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Exam (elaborations)

A Level Further Mathematics B (MEI) Y422/01 Statistics Major Exam Paper 2026 – Mark Scheme & Practice Questions

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Access the A Level Further Mathematics B (MEI) Y422/01 Statistics major exam paper including probability distributions, hypothesis testing, regression analysis, and Normal approximation. Includes official mark scheme for 2026. A Level Further Maths, MEI Y422, Statistics major exam, probability distribution, hypothesis testing, regression analysis, Normal approximation, chi-squared test, Spearman rank correlation, Kolmogorov-Smirnov test, exam mark scheme, A Level practice paper, maths revision notes, further maths past paper, statistics exam questions, A Level 2026

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Section A (27 marks)


1 The number of insurance policy sales made per month by a salesperson is modelled by the random
variable X, with probability distribution shown in the table.
r 0 1 2 3 4 5 6

P(X = r) 0.05 0.1 0.25 0.3 0.15 0.1 0.05
(a) Find each of the following.
• E(X )
• Var(X ) [2]


The salesperson is paid a basic salary of £1000 per month plus £500 for each policy that is sold.

(b) Find the mean and standard deviation of the salesperson’s monthly salary. [3]




2 The number of cars arriving per minute to queue at a drive-through fast-food restaurant is
modelled by the random variable X. The standard deviation of X is 0.6. You should assume that
arrivals are random and independent and occur at a constant average rate.

(a) Find the mean of X. [2]

(b) (i) Calculate P(X = 1). [1]

(ii) Calculate P(X 2 1). [2]

(c) Find the probability that fewer than 5 cars arrive in a randomly chosen 20-minute period.
[2] 3 At a launderette the process of cleaning a load of clothes consists of three
stages: washing, drying and folding. The times in minutes for each process are modelled
by independent Normal distributions with means and standard deviations as shown in the
table.
Mean Standard deviation
Washing 35 2.4
Drying 46 3.1
Folding 12 2.2
(a) Find the probability that drying a randomly chosen load of clothes takes more than 50
minutes.
[1]

(b) It is given that for 99% of loads of clothes the washing time is less than k minutes.


© OCR 2024 Y422/01 Jun24

, 3

Find the value of k. [1]

(c) Determine the probability that the drying time for a randomly chosen load of clothes is
less
than the total of the washing and folding times. [3]

(d) Determine the probability that the mean time for cleaning 5 randomly chosen loads of
clothes is less than 90 minutes. You should assume that the time for cleaning any load is
independent of the time for cleaning any other load. [3]




4 An archer fires arrows at a circular target of radius 50 cm. The distance in cm that an arrow lands
from the centre of the target is modelled by the random variable X, with probability density
function given by
ax 0
f( )x =
'0 otherwiseG Gx 50, ,

where a is a constant.

(a) Determine the value of a. [2]

(b) Determine the probability that an arrow will land within 5 cm of the centre of the target.
[2]

(c) Determine the median distance from the centre of the target that an arrow will land.
[3] Section B (93 marks)


5 A researcher is investigating whether doing yoga has any effect on quality of sleep in older
people. The researcher selects a random sample of 40 older people, who then complete a yoga
course. Before they start the course and again at the end, the 40 people fill in a questionnaire
which measures their perceived sleep quality. The higher the score, the better is the perceived
quality of sleep.

The researcher uses software to produce a 90% confidence interval for the difference in mean
sleep quality (sleep quality after the course minus sleep quality before the course). The output
from the software is shown below.
Z Estimate of a Mean

Confidence level 0.9




© OCR 2024 Y422/01 Jun24 Turn over

, 4
Sample

Mean 0.586
s 2.14

N 40

Result

Z Estimate of a Mean


Mean 0.586
s 2.14
SE 0.3384
N 40
Lower limit 0.029
Upper limit 1.143

Interval 0.586 ± 0.557
(a) Explain why the confidence interval is based on the Normal distribution even though the
distribution of the population of differences is not known. [2]

(b) Explain whether the confidence interval suggests that the mean sleep qualities before
and
after completing a yoga course are different. [2]

(c) In the output from the software, SE stands for ‘standard error’.

(i) Explain what standard error is. [1]

(ii) Show how the standard error was calculated in this case. [1]
(d) A colleague of the researcher suggests that the confidence level should have been 95%
rather than 90%.

Determine whether this would have made a difference to your answer to part (b). [4] 6
A student is investigating the relationship between age and grip strength in adults. The student selects
10 people and records their ages in years and the grip strengths of their dominant hand, measured in kg.
The data are shown in the table below, together with a scatter diagram to illustrate the data.
Age 22 29 36 39 53 57 60 71 76 82
Grip strength 38 46 42 49 37 47 36 33 34 24

60


40
Grip
strength
20


0
0 20 40 60 80
Age

© OCR 2024 Y422/01 Jun24

, 5

The student decides to carry out a hypothesis test to investigate whether there is negative
association between age and grip strength.

(a) Explain why the student decides to carry out a test based on Spearman’s rank correlation
coefficient. [2]

(b) State what property of the sample is required in order for it to be valid to carry out a
hypothesis test. [1]

(c) In this question you must show detailed reasoning.

Assuming that the property in part (b) holds, carry out the test at the 5% significance
level. [8] 7 An environmental investigator wants to check whether the level of selenium in carrots in
fields near a mine is different from the usual level in the country, which is 9.4 ng / g (nanograms per
gram). She takes a random sample of 10 carrots from fields near the mine and measures the selenium
level of each of them in ng / g, with results as follows.

6.20 10.72 11.42 16.32 15.33 10.56 8.83 9.21 7.78 14.32

(a) Find estimates of each of the following.
• The population mean
• The population standard deviation [2]


The investigator produces a Normal probability plot and carries out a Kolmogorov-Smirnov test
for these data as shown in the diagram.
2

1.5 Kolmogorov-Smirnov
1 test for Normality p-
value = 0.68
0.5

0

−0.5

−1

−1.5

−2
0.00 5.00 10.00
(b) Comment on what the Normal probability plot and the p-value of the test suggest about
the data. [3]

(c) State the null hypothesis for the Kolmogorov-Smirnov test for Normality. [1] (d)

In this question you must show detailed reasoning.

Carry out a test at the 5% significance level to investigate whether the mean selenium level in
carrots from fields near the mine is different from 9.4 ng / g. [8]


© OCR 2024 Y422/01 Jun24 Turn over

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