award A1 also for \(c=200-(11.2+38.4+68.4+56.4+21.5)=4.1\).
66AASL4-q00984.10 Regression of x on yPaper 2Easy6 marks
A forensic scientist records the height, \(x\) cm, and the shoe size, \(y\) (European sizes), of ten adults.
\(x\) (cm)
158
163
167
170
172
175
178
181
184
190
\(y\)
42.1
43.9
44.6
45.8
45.9
47.2
47.6
48.9
49.1
51.0
(a)
Write down the value of Pearson’s product-moment correlation coefficient, \(r\).
(1)
(b)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(c)
A shoe print of size 46.5 is found. Estimate the height of the person who made it.
(2)
(d)
Explain why the regression line of \(x\) on \(y\), rather than \(y\) on \(x\), is used in part (c).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.996\)
A1
\(0.996104\ldots\)
(b)
gradient \(3.64\)
A1
\(3.64120\ldots\)
\(x=3.64y+4.08\)
A1
\(4.08329\ldots\)
(c)
\(x=3.64120(46.5)+4.08329\)
(M1)
\(173\) cm
A1
\(173.399\ldots\) Accept 173.4.
(d)
the shoe size \(y\) is known and the height \(x\) is to be estimated, so the line of \(x\) on \(y\) is required
R1
Answers: (a) 0.996 · (b) \(x=3.64y+4.08\) · (c) 173 cm
67AASL4-q00044.1 Sampling, data and outliersPaper 2Medium8 marks
The masses, in grams, of 15 apples picked from one tree were recorded.
Apple
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Mass (g)
238
251
212
263
244
397
231
256
270
238
247
284
225
263
259
(a)
Write down the median and the lower and upper quartiles of the masses.
(2)
(b)
Show that the mass of 397 g is an outlier.
(2)
(c)
It is discovered that the value 397 was a recording error and the mass of that apple was 297 g. Using the corrected data, find the mean and the standard deviation of the masses.
(2)
(d)
Explain why correcting the error does not change the median or the quartiles.
(1)
(e)
Determine whether 297 g is an outlier.
(1)
Show mark scheme
Working
Mark
Notes
(a)
median \(=251\) g
A1
\(Q_{1}=238\) g and \(Q_{3}=263\) g
A1
Both required.
(b)
\(Q_{3}+1.5\times\text{IQR}=263+1.5\times25\)
M1
Award for 1.5 × their IQR added to their upper quartile.
\(=300.5\) and \(397\gt 300.5\)
A1
Must see 300.5 and a comparison.
(c)
mean \(=251.87\) g, i.e. \(252\) g
A1
\(251.866\ldots\)
standard deviation \(=21.5\) g
A1
\(21.4968\ldots\) (GDC \(\sigma_{x}\)). Accept \(22.3\) (\(s_{x}\)) only if the data are explicitly treated as a sample.
(d)
397 and 297 are both the largest value, so the order (positions) of all the data values is unchanged
R1
(e)
\(297\lt 300.5\), so 297 g is not an outlier
A1
FT from their upper boundary in (b).
Answers: (a) median 251 g, \(Q_{1}=238\) g, \(Q_{3}=263\) g · (c) mean 252 g, sd 21.5 g · (e) not an outlier
68AASL4-q00054.1 Sampling, data and outliersPaper 2Medium7 marks
The table shows the members of a tennis club, classified by age group and sex.
Age group
Female
Male
Under 18
46
39
18 to 59
118
97
60 and over
63
52
The club secretary wants a stratified sample of 50 members, using all six age-group and sex categories as strata.
(a)
Find the number of female members aged 18 to 59 who should be in the sample.
(2)
(b)
Show that, if the number for each of the six strata is rounded to the nearest integer, the total sample size is not 50.
(2)
(c)
Suggest how the secretary could adjust the numbers so that the sample size is exactly 50.
(1)
(d)
Once the number from each stratum is known, state how the members from that stratum should be chosen.
(1)
(e)
State one advantage of a stratified sample compared with a simple random sample of 50 members.
(1)
Show mark scheme
Working
Mark
Notes
(a)
total \(=415\); \(\dfrac{118}{415}\times50\)
(M1)
\(14.2\ldots\), so \(14\) members
A1
\(14.2168\ldots\) Award A1 for 14 only.
(b)
\(5.54,\ 4.70,\ 14.22,\ 11.69,\ 7.59,\ 6.27\) (at least four correct)
M1
Female under 18, male under 18, female 18–59, male 18–59, female 60+, male 60+.
rounded values \(6+5+14+12+8+6=51\neq50\)
A1
Must see the sum 51.
(c)
reduce by one the stratum whose rounding up was greatest relative to its value, e.g. female under 18: \(5.54\to5\) (so \(5,5,14,12,8,6\))
R1
Accept any sensible adjustment that removes one member and is justified, e.g. reduce the value with the smallest decimal part among those rounded up.
(d)
by simple random sampling within the stratum (e.g. a random number generator applied to a numbered list of that stratum)
A1
(e)
each age group and sex is represented in proportion to its size in the club, so the sample is more representative
R1
Answers: (a) 14 · (b) total 51 · (c) e.g. take 5 females under 18
69AASL4-q00064.1 Sampling, data and outliersPaper 2Medium8 marks
Eleven students recorded the number of hours they slept on one night. The data were entered into a spreadsheet as follows.
Student
1
2
3
4
5
6
7
8
9
10
11
Hours
7.5
8
6.5
9
75
7
8.5
6
7
8
5.5
(a)
Find the mean of the values in the spreadsheet.
(1)
(b)
Explain why the value 75 must be an error.
(1)
(c)(i)
The student who recorded 75 confirms that the correct value is 7.5.
Find the mean of the corrected data.
(1)
(c)(ii)
Find the median of the corrected data.
(1)
(d)
Instead of correcting the value, a second analyst removes it. Find the mean and the median of the remaining ten values.
(2)
(e)
State which of the mean and the median of the data in the spreadsheet was less affected by the error, giving a reason.
(1)
(f)
A twelfth student did not answer the question. Explain why this student’s value should not be entered as 0.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(13.5\) hours
A1
\(13.4545\ldots\)
(b)
75 hours is more than the 24 hours in a day (it is likely 7.5 with the decimal point missing)
R1
(c)(i)
\(7.32\) hours
A1
\(7.31818\ldots\)
(c)(ii)
\(7.5\) hours
A1
(d)
mean \(=7.3\) hours
A1
median \(=7.25\) hours
A1
(e)
the median: with the error the median is still 7.5, the same as for the corrected data, whereas the mean almost doubles (13.5 compared with 7.32)
R1
Must give the median with a reason referring to the values or to the median not depending on the size of extreme values.
(f)
0 would be treated as a real value (a student who did not sleep) and would lower the mean; the missing value should be left out of the calculations
R1
Answers: (a) 13.5 · (c)(i) 7.32 (ii) 7.5 · (d) mean 7.3, median 7.25 · (e) median
70AASL4-q00084.1 Sampling, data and outliersPaper 2Medium9 marks
The resting heart rates, in beats per minute (bpm), of 18 members of a cycling team are shown.
bpm
64
71
58
67
73
62
69
75
41
bpm
66
70
63
68
77
60
72
65
79
(a)
Find the median and the interquartile range of the heart rates.
(2)
(b)
Show that the heart rate of 41 bpm is an outlier.
(2)
(c)
The heart rate of 41 bpm belongs to an Olympic cyclist. Very fit athletes often have resting heart rates below 45 bpm. State, with a reason, whether this value should be removed from the data.
(1)
(d)
Find the mean and the standard deviation of all 18 heart rates.
(2)
(e)
Find the standard deviation of the heart rates when the value 41 is removed.
(1)
(f)
Comment on the effect of the value 41 on the standard deviation and on the interquartile range.
(1)
Show mark scheme
Working
Mark
Notes
(a)
median \(=67.5\) bpm
A1
\(\text{IQR}=72-63=9\) bpm
A1
\(Q_{1}=63,\ Q_{3}=72\)
(b)
\(Q_{1}-1.5\times\text{IQR}=63-1.5\times9\)
M1
\(=49.5\) and \(41\lt 49.5\)
A1
Must see 49.5 and a comparison.
(c)
it should not be removed: it is a genuine (valid) value for a member of the team, not an error
R1
Award R0 for “remove it because it is an outlier”.
(d)
mean \(=66.7\) bpm
A1
\(66.666\ldots\)
\(\sigma=8.36\) bpm
A1
\(8.35995\ldots\)
(e)
\(5.74\) bpm
A1
\(5.74185\ldots\)
(f)
the standard deviation is much larger with 41 included (8.36 compared with 5.74), whereas the IQR is unchanged (9 in both cases), because quartiles do not depend on the extreme values
R1
Accept IQR \(72.5-63.5=9\) without 41. Must compare both measures.
Answers: (a) 67.5, 9 · (d) 66.7, 8.36 · (e) 5.74
71AASL4-q00154.2 Presentation of dataPaper 2Medium9 marks
The lengths, \(l\) cm, of 200 leaves from a tree were measured. The cumulative frequency graph shows the results. The shortest leaf is 5 cm long and the longest is 16 cm long.
Graph drawn to scale
(a)
Use the graph to write down the median length.
(1)
(b)
Use the graph to find the interquartile range.
(2)
(c)
Find the 15th percentile of the lengths.
(2)
(d)
Estimate the number of leaves that are longer than 13 cm.
(2)
(e)
Determine whether any of the leaves have lengths that are outliers.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(10.5\) cm
A1
Accept \(10.4\)–\(10.6\).
(b)
\(Q_{1}=9\) and \(Q_{3}=12\) (reading at 50 and 150)
(M1)
Accept \(8.9\)–\(9.1\) and \(11.9\)–\(12.1\).
\(\text{IQR}=3\) cm
A1
(c)
reading at cumulative frequency \(0.15\times200=30\)
(M1)
\(8\) cm
A1
Accept \(7.9\)–\(8.1\).
(d)
cumulative frequency at 13 cm is about 171
(M1)
Accept \(169\)–\(174\).
\(200-171=29\) leaves
A1
Accept \(26\)–\(31\).
(e)
boundaries \(9-1.5\times3=4.5\) and \(12+1.5\times3=16.5\)
M1
FT from their quartiles.
all lengths are between 5 cm and 16 cm, which lie inside \(4.5\) to \(16.5\), so there are no outliers
A1
Answers: (a) 10.5 cm · (b) 3 cm · (c) 8 cm · (d) about 29 leaves · (e) no outliers
72AASL4-q00164.2 Presentation of dataPaper 2Medium9 marks
The amounts, \(x\) euros (€), spent by 200 customers in a supermarket on one day are summarised below.
Amount, \(x\) (€)
\(0\le x\lt 10\)
\(10\le x\lt 20\)
\(20\le x\lt 30\)
\(30\le x\lt 40\)
\(40\le x\lt 50\)
\(50\le x\lt 60\)
\(60\le x\lt 70\)
Frequency
8
30
52
46
34
20
10
The cumulative frequency table is
Amount, \(x\) (€)
\(x\lt 10\)
\(x\lt 20\)
\(x\lt 30\)
\(x\lt 40\)
\(x\lt 50\)
\(x\lt 60\)
\(x\lt 70\)
Cumulative frequency
8
38
90
\(a\)
170
\(b\)
200
(a)
Write down the value of \(a\) and the value of \(b\).
(2)
(b)
Write down the class that contains the median amount.
(1)
(c)
Find the percentage of the customers who spent at least €50.
(1)
(d)
Find an estimate of the mean and an estimate of the standard deviation of the amounts spent.
(3)
(e)
The 25% of customers who spent the most are sent a discount voucher. Determine the class that contains the least amount spent by a customer who is sent a voucher.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(a=136\)
A1
\(b=190\)
A1
(b)
\(30\le x\lt 40\)
A1
100th/100.5th value; cumulative frequency passes 100 in this class.
(c)
\(\dfrac{20+10}{200}=15\%\)
A1
(d)
use of mid-interval values \(5,15,\ldots,65\) with the frequencies
(M1)
mean \(=\)€33.4
A1
\(\sigma=\)€14.9
A1
\(14.8808\ldots\)
(e)
\(25\%\) of \(200=50\) customers, so the boundary is at the 150th/151st amount (cumulative frequency \(150\)), i.e. the upper quartile
(M1)
\(40\le x\lt 50\) (cumulative frequency \(136\) for \(x\lt 40\) and \(170\) for \(x\lt 50\))
73AASL4-q00274.3 Measures of central tendency and dispersionPaper 2Medium7 marks
The number of people, \(n\), living in each of 85 households in a village is shown in the table.
Number of people, \(n\)
1
2
3
4
5
Number of households
23
31
18
9
4
(a)
Find the mean of \(n\).
(1)
(b)
Find the standard deviation and the variance of \(n\).
(2)
(c)
Find the number of households for which \(n\) is within one standard deviation of the mean.
(2)
(d)
Find the median of \(n\).
(1)
(e)
With reference to the shape of the distribution, explain why the mean is greater than the median.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(2.29\)
A1
\(\tfrac{195}{85}=2.29411\ldots\)
(b)
\(\sigma=1.11\)
A1
\(1.11485\ldots\)
variance \(=1.24\)
A1
\(1.24290\ldots\)
(c)
\(2.294-1.115=1.18\) to \(2.294+1.115=3.41\), so \(n=2\) or \(n=3\)
(M1)
\(31+18=49\)
A1
(d)
\(2\)
A1
43rd value; cumulative frequencies 23, 54.
(e)
the distribution is positively skewed (most households have 1 or 2 people, with a long tail up to 5), so the few large values pull the mean above the median
74AASL4-q00324.4 Correlation and regression of y on xPaper 2Medium8 marks
The table shows the altitude, \(x\) metres, of ten weather stations in a mountain region and the mean July temperature, \(y\) °C, recorded at each station.
75AASL4-q00354.4 Correlation and regression of y on xPaper 2Medium10 marks
Nine teenagers recorded their average daily screen time, \(x\) hours, and their average night’s sleep, \(y\) hours, over a month.
\(x\) (hours)
1.5
2
2.5
3
3.5
4.5
5
6
6.5
\(y\) (hours)
8.2
7.6
8
7.1
7.4
6.8
6.2
6.5
5.9
(a)
Find the value of \(r\) and describe the correlation between \(x\) and \(y\).
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Use your equation to estimate the average sleep of a teenager whose average daily screen time is 4 hours.
(2)
(d)
A teenager sleeps an average of 7.5 hours per night. Explain why your equation should not be used to estimate this teenager’s screen time.
(1)
(e)
The sleep times are converted to minutes, \(Y=60y\). Write down the equation of the regression line of \(Y\) on \(x\).
(2)
(f)
A newspaper states that “screen time causes teenagers to lose sleep”. Comment on whether the data support this claim.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.937\)
A1
\(-0.936939\ldots\)
strong negative (linear) correlation
A1
(b)
\(a=-0.421\)
A1
\(-0.421333\ldots\)
\(y=-0.421x+8.69\)
A1
\(8.69289\ldots\)
(c)
\(-0.421333(4)+8.69289\)
(M1)
\(7.01\) hours
A1
\(7.00756\ldots\) Accept 7.006 or 7.0 from 3 s.f. coefficients.
(d)
the line of \(y\) on \(x\) is for predicting \(y\) from \(x\); it cannot reliably be used to predict \(x\) from a value of \(y\)
R1
(e)
gradient \(-25.3\)
A1
\(60\times(-0.421333)=-25.28\)
\(Y=-25.3x+522\)
A1
\(521.573\ldots\) FT from (b).
(f)
no: the strong correlation shows an association, but correlation does not imply causation (other factors, such as stress or schoolwork, could affect both)
77AASL4-q00384.4 Correlation and regression of y on xPaper 2Medium8 marks
For nine towns, the table shows the number of public libraries, \(L\), the number of car thefts reported in one year, \(C\), and the population, \(P\) thousand.
Town
A
B
C
D
E
F
G
H
I
\(L\)
3
5
4
8
6
11
9
14
12
\(C\)
410
620
540
1080
700
1390
1210
1720
1450
\(P\)
48
71
60
118
85
152
131
190
163
(a)
Find Pearson’s product-moment correlation coefficient between \(L\) and \(C\).
(1)
(b)
A politician claims that closing libraries would reduce car thefts. Comment on this claim.
(2)
(c)
Find Pearson’s product-moment correlation coefficient between \(P\) and \(C\).
(1)
(d)
Find the equation of the regression line of \(C\) on \(P\).
(2)
(e)
Hence estimate the number of car thefts in a town with a population of 100 thousand.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.994\)
A1
\(0.994420\ldots\)
(b)
correlation does not imply causation
R1
both \(L\) and \(C\) increase with the size of the population: larger towns have more libraries and more cars to steal
R1
Must identify population (town size) as a third variable.
78AASL4-q00454.5 Probability and expected occurrencesPaper 2Medium7 marks
A town has 8000 residents. The table shows the number of residents in each age group and, from previous years, the probability that a resident in that group catches influenza during a winter.
Age group
Number of residents
Probability of catching influenza
0–17
1 200
0.084
18–64
5 300
0.031
65 and over
1 500
0.047
(a)
Find the expected number of residents aged 18–64 who catch influenza during the winter.
(2)
(b)
Find the expected total number of residents who catch influenza during the winter.
(2)
(c)
Hence find the probability that a resident chosen at random from the town catches influenza during the winter.
(1)
(d)
During one winter, 362 residents caught influenza. Find the relative frequency of catching influenza that winter and comment on your answer, with reference to part (c).
this is higher than the expected probability of 0.0420, so more residents than expected caught influenza that winter (the past probabilities may underestimate that winter’s risk)
R1
Accept any valid comment that compares the two values.
79AASL4-q00554.6 Combined events and diagramsPaper 2Medium8 marks
A system contains three components, \(A\), \(B\) and \(C\), which work independently of each other. The system works if \(A\) works and at least one of \(B\) and \(C\) works. The diagram shows the probability that each component works.
Probability that each component works is shown above it
(a)
Find the probability that both \(B\) and \(C\) fail.
(1)
(b)
Find the probability that the system works.
(2)
(c)
Component \(C\) is replaced by a new component, which works with probability \(q\). The system now works with probability 0.92. Find the value of \(q\).
(3)
(d)
Three of these systems operate independently. Find the probability that at least one of them does not work.
81AASL4-q00754.8 The binomial distributionPaper 2Medium8 marks
The probability that a randomly chosen visitor to a website clicks on an advertisement is 0.08, independently of other visitors.
(a)
Find the probability that at least one of 20 visitors clicks on the advertisement.
(2)
(b)
Find the least number of visitors needed for the probability that at least one of them clicks on the advertisement to be at least 0.99.
(3)
(c)
Find the least number of visitors needed for the probability that at least two of them click on the advertisement to be greater than 0.9.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(1-0.92^{20}\)
(M1)
\(0.811\)
A1
\(0.811306\ldots\)
(b)
\(1-0.92^{n}\ge 0.99\), or \(0.92^{n}\le 0.01\)
(M1)
\(n\ge 55.2\), or \(n=55:\ 0.98981\), \(n=56:\ 0.99062\)
(A1)
\(56\)
A1
(c)
\(1-\mathrm{P}(X\le 1)\gt 0.9\) for \(X\sim\mathrm{B}(n,\,0.08)\), using a table of values
(M1)
\(n=47:\ 0.8990\), \(n=48:\ 0.9055\)
(A1)
\(48\)
A1
Answers: (a) 0.811 · (b) 56 · (c) 48
82AASL4-q00764.8 The binomial distributionPaper 2Medium8 marks
At a supermarket, 36% of customers use a loyalty card, independently of each other. A random sample of 25 customers is taken. Let \(X\) be the number of customers in the sample who use a loyalty card.
(a)
Find the mean and the standard deviation of \(X\).
(3)
(b)
Find the probability that \(X\) is equal to its mean.
(2)
(c)
Find the probability that \(X\) lies within one standard deviation of its mean.
Answers: (a) mean 9, standard deviation 2.4 · (b) 0.164 · (c) 0.703
83AASL4-q00844.9 The normal distributionPaper 2Medium7 marks
The battery life, \(H\) hours, of a model of mobile phone is normally distributed with mean 14.2 hours and standard deviation 2.6 hours.
(a)
Find the probability that a randomly chosen phone has a battery life of less than 10 hours.
(2)
(b)
Find the lower quartile and the upper quartile of \(H\).
(2)
(c)
A battery life is classed as an outlier if it is more than \(1.5\times\) IQR from the nearest quartile. Find the probability that a randomly chosen phone has a battery life that is an outlier.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(H\lt 10)\) with \(\mu=14.2,\ \sigma=2.6\)
(M1)
\(0.0531\)
A1
\(0.0531137\ldots\)
(b)
\(Q_{1}=12.4\) hours
A1
\(12.4463\ldots\), from invNorm(0.25, 14.2, 2.6)
\(Q_{3}=16.0\) hours
A1
\(15.9537\ldots\)
(c)
boundaries \(Q_{1}-1.5\,\text{IQR}\) and \(Q_{3}+1.5\,\text{IQR}\), with IQR \(=3.5073\ldots\)
84AASL4-q00864.9 The normal distributionPaper 2Medium7 marks
The times, \(R\) minutes, taken by runners to complete a 10 km charity run are normally distributed with mean 52 minutes and standard deviation 6.5 minutes.
(a)
Find the probability that a randomly chosen runner finishes in less than 45 minutes.
(2)
(b)
Find the value of \(k\) such that \(\mathrm{P}(52-k\lt R\lt 52+k)=0.9\).
(3)
(c)
The slowest 10% of runners are given a free training plan. Find the shortest time for which a runner receives the plan.
(2)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(−large, 45, 52, 6.5)
(M1)
\(0.141\)
A1
\(0.140757\ldots\)
(b)
by symmetry \(\mathrm{P}(R\lt 52+k)=0.95\) (or \(\mathrm{P}(R\lt 52-k)=0.05\))
\(\mathrm{P}(R\gt t)=0.1\), i.e. invNorm(0.9, 52, 6.5)
(M1)
\(60.3\) minutes
A1
\(60.3300\ldots\) Award (M1)A0 for 43.7 (from invNorm(0.1, …)).
Answers: (a) 0.141 · (b) \(k=10.7\) · (c) 60.3 min
85AASL4-q00884.9 The normal distributionPaper 2Medium9 marks
The masses, \(M\) grams, of apricots picked on a farm are normally distributed with mean 210 g and standard deviation 25 g. Apricots are graded and sold as follows.
Grade
Small
Medium
Large
Mass (g)
\(M\lt 195\)
\(195\le M\le 230\)
\(M\gt 230\)
Price per apricot (€)
0.25
0.35
0.50
(a)
Find the probability that a randomly chosen apricot is graded medium.
(2)
(b)
Find the probability that a randomly chosen apricot is graded small and the probability that it is graded large.
(2)
(c)
Find the expected income from 1000 apricots. Give your answer to the nearest euro.
(3)
(d)
The heaviest 15% of apricots are also labelled “premium”. Find the minimum mass of a premium apricot.
86AASL4-q00894.9 The normal distributionPaper 2Medium8 marks
The delivery times, \(D\) minutes, of a pizza company are normally distributed with mean 26 minutes and standard deviation 4 minutes. A pizza is free if its delivery takes more than 32 minutes. Assume that delivery times are independent.
(a)
Find the probability that a delivery is free.
(2)
(b)
On one evening there are 20 deliveries. Find the probability that at least two of them are free.
(2)
(c)
Find the probability that the first free delivery of the evening is the fifth delivery.
(2)
(d)
The company wants only 2% of deliveries to be free, without changing the distribution of delivery times. Find the new time limit, in minutes.
87AASL4-q00904.9 The normal distributionPaper 2Medium7 marks
The lifetimes of two brands of LED lamp are normally distributed. Brand A lamps have mean 1200 hours and standard deviation 150 hours. Brand B lamps have mean 1300 hours and standard deviation 250 hours.
(a)
For each brand, find the probability that a lamp lasts more than 1500 hours.
(2)
(b)
For each brand, find the probability that a lamp lasts more than 1000 hours.
(2)
(c)
Show that the probability that a lamp lasts more than \(t\) hours is the same for both brands when \(t=1050\).
(2)
(d)
A hospital needs lamps that are as likely as possible to last at least 1000 hours. State, with a reason, which brand it should buy.
(1)
Show mark scheme
Working
Mark
Notes
(a)
A: \(0.0228\)
A1
\(0.0227501\ldots\)
B: \(0.212\)
A1
\(0.211855\ldots\)
(b)
A: \(0.909\)
A1
\(0.908789\ldots\)
B: \(0.885\)
A1
\(0.884930\ldots\)
(c)
equal probabilities when the standardized values are equal: \(\dfrac{t-1200}{150}=\dfrac{t-1300}{250}\)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Use your equation to estimate the height of the son of a father who is 172 cm tall.
(1)
(d)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(e)
A son is 165 cm tall. Estimate the height of his father.
(2)
(f)
Explain why you did not use the equation from part (b) in part (e).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.982\)
A1
\(0.98182\ldots\)
(b)
gradient \(0.937\)
A1
\(0.93690\ldots\)
\(y=0.937x+11.0\)
A1
\(11.0011\ldots\)
(c)
\(172\) cm
A1
\(172.148\ldots\) Accept 172.1 or 172.2.
(d)
gradient \(1.03\)
A1
\(1.02890\ldots\)
\(x=1.03y-5.21\)
A1
\(-5.21099\ldots\) Award A0A0 for rearranging the equation of \(y\) on \(x\).
(e)
\(x=1.02890(165)-5.21099\)
(M1)
FT from their line of x on y only.
\(165\) cm
A1
\(164.558\ldots\) Accept answers from 164.5 to 164.8 (from rounded coefficients). Award M0A0 for \(164.4\), from rearranging the line of \(y\) on \(x\).
(f)
the line of \(y\) on \(x\) should only be used to predict \(y\) from \(x\); to predict \(x\) from \(y\) the regression line of \(x\) on \(y\) must be used
R1
Answers: (a) 0.982 · (b) \(y=0.937x+11.0\) · (c) 172 cm · (d) \(x=1.03y-5.21\) · (e) 165 cm (164.6)
89AASL4-q00954.10 Regression of x on yPaper 2Medium9 marks
On a long journey, a driver records the distance, \(x\) km, travelled since the fuel tank was filled, and the volume of fuel, \(y\) litres, remaining in the tank.
Distance, \(x\) (km)
50
120
180
240
310
370
450
520
Fuel, \(y\) (litres)
50.4
47.5
40.6
38.3
30.9
28.8
20.5
17.6
(a)
Find Pearson’s product-moment correlation coefficient, \(r\), and interpret its value in context.
(2)
(b)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(c)
On a later journey, the trip counter fails but the fuel gauge shows 34 litres remaining. Use your answer to part (b) to estimate the distance travelled since the tank was filled.
(2)
(d)
Find the equation of the regression line of \(y\) on \(x\), and use it to estimate the fuel remaining after 400 km.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.995\)
A1
\(-0.994907\ldots\)
very strong negative linear correlation: the further the car has travelled, the less fuel remains
R1
(b)
gradient \(-13.6\)
A1
\(-13.5522\ldots\)
\(x=-13.6y+745\)
A1
\(745.181\ldots\)
(c)
\(x=-13.5522(34)+745.181\)
(M1)
Award M0 for rearranging the line of \(y\) on \(x\).
\(284\) km
A1
\(284.404\ldots\) Accept 282 to 285 (from rounded coefficients).
90AASL4-q00974.10 Regression of x on yPaper 2Medium8 marks
The scatter diagram and table show the floor area, \(x\) m2, and the monthly energy bill, \(y\) euros, for ten apartments in one building.
Graph drawn to scale
\(x\) (m2)
62
75
80
88
95
104
112
120
135
148
\(y\) (€)
48
55
61
63
70
74
81
84
140
101
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), for all ten apartments.
(1)
(b)
One bill was recorded incorrectly. Identify this apartment, giving a reason.
(1)
(c)
The correct bill for this apartment was €94. Using the corrected data for all ten apartments, find the new value of \(r\) and the equation of the regression line of \(x\) on \(y\).
(3)
(d)
Another apartment in the building has a monthly bill of €90. Use your equation from part (c) to estimate its floor area.
(2)
(e)
Comment on whether your equation would give a reliable estimate of the floor area of an apartment with a bill of €30.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.868\)
A1
\(0.867525\ldots\)
(b)
the apartment with \(x=135,\ y=140\); its bill is far above the pattern of the other nine points
R1
Must identify the point and give a reason.
(c)
\(r=0.998\)
A1
\(0.997817\ldots\)
gradient \(1.60\)
A1
\(1.59673\ldots\)
\(x=1.60y-14.8\)
A1
\(-14.8210\ldots\)
(d)
\(x=1.59673(90)-14.8210\)
(M1)
Must use their line of x on y.
\(129\) m2
A1
\(128.884\ldots\) Accept 129.2 from rounded coefficients.
(e)
not reliable: €30 is outside the range of the bills in the data (€48 to €101), so this would be extrapolation
R1
Answers: (a) 0.868 · (b) area 135 m2, bill €140 · (c) \(r=0.998\), \(x=1.60y-14.8\) · (d) 129 m2
91AASL4-q00994.10 Regression of x on yPaper 2Medium7 marks
On ten days, an engineer records the mean wind speed, \(x\) m s−1, at a small wind turbine and its mean power output, \(y\) kW.
\(x\) (m s−1)
3.2
4.0
4.5
5.1
5.8
6.3
7.0
7.6
8.4
9.1
\(y\) (kW)
21
30
33
42
47
55
61
66
77
82
(a)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(b)
Interpret the gradient of this line in context.
(1)
(c)
On one day the anemometer failed, but the mean power output was 50 kW. Estimate the mean wind speed on that day.
(2)
(d)
The wind speed \(K\) in km h−1 is given by \(K=3.6x\). Find the equation of the regression line of \(K\) on \(y\), giving your answer in the form \(K=ay+b\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(0.0950\)
A1
\(0.0950139\ldots\)
\(x=0.0950y+1.22\)
A1
\(1.21628\ldots\)
(b)
each additional kW of power output is associated with an increase of about \(0.095\) m s−1 in the mean wind speed
R1
Accept equivalent, e.g. 10 kW more output corresponds to about 0.95 m s⁻¹ more wind speed.
(c)
\(x=0.0950139(50)+1.21628\)
(M1)
Award M0 for rearranging a line of y on x.
\(5.97\) m s−1
A1
\(5.96698\ldots\) Accept 5.96 to 5.97.
(d)
\(K=3.6(0.0950139y+1.21628)\)
(M1)
Or by finding the line of K on y directly with technology.
Answers: (a) 0.998 · (b) \(y=2.61x+7.91\) · (c) 60.1 cm · (d) \(x=0.381y-2.89\) · (e)(i) 0.0912 (ii) 89.2 cm (iii) 31.1 years · (f) 0.273
93AASL4-q01084.11 Conditional probability and independencePaper 2Medium7 marks
A survey asked 200 students in Grades 11 and 12 how they usually travel to school. The results are shown in the table.
Walk (W)
Bus (B)
Car (C)
Total
Grade 11 (G)
30
50
20
100
Grade 12
30
40
30
100
Total
60
90
50
200
A student is chosen at random from the 200 students.
(a)
Find the probability that the student travels by bus.
(1)
(b)
Given that the student travels by car, find the probability that the student is in Grade 12.
(2)
(c)
Show that the events \(W\) and \(G\) are independent.
(2)
(d)
Determine whether the events \(B\) and \(G\) are independent.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{90}{200}=0.45\)
A1
(b)
\(\dfrac{30}{50}\)
(M1)
\(0.6\)
A1
(c)
\(\mathrm{P}(W\mid G)=\dfrac{30}{100}=0.3\) and \(\mathrm{P}(W)=\dfrac{60}{200}=0.3\)
M1
Or \(\mathrm{P}(W\cap G)=0.15=\mathrm{P}(W)\,\mathrm{P}(G)=0.3\times0.5\).
\(\mathrm{P}(W\mid G)=\mathrm{P}(W)\), so \(W\) and \(G\) are independent
R1
(d)
\(\mathrm{P}(B\mid G)=\dfrac{50}{100}=0.5\) and \(\mathrm{P}(B)=0.45\)
M1
Or \(\mathrm{P}(B\cap G)=0.25\) and \(\mathrm{P}(B)\,\mathrm{P}(G)=0.225\).
\(0.5\neq0.45\), so \(B\) and \(G\) are not independent
A1
Answers: (a) 0.45 · (b) 0.6 · (d) not independent
94AASL4-q01134.12 Standardization and inverse normalPaper 2Medium8 marks
In a year group, the marks in a mathematics test are normally distributed with mean 62 and standard deviation 8, and the marks in a physics test are normally distributed with mean 55 and standard deviation 6. Ana scored 74 in mathematics and 65 in physics.
(a)
Find Ana’s standardized score, \(z\), for each test.
(2)
(b)
Hence state, with a reason, in which test Ana performed better relative to the year group.
(1)
(c)
Find the proportion of the year group who scored more than Ana in physics.
(2)
(d)
The marks in a chemistry test are normally distributed with mean 60. Exactly 10% of the year group scored more than 75. Find the standard deviation of the chemistry marks.
(3)
Show mark scheme
Working
Mark
Notes
(a)
mathematics: \(z=\dfrac{74-62}{8}=1.5\)
A1
physics: \(z=\dfrac{65-55}{6}=1.67\)
A1
\(1.666\ldots\)
(b)
physics, because her \(z\)-score is higher (she is more standard deviations above the mean)
R1
FT from (a).
(c)
\(\mathrm{P}(Z\gt 1.666\ldots)\) or normalcdf(65, large, 55, 6)
95AASL4-q01154.12 Standardization and inverse normalPaper 2Medium8 marks
A machine fills boxes with cereal. The mass of cereal in a box is normally distributed with mean \(\mu\) grams and standard deviation 4 grams. The label states 500 g, and only 1% of boxes may contain less than 500 g.
(a)
Find the least value of \(\mu\) that meets this requirement.
(3)
(b)
With \(\mu\) equal to the value found in part (a), find the probability that a box contains more than 515 g.
(2)
(c)
A new machine has mean 506 g. Find the greatest standard deviation for which it meets the 1% requirement.
Answers: (a) 40.6 mm · (b) 0.890 mm · (c) 0.0123 · (d) 2.25
97AASL4-q01194.12 Standardization and inverse normalPaper 2Medium7 marks
At a regional athletics meeting, the 50 m swimming times of competitors are normally distributed with mean 31.5 seconds and standard deviation 2.2 seconds, and the shot-put distances are normally distributed with mean 9.4 m and standard deviation 1.1 m. Kofi swims 50 m in 28.9 seconds and puts the shot 10.8 m.
(a)
Find Kofi’s standardized value, \(z\), for each event.
(2)
(b)
Justify which of Kofi’s two performances is better relative to the other competitors.
(1)
(c)
Find the shot-put distance that represents the same relative standing as Kofi’s swimming time.
(2)
(d)
Find the proportion of swimmers who are faster than Kofi.
(2)
Show mark scheme
Working
Mark
Notes
(a)
swimming: \(z=\dfrac{28.9-31.5}{2.2}=-1.18\)
A1
\(-1.18181\ldots\)
shot put: \(z=\dfrac{10.8-9.4}{1.1}=1.27\)
A1
\(1.27272\ldots\)
(b)
the shot put: a lower time is better in swimming, and his shot put is \(1.27\) standard deviations better than the mean, more than the \(1.18\) for swimming
R1
Must refer to the direction of a good performance in each event. FT from (a).
(c)
\(9.4+1.18181\ldots(1.1)\)
(M1)
Sign must reflect that a fast time corresponds to a long distance.
\(10.7\) m
A1
\(10.7\) exactly.
(d)
\(\mathrm{P}(S\lt 28.9)\) with \(\mu=31.5,\ \sigma=2.2\)
(M1)
\(0.119\)
A1
\(0.118641\ldots\)
Answers: (a) −1.18 and 1.27 · (b) shot put · (c) 10.7 m · (d) 0.119
98AASL4-q00074.1 Sampling, data and outliersPaper 2Hard7 marks
The times, in minutes, that ten employees take to travel to work are
Employee
1
2
3
4
5
6
7
8
9
10
Time (min)
22
14
26
18
29
21
17
25
20
23
An eleventh employee joins the company. Their travel time is \(x\) minutes, where \(x\in\mathbb{Z}^{+}\). Quartiles are found using the method of your GDC.
(a)
Given that \(x\ge 29\), find the lower quartile and the upper quartile of the eleven times.
(2)
(b)
Hence find the least value of \(x\) for which \(x\) is an outlier.
(2)
(c)
Given instead that \(x\le 17\), find all the values of \(x\) for which \(x\) is an outlier.
(2)
(d)
Explain why \(x\) cannot be an outlier if \(17\lt x\lt 29\).
for these values \(17\le Q_{1}\le 18\) and \(25\le Q_{3}\le 26\), so \(\text{IQR}\ge 7\) and the outlier boundaries are at most \(18-10.5=7.5\) and at least \(25+10.5=35.5\); every \(x\) with \(17\lt x\lt 29\) lies between them
R1
Accept a clear argument that \(x\) lies within the interval \(\left[Q_{1}-1.5\,\text{IQR},\ Q_{3}+1.5\,\text{IQR}\right]\).
99AASL4-q00174.2 Presentation of dataPaper 2Hard16 marks
The cumulative frequency graph shows the time, \(t\) minutes, that 160 people take to travel to work.
Graph drawn to scale
(a)
Use the graph to estimate the median time.
(1)
(b)
Use the graph to estimate the lower quartile and the upper quartile, and hence find the interquartile range.
(3)
(c)
Ninety per cent of the people take less than \(k\) minutes to travel to work. Use the graph to estimate \(k\).
(2)
(d)(i)
The information is also shown in the following frequency table.
Time, \(t\) (minutes)
\(0\lt t\le 10\)
\(10\lt t\le 20\)
\(20\lt t\le 30\)
\(30\lt t\le 40\)
\(40\lt t\le 50\)
\(50\lt t\le 60\)
Frequency
12
28
52
40
\(p\)
\(q\)
Write down the value of \(p\) and the value of \(q\).
(2)
(d)(ii)
Write down the modal class.
(1)
(d)(iii)
Find an estimate of the mean and an estimate of the standard deviation of the times.
(3)
(e)(i)
A researcher suggests that the times can be modelled by a normal distribution with the mean and standard deviation found in part (d)(iii).
Use this model to find the probability that a person takes more than 50 minutes, and compare your answer with the proportion of the 160 people who took more than 50 minutes.
(2)
(e)(ii)
Use the model to find \(\mathrm{P}(T\lt 0)\) and comment on what this shows about the model.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(28\) minutes
A1
Accept 27 to 28.5. (Read at cumulative frequency 80.)
(b)
\(Q_{1}=20\) minutes
A1
Accept 19 to 21 (read at 40).
\(Q_{3}=36\) minutes
A1
Accept 35.5 to 37.5 (read at 120).
IQR \(=16\) minutes
A1
Accept 15 to 17. FT from their quartiles.
(c)
reading at cumulative frequency \(0.9\times160=144\)
(M1)
\(k=45\)
A1
Accept 44 to 46.
(d)(i)
\(p=152-132=20\)
A1
\(q=160-152=8\)
A1
(d)(ii)
\(20\lt t\le 30\)
A1
(d)(iii)
use of mid-interval values 5, 15, 25, 35, 45, 55 with their frequencies
(M1)
mean \(=28.25\) minutes
A1
Accept 28.3. FT from their p and q.
standard deviation \(=12.5\) minutes
A1
\(12.5274\ldots\) (\(\sigma_{x}\)).
(e)(i)
\(\mathrm{P}(T\gt 50)=0.0413\)
A1
\(0.041265\ldots\) Accept 0.0409 to 0.0413 (from 12.5).
from the data, \(\dfrac{8}{160}=0.05\); the model gives a slightly lower value but the two are close
R1
FT from their q. Must see 0.05 and a comparison.
(e)(ii)
\(\mathrm{P}(T\lt 0)=0.0121\)
A1
\(0.012065\ldots\) Accept 0.0119 to 0.0121.
times cannot be negative, so the model is not perfect (it gives a non-zero probability to impossible values); it is only an approximation
R1
Answers: (a) 28 min · (b) 20, 36, IQR 16 · (c) 45 · (d)(i) \(p=20,\ q=8\), (ii) \(20\lt t\le 30\), (iii) 28.25, 12.5 · (e)(i) 0.0413 vs 0.05, (ii) 0.0121
100AASL4-q00184.2 Presentation of dataPaper 2Hard13 marks
Two groups of 100 people, X and Y, completed the same task. Group X had a short training session first; group Y did not. The cumulative frequency curves show the times, \(t\) minutes, taken to complete the task. In each group the longest time was 12 minutes.
Graph drawn to scale
(a)
Write down the median time for each group.
(2)
(b)
Find the interquartile range for each group.
(2)
(c)
Compare the times taken by the two groups.
(2)
(d)
Estimate the percentage of group Y who took no longer than the median time of group X.
(2)
(e)
The slowest 10% of group Y took longer than \(k\) minutes. Find the value of \(k\).
(2)
(f)
Determine, for each group, whether the time of 12 minutes is an outlier.
(3)
Show mark scheme
Working
Mark
Notes
(a)
X: \(5.5\) minutes
A1
Accept \(5.4\)–\(5.6\).
Y: \(7.5\) minutes
A1
Accept \(7.4\)–\(7.6\).
(b)
X: \(7-4=3\) minutes
A1
Y: \(9-6=3\) minutes
A1
(c)
group X has the lower median (5.5 compared with 7.5), so the people with training generally completed the task faster
R1
Must compare a measure of centre.
the IQRs are equal (3 minutes), so the spread of the middle half of the times is the same
R1
Must compare a measure of spread.
(d)
reading the curve for Y at \(t=5.5\)
(M1)
FT from their median of X.
about \(20\%\)
A1
Accept \(18\%\)–\(23\%\).
(e)
reading the curve for Y at cumulative frequency \(90\)
X: \(12\gt 11.5\), so 12 minutes is an outlier in group X
A1
Y: \(12\lt 13.5\), so 12 minutes is not an outlier in group Y
A1
Answers: (a) X 5.5, Y 7.5 · (b) 3 and 3 · (d) about 20% · (e) \(k=10.5\) · (f) outlier in X only
101AASL4-q00264.3 Measures of central tendency and dispersionPaper 2Hard10 marks
The waiting times, in minutes, of nine patients at a clinic were
Time (min)
12
15
17
20
22
25
29
31
49
(a)
Find the standard deviation and the variance of the waiting times.
(2)
(b)
Use your GDC to write down the lower quartile and the upper quartile.
(2)
(c)
Using your answers to part (b), show that the time of 49 minutes is not an outlier.
(2)
(d)
A receptionist finds the quartiles by hand. He includes the median in both halves of the data and finds the median of each half. Find the quartiles obtained by this method, and determine whether 49 minutes is then an outlier.
(3)
(e)
Comment on your answers to parts (c) and (d).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\sigma=10.5\) minutes
A1
\(10.49985\ldots\)
variance \(=110\) minutes2
A1
\(110.247\ldots\) Accept \(110.2\); accept 110.25 from 10.5².
102AASL4-q00284.3 Measures of central tendency and dispersionPaper 2Hard15 marks
The annual salaries, in thousands of euros, of the 11 employees of a small company are
Salary (€000)
28
31
29
34
30
27
33
32
36
30
118
The salary of 118 thousand euros is that of the company director.
(a)
Find the mean and the median salary.
(2)
(b)
Find the standard deviation of the salaries.
(1)
(c)
Show that the director’s salary is an outlier.
(3)
(d)
In an advertisement for a job at the company, the director states that the “average salary” is €38 900. State, with a reason, whether this gives a fair picture of a typical salary at the company.
(2)
(e)
Find the mean and the standard deviation of the salaries of the 10 employees other than the director.
(2)
(f)(i)
Two pay rises are proposed for these 10 employees.
Under proposal 1, each salary increases by €1500. Write down the new mean and the new standard deviation of their salaries.
(2)
(f)(ii)
Under proposal 2, each salary increases by 5%. Find the new mean and the new standard deviation of their salaries.
(2)
(g)
Using your answers to part (f), state which proposal increases the spread of the salaries, and explain why.
(1)
Show mark scheme
Working
Mark
Notes
(a)
mean \(=38.9\) thousand euros
A1
\(38.9090\ldots\)
median \(=31\) thousand euros
A1
(b)
\(25.1\) thousand euros
A1
\(25.1376\ldots\)
(c)
\(Q_{1}=29,\ Q_{3}=34\)
A1
\(34+1.5\times5\)
M1
\(=41.5\) and \(118\gt 41.5\)
A1
(d)
no, it does not give a fair picture
A1
Do not award A1 without R1.
the mean is pulled up by the director’s (outlying) salary; 10 of the 11 employees earn less than €38 900 (the median is €31 000)
R1
(e)
mean \(=31\) thousand euros
A1
standard deviation \(=2.65\) thousand euros
A1
\(2.64575\ldots\)
(f)(i)
mean \(=32.5\) thousand euros
A1
FT from (e).
standard deviation \(=2.65\) thousand euros (unchanged)
A1
FT from (e).
(f)(ii)
mean \(=1.05\times31=32.55\) thousand euros
A1
Accept 32.6. FT from (e).
standard deviation \(=1.05\times2.6458=2.78\) thousand euros
A1
\(2.77804\ldots\) FT from (e).
(g)
proposal 2: multiplying every salary by 1.05 multiplies the standard deviation by 1.05, whereas adding the same amount to every salary does not change the spread
103AASL4-q00344.4 Correlation and regression of y on xPaper 2Hard8 marks
In school A, ten students recorded the number of hours, \(x\), they spent practising for a music examination and their examination score, \(y\).
\(x\)
2
3
4
5
7
9
10
12
14
15
\(y\)
43
52
63
67
69
53
47
75
75
85
In school B, six students recorded the same information.
\(x\)
3
5
6
8
12
13
\(y\)
38
57
68
51
81
68
The table gives critical values of \(r\) for a sample of size \(n\). If \(\lvert r\rvert\) is greater than the critical value, there is significant evidence of linear correlation.
\(n\)
6
8
10
12
16
Critical value
0.811
0.707
0.632
0.576
0.497
(a)
Find the value of \(r\) for school A and for school B.
(2)
(b)
Determine, for each school, whether there is significant evidence of linear correlation between \(x\) and \(y\).
(2)
(c)
Explain why the data for school B give weaker evidence of correlation than the data for school A, even though the value of \(r\) is greater.
(1)
(d)
The data for the two schools are combined. Find \(r\) for the combined data and determine whether there is significant evidence of linear correlation.
(2)
(e)
A teacher concludes that practising more causes students to score higher. Comment on this conclusion.
(1)
Show mark scheme
Working
Mark
Notes
(a)
school A: \(r=0.679\)
A1
\(0.679290\ldots\)
school B: \(r=0.758\)
A1
\(0.758382\ldots\)
(b)
school A: \(0.679\gt 0.632\) (\(n=10\)), so there is significant evidence of correlation
R1
FT from their r.
school B: \(0.758\lt 0.811\) (\(n=6\)), so there is not significant evidence of correlation
R1
FT from their r.
(c)
school B has a smaller sample, so a strong-looking value of \(r\) is more likely to occur by chance; the critical value is larger for smaller \(n\)
R1
(d)
\(r=0.701\)
A1
\(0.701325\ldots\)
\(0.701\gt 0.497\) (\(n=16\)), so there is significant evidence of linear correlation
R1
FT from their r.
(e)
correlation does not imply causation; other factors (e.g. ability, or the teacher) may affect both practice time and score
R1
Answers: (a) A 0.679, B 0.758 · (b) A significant, B not · (d) 0.701, significant
104AASL4-q00374.4 Correlation and regression of y on xPaper 2Hard10 marks
The fuel consumption, \(y\) litres per 100 km, of a car was measured at ten different constant speeds, \(x\) km h−1.
\(x\)
30
40
50
60
70
80
90
100
110
120
\(y\)
9.1
7.6
6.5
5.8
5.5
5.7
6.2
7
8.1
9.4
Graph drawn to scale
(a)
Find the value of \(r\) for all ten data pairs.
(1)
(b)
A student says that, since \(r\) is close to zero, there is no relationship between speed and fuel consumption. Comment on this statement.
(2)
(c)
Using only the data for speeds from 70 to 120 km h−1, find the value of \(r\) and the equation of the regression line of \(y\) on \(x\).
(3)
(d)
Use the regression line from part (c) to estimate the fuel consumption at a speed of 85 km h−1.
(2)
(e)
Use the regression line from part (c) to estimate the fuel consumption at 40 km h−1. Comment on your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.131\)
A1
\(0.131307\ldots\)
(b)
the scatter diagram shows a strong relationship: consumption decreases and then increases (a U-shaped curve)
R1
\(r\) only measures linear correlation, so a value near zero only shows that there is little linear relationship
R1
(c)
\(r=0.967\)
A1
\(0.967217\ldots\)
\(a=0.0786\)
A1
\(0.0785714\ldots\)
\(y=0.0786x-0.481\)
A1
\(-0.480952\ldots\)
(d)
\(0.0785714(85)-0.480952\)
(M1)
\(6.20\) litres per 100 km
A1
\(6.19761\ldots\)
(e)
\(2.66\) litres per 100 km
A1
\(2.66190\ldots\)
unreliable: 40 is outside the range 70 to 120 used for the line (extrapolation), and the data show that consumption at 40 km h−1 is actually about 7.6, so the linear model does not apply there
105AASL4-q00394.4 Correlation and regression of y on xPaper 2Hard15 marks
An online shop recorded the amount it spent on advertising, \(x\) hundred euros, and the number of orders it received, \(y\), in each of ten weeks.
\(x\) (€00)
18
20
21
23
24
26
27
29
30
32
\(y\)
64
75
73
88
91
104
103
119
121
134
(a)
Find the mean point \((\bar{x},\,\bar{y})\).
(2)
(b)
Find the value of \(r\) and describe the correlation.
(2)
(c)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(d)
Interpret the gradient of the regression line in context.
(1)
(e)
Use the regression line to estimate the number of orders in a week when €2800 is spent on advertising.
(2)
(f)
Use the regression line to find the number of orders it predicts for a week when €500 is spent on advertising. Comment on your answer.
(2)
(g)(i)
In week 11 the shop spent €2500 on advertising and received 110 orders. The data for week 11 are added to the data for the first ten weeks.
Find the value of \(r\) for the eleven weeks.
(1)
(g)(ii)
Find the equation of the regression line of \(y\) on \(x\) for the eleven weeks.
(2)
(g)(iii)
Compare the regression lines in parts (c) and (g)(ii), and explain why the gradient has not changed.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\bar{x}=25\)
A1
\(\bar{y}=97.2\)
A1
(b)
\(r=0.994\)
A1
\(0.993750\ldots\)
strong positive (linear) correlation
A1
(c)
\(a=4.98\)
A1
\(4.98421\ldots\)
\(y=4.98x-27.4\)
A1
\(-27.4052\ldots\)
(d)
for each extra €100 spent on advertising, the shop receives about 5 more orders per week
R1
(e)
\(4.98421(28)-27.4052\)
(M1)
\(112\) orders
A1
\(112.152\ldots\) Accept 112.1 or 112.2.
(f)
\(-2.48\)
A1
\(-2.48421\ldots\) Accept \(-2.5\).
a negative number of orders is impossible: €500 \((x=5)\) is far outside the range of the data (€1800 to €3200), so this is extrapolation and is not reliable
R1
(g)(i)
\(r=0.979\)
A1
\(0.978619\ldots\)
(g)(ii)
\(a=4.98\)
A1
\(4.98421\ldots\)
\(y=4.98x-26.2\)
A1
\(-26.2416\ldots\)
(g)(iii)
the new line is the line in (c) moved up by about 1.16 (same gradient, larger intercept): the new point has \(x=25=\bar{x}\), so \(\bar{x}\) is unchanged and the point does not tilt the line; it only raises the mean of \(y\) (to 98.4) through which the line must pass
R1
Must refer to the new \(x\)-value being equal to the mean of \(x\).
106AASL4-q00574.6 Combined events and diagramsPaper 2Hard16 marks
On any working day, the probability that it rains is 0.3. If it rains, the probability that Mia is late for work is 0.25. If it does not rain, the probability that Mia is late for work is 0.1. The tree diagram shows this information.
R = rains, L = Mia is late
(a)
Write down the values of \(p\), \(q\) and \(r\).
(2)
(b)
Find the probability that Mia is late for work on a given day.
(2)
(c)
Given that Mia is late for work, find the probability that it rained.
(3)
(d)(i)
Assume that whether Mia is late is independent from day to day. Let \(L\) be the number of days on which Mia is late in a period of 20 working days.
Find \(\mathrm{E}(L)\).
(1)
(d)(ii)
Find the probability that Mia is late on at most 2 of the 20 days.
(2)
(d)(iii)
Find the probability that Mia is late on more than 4 of the 20 days.
(2)
(e)
A working week has 5 days. A week is called “bad” if Mia is late on at least 3 days. Find the probability that at least one of the next 4 weeks is bad.
(3)
(f)
Suggest a reason why the assumption of independence in part (d) may not be valid.
107AASL4-q00674.7 Discrete random variablesPaper 2Hard15 marks
A bakery makes a speciality cake each morning. The daily demand, \(D\), for the cake has the probability distribution shown. Any cake not sold that day is thrown away.
\(d\)
0
1
2
3
4
\(\mathrm{P}(D=d)\)
0.1
0.2
0.35
0.25
0.1
Each cake costs $6 to make and is sold for $15. Demand is never more than 4 cakes and is independent from day to day.
(a)
Find \(\mathrm{E}(D)\).
(2)
(b)
Find the expected number of days in a 30-day month on which there is no demand for the cake.
(1)
(c)(i)
The bakery makes 3 cakes each day. Let \(Y\) be the daily profit, in dollars.
Find the probability distribution of \(Y\).
(3)
(c)(ii)
Find \(\mathrm{E}(Y)\).
(2)
(d)
Show that the expected daily profit is $12 if the bakery makes 2 cakes each day.
(2)
(e)
Find the expected daily profit if the bakery makes 4 cakes each day, and hence determine the number of cakes the bakery should make each day to maximize its expected profit.
(2)
(f)
The bakery makes 2 cakes each day for 30 days. Find the probability that it sells all of its cakes on at least 20 of these days.
108AASL4-q00714.8 The binomial distributionPaper 2Hard9 marks
A basketball player takes 15 free throws in each training session. The probability that she scores with any free throw is 0.78. Let \(X\) be the number of free throws she scores in a session.
(a)
State one assumption needed for \(X\) to be modelled by a binomial distribution.
(1)
(b)
Find the probability that she scores at least 12 free throws in a session.
(2)
(c)
Given that she scores at least 12 free throws in a session, find the probability that she scores exactly 12.
(3)
(d)
Find the probability that, in 8 sessions, she scores at least 12 free throws in exactly 5 of the sessions.
(3)
Show mark scheme
Working
Mark
Notes
(a)
the outcomes of different free throws are independent (or the probability of scoring is the same, 0.78, for every throw)
recognizing a second binomial: \(Y\sim\mathrm{B}(8,\,0.572625\ldots)\)
(M1)
\(\mathrm{P}(Y=5)\)
(M1)
\(0.269\)
A1
\(0.269133\ldots\) Using the 3 s.f. value 0.573 also gives 0.269.
Answers: (b) 0.573 · (c) 0.429 · (d) 0.269
109AASL4-q00724.8 The binomial distributionPaper 2Hard8 marks
A tour company uses a minibus with 20 passenger seats. From past records, the probability that a passenger who has bought a ticket turns up for a tour is 0.85, independently of other passengers. The company sells \(n\) tickets for each tour, where \(n\ge 20\).
(a)
For \(n=22\), find the probability that more passengers turn up than there are seats.
(2)
(b)
Find the greatest value of \(n\) for which the probability that more passengers turn up than there are seats is less than 0.25.
(3)
(c)
For \(n=22\), the company pays $40 to each passenger who turns up but cannot get a seat. Find the expected amount the company pays per tour.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(22,\,0.85)\) and \(\mathrm{P}(X\ge 21)\)
(M1)
\(0.137\)
A1
\(0.136724\ldots\)
(b)
\(\mathrm{P}(X\ge 21)\lt 0.25\) for \(X\sim\mathrm{B}(n,\,0.85)\); table of values
111AASL4-q00784.8 The binomial distributionPaper 2Hard9 marks
In each of 80 trays, 5 seeds are planted. The number of seeds that germinate in each tray is recorded.
Number germinating
0
1
2
3
4
5
Number of trays
1
3
9
21
28
18
(a)
Find the mean number of seeds germinating per tray.
(2)
(b)
Hence estimate the probability, \(p\), that a seed germinates.
(1)
(c)
The number of seeds germinating in a tray is modelled by \(\mathrm{B}(5,\,p)\), using your value of \(p\). Find the expected number of trays, out of 80, in which all 5 seeds germinate.
(2)
(d)
Find the variance of the data and the variance predicted by the model.
(2)
(e)
Use your answers to part (d) to comment on the suitability of the binomial model, suggesting a reason in context.
112AASL4-q00794.8 The binomial distributionPaper 2Hard14 marks
Eva plays tennis matches against Lina. In each set, the probability that Eva wins the set is 0.7, independently of all other sets. In a best-of-three match, the first player to win two sets wins the match.
(a)
Find the probability that Eva wins a best-of-three match in exactly two sets.
(1)
(b)
Show that the probability that Eva wins a best-of-three match is 0.784.
(3)
(c)
Let \(Y\sim\mathrm{B}(3,\,0.7)\). Explain why the probability that Eva wins a best-of-three match is equal to \(\mathrm{P}(Y\ge 2)\).
(2)
(d)
In a best-of-five match, the first player to win three sets wins the match. Find the probability that Eva wins a best-of-five match.
(2)
(e)
Comment on your answers to parts (b) and (d).
(1)
(f)(i)
Eva and Lina play 20 best-of-three matches. Let \(W\) be the number of matches that Eva wins.
Find \(\mathrm{E}(W)\).
(1)
(f)(ii)
Find the probability that Eva wins at least 16 matches.
(2)
(g)
Find the least probability of winning a set that Eva would need for the probability that she wins a best-of-five match to be at least 0.9.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.7^{2}=0.49\)
A1
(b)
Eva wins in three sets: LWW or WLW
M1
\(2\times0.7^{2}\times0.3=0.294\)
A1
\(0.49+0.294\)
A1
\(=0.784\)
AG
(c)
imagine all three sets are always played; the extra set does not change who wins
R1
Eva wins the match exactly when she wins at least 2 of the 3 sets, and the number of sets she wins is \(\mathrm{B}(3,\,0.7)\)
R1
(d)
\(\mathrm{P}(Z\ge 3)\) for \(Z\sim\mathrm{B}(5,\,0.7)\)
(M1)
\(0.837\)
A1
\(0.83692\)
(e)
the longer match gives the stronger player (Eva) a greater probability of winning
113AASL4-q00914.9 The normal distributionPaper 2Hard14 marks
The masses, \(M\) grams, of eggs from a farm are normally distributed with mean 62 g and standard deviation 4.5 g. An egg is classed as large if its mass is more than 65 g.
(a)
Find \(\mathrm{P}(55\lt M\lt 65)\).
(2)
(b)
Find the mass \(m\) such that 95% of eggs have a mass less than \(m\) grams.
(2)
(c)
An egg is chosen at random and is found to be large. Find the probability that its mass is more than 70 g.
(3)
(d)(i)
Eggs are packed in boxes of 12. Assume the masses of the eggs in a box are independent.
Find the probability that a box contains exactly 4 large eggs.
(2)
(d)(ii)
Find the probability that a box contains at least 3 large eggs.
(2)
(e)
The farmer changes the feed. The mean mass is still 62 g, but now 30% of eggs are large. Find the new standard deviation.
(3)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(55, 65, 62, 4.5)
(M1)
\(0.688\)
A1
\(0.687600\ldots\)
(b)
\(\mathrm{P}(M\lt m)=0.95\), inverse normal
(M1)
\(m=69.4\) g
A1
\(69.4018\ldots\)
(c)
\(\mathrm{P}(M\gt 65)=0.252492\ldots\) and \(\mathrm{P}(M\gt 70)=0.0377201\ldots\)
Answers: (a) 0.688 · (b) 69.4 g · (c) 0.149 · (d)(i) 0.196, (ii) 0.617 · (e) 5.72 g
114AASL4-q00964.10 Regression of x on yPaper 2Hard12 marks
A manager records, for eight sales staff, the number of years of experience, \(x\), and the number of sales made in one week, \(y\).
\(x\)
2
4
5
7
8
10
12
13
\(y\)
9
6
14
8
15
11
18
13
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), and describe the correlation.
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(d)
A new member of staff made 15 sales in one week. Estimate his number of years of experience using the appropriate regression line.
(2)
(e)
A colleague estimates the experience by rearranging the equation from part (b) and obtains a different answer. Find this answer, and explain why the two estimates are so different for these data.
(3)
(f)
Write down the coordinates of the point where the two regression lines intersect.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.610\)
A1
\(0.609749\ldots\)
moderate positive (linear) correlation
A1
(b)
gradient \(0.626\)
A1
\(0.625738\ldots\)
\(y=0.626x+6.98\)
A1
\(6.97874\ldots\)
(c)
gradient \(0.594\)
A1
\(0.594170\ldots\)
\(x=0.594y+0.643\)
A1
\(0.643497\ldots\)
(d)
\(x=0.594170(15)+0.643497\)
(M1)
Must use the line of x on y.
\(9.56\) years
A1
\(9.55605\ldots\) Accept 9.55 to 9.56.
(e)
\(15=0.625738x+6.97874\)
(M1)
\(x=12.8\) years
A1
\(12.8188\ldots\)
the correlation is only moderate (\(|r|\) not close to 1), so the two regression lines are far apart; they coincide only when \(|r|=1\)
R1
(f)
\((7.625,\ 11.75)\)
A1
The mean point \((\bar{x},\,\bar{y})\). Accept \((7.63,\ 11.8)\).
Answers: (a) 0.610 · (b) \(y=0.626x+6.98\) · (c) \(x=0.594y+0.643\) · (d) 9.56 years · (e) 12.8 years · (f) \((7.625,\,11.75)\)
115AASL4-q01094.11 Conditional probability and independencePaper 2Hard7 marks
In a country, the heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm, and the heights of adult women are normally distributed with mean 165 cm and standard deviation 6.5 cm. A man is described as tall if his height is more than 185 cm, and a woman is described as tall if her height is more than 170 cm. One man and one woman are chosen independently at random.
(a)
Find the probability that the man is tall and the probability that the woman is tall.
(2)
(b)
Find the probability that exactly one of them is tall.
(3)
(c)
Given that exactly one of them is tall, find the probability that it is the man.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(\text{man tall})=0.159\)
A1
\(0.158655\ldots\)
\(\mathrm{P}(\text{woman tall})=0.221\)
A1
\(0.220878\ldots\)
(b)
\(m(1-w)+w(1-m)\) with their values
M1
Two products, added.
\(0.123611\ldots+0.185833\ldots\)
(A1)
\(0.309\)
A1
\(0.309446\ldots\)
(c)
\(\dfrac{0.123611\ldots}{0.309446\ldots}\)
(M1)
\(0.399\)
A1
\(0.399460\ldots\)
Answers: (a) 0.159, 0.221 · (b) 0.309 · (c) 0.399
116AASL4-q01114.11 Conditional probability and independencePaper 2Hard15 marks
In a youth athletics squad, long-jump distances, \(L\) metres, are normally distributed with mean 5.6 m and standard deviation 0.32 m, and 100 m sprint times, \(T\) seconds, are normally distributed with mean 12.8 s and standard deviation 0.45 s. An athlete qualifies for a regional final if she jumps more than 6.0 m or runs 100 m in less than 12.2 s (or both). For any one athlete, \(L\) and \(T\) are independent.
(a)
Find \(\mathrm{P}(L\gt 6.0)\).
(2)
(b)
Find \(\mathrm{P}(T\lt 12.2)\).
(2)
(c)
Find the probability that a randomly chosen athlete qualifies.
(3)
(d)
Given that an athlete qualifies, find the probability that she achieves both standards.
(2)
(e)
Eight athletes from the squad are chosen at random. Find the probability that at least two of them qualify.
(3)
(f)
The organisers change the long-jump standard to \(d\) metres, keeping the sprint standard at 12.2 s, so that the probability that an athlete qualifies is 0.25. Find \(d\).
117AASL4-q01144.12 Standardization and inverse normalPaper 2Hard8 marks
The masses of bags of flour filled by a machine are normally distributed with mean \(\mu\) grams and standard deviation \(\sigma\) grams. It is known that 8% of bags have a mass less than 990 g and 15% of bags have a mass greater than 1030 g.
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(5)
(b)
The machine is adjusted so that the mean changes but the standard deviation does not. After the adjustment, only 2% of bags have a mass less than 990 g. Find the new mean.
(3)
Show mark scheme
Working
Mark
Notes
(a)
use of inverse normal on \(Z\): \(\mathrm{P}(Z\lt z_{1})=0.08\) or \(\mathrm{P}(Z\lt z_{2})=0.85\)
(M1)
\(z_{1}=-1.40507\ldots\) and \(z_{2}=1.03643\ldots\)
A1
\(\dfrac{990-\mu}{\sigma}=-1.40507\ldots\) and \(\dfrac{1030-\mu}{\sigma}=1.03643\ldots\)
Answers: (a) \(\mu=1013\) g, \(\sigma=16.4\) g · (b) 1024 g (1023.6)
118AASL4-q01174.12 Standardization and inverse normalPaper 2Hard8 marks
The masses, \(M\) grams, of fish in a fish farm are modelled by a normal distribution in which the standard deviation is 25% of the mean. It is known that 5% of the fish have a mass greater than 120 g.
(a)
Find the mean and the standard deviation of \(M\).
(4)
(b)
Find the probability that a fish has a mass less than 50 g.
(2)
(c)
Find \(\mathrm{P}(M\lt 0)\) and comment on what your answer shows about the suitability of the model.
119AASL4-q01184.12 Standardization and inverse normalPaper 2Hard8 marks
Metal rods are cut to a target length of 250 mm. A rod is accepted if its length is between 245 mm and 255 mm. The lengths of the rods are normally distributed with mean \(\mu\) mm and standard deviation \(\sigma\) mm.
(a)
For \(\mu=251\) and \(\sigma=1.8\), find the probability that a rod is accepted.
(2)
(b)
With \(\sigma=1.8\), the probability that a rod is accepted must be at least 0.99. Find the set of possible values of \(\mu\), giving the end points correct to two decimal places.
(3)
(c)
The mean is set to \(\mu=250\). Find the greatest value of \(\sigma\) for which at least 99.5% of rods are accepted.
(3)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(245, 255, 251, 1.8)
(M1)
\(0.986\)
A1
\(0.986436\ldots\)
(b)
\(\mathrm{P}(245\lt X\lt 255)=0.99\) solved for \(\mu\) with technology (graph or solver)
M1
\(\mu=249.23\) or \(\mu=250.77\)
A1
\(249.234\ldots,\ 250.765\ldots\), symmetric about 250.
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