A Level Further Mathematics 9231 Paper 41, May/June 2026

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9231/41 · May/June 2026

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The manager of a village store is interested in the shopping habits of his customers. From previous records, the median number of items bought per customer was found to be 8. The manager believes the median number of items bought per customer has increased. He selects a random sample of 15 customers and records the number of items each customer buys at his store. The results are shown in the stem-and-leaf diagram. key 2 | 3 represents 23 items 0 | 6 6 7 9 9 1 | 0 5 7 7 2 | 3 6 3 | 4 4 | 0 5 | 6 6 | 7 | 4

Suggest a reason why using a Wilcoxon signed-rank test to examine the manager’s belief might not be appropriate in this situation. [1]

[1]

Use a sign test at the 5% significance level to investigate the manager’s belief. [6]

[6]

Three different bus companies, $A$ , $B$ and $C$ , record the reliability of their service. Each bus journey is recorded as being on time, late or cancelled. A reliable bus service is considered to be one that is on time. The records from a random sample of buses for each company are taken. These records are summarised in the table.

	company $A$	company $B$	company $C$	total
on time	7	22	20	49
late	14	11	12	37
cancelled	4	6	4	14
total	25	39	36	100

Test at the 5% significance level whether the reliability of the service is independent of the bus company. [8]

[8]

Further evidence indicates that one company may be particularly unreliable compared with the other two companies. Use your calculations in 2(a) to suggest which company this is likely to be. [2]

[2]

Students studying geography at a college choose either human geography or physical geography. A teacher is investigating the times taken by students to complete assignments of equal difficulty in human geography and physical geography. He selects a random sample of 7 human geography students and records the times taken, $x$ minutes, to complete an assignment. He also selects a random sample of 9 physical geography students and records the times taken, $y$ minutes, to complete an assignment. The results are summarised as follows. $Σ x = 445 Σ x^{2} = 29144 Σ y = 310 Σ y^{2} = 11622$ You should assume that the underlying distributions are normal and have equal population variances. Find a 95% confidence interval for the difference in population mean times taken to complete assignments in human geography and physical geography. [6]

[6]

The continuous random variable $X$ has cumulative distribution function $F$ given by $F (x) = {\begin{cases} 0 & x < 0, \\ - \frac{1}{108} x^{3} + \frac{1}{12} x^{2} & 0 ⩽ x ⩽ 3, \\ - \frac{1}{32} x^{2} + \frac{7}{16} x + a & 3 < x ⩽ b, \\ 1 & x > b . \end{cases}$

Verify that the median of $X$ is 3. [1]

[1]

Show that $a = - \frac{17}{32}$ and find the value of $b$ . [3]

[3]

Find the probability density function of $X$ . [2]

[2]

Find $E (\frac{1}{X})$ . [3]

[3]

Margaret has developed a new revision course to help students improve their ability in mathematics. The students each sit a mathematics test before attending the course and a test of similar difficulty after attending the course. Margaret claims that her course will increase a student’s score by more than 5 marks on average. For a random sample of 8 students, the test scores before and after the revision course are given in the table.

student	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$
test score before	43	29	67	54	34	29	52	60
test score after	49	53	79	62	39	55	65	62

Use a $t$ -test at the 10% significance level to investigate Margaret’s claim. [7]

[7]

State an assumption that is necessary for the test in 5(a) to be valid. [1]

[1]

The discrete random variable $X$ has probability generating function $G_{X} (t)$ defined by $G_{X} (t) = \frac{k}{t} (a t + 1)^{2} .$ It is given that $E (X) = \frac{1}{3}$ .

Show that $a = 2$ and find the value of $k$ . [5]

[5]

Find $Var (X)$ . [2]

[2]

Three independent observations of $X$ are taken. The random variable $Y$ is the sum of these observations. Find $P (Y = 0)$ . [3]

[3]

Mark scheme

9231/41 · May/June 2026

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1a · 1 mark

B1
Data has outliers / is skewed / lacks symmetry, suggesting population / distribution may not be symmetrical.

1b · 6 marks

B1
H0: population median is 8. H1: population median is greater than 8.
B1
$X \sim B (15, 0.5)$
M1
$P (X ⩽ 3) = (0.5)^{15} + (\begin{array}{c} 15 \\ 1 \end{array}) (0.5)^{15} + (\begin{array}{c} 15 \\ 2 \end{array}) (0.5)^{15} + (\begin{array}{c} 15 \\ 3 \end{array}) (0.5)^{15}$ OR $\pm (\frac{3 (\pm 0.5) - 7.5}{\sqrt{3.75}})$
A1
0.0176 OR $\pm 2.07$
M1
‘0.0176’ < 0.05, reject H0 / significant. OR 2.07 > 1.645, reject H0 / significant.
A1
Sufficient evidence to support the manager’s belief.

2a · 8 marks

B1
H0: reliability of service is independent of bus company. H1: reliability of service is not independent of bus company.
M1
Calculate expected frequencies (must be seen).
7 (12.25) 22 (19.11) 20 (17.64)
14 (9.25) 11 (14.43) 12 (13.32)
4 (3.50) 6 (5.46) 4 (5.04)
A1
All expected frequencies correct to at least 3 significant figures.
M1
Combine rows (for example rows 2 and 3):
7 (12.25) 22 (19.11) 20 (17.64)
18 (12.75) 17 (19.89) 16 (18.36)
OR Combine columns (for example columns A and B):
29 (31.36) 20 (17.64)
25 (23.68) 12 (13.32)
10 (8.96) 4 (5.04)
M1
Combining rows 2 and 3: $\frac{(7 - 12.25)^{2}}{12.25} + \frac{(22 - 19.11)^{2}}{19.11} + \dots + \frac{(16 - 18.36)^{2}}{18.36}$ $[= 2.25 + 0.43705 + 0.31574 + 2.16176 + 0.41991 + 0.30336]$ OR $\frac{7^{2}}{12.25} + \frac{22^{2}}{19.11} + \dots + \frac{16^{2}}{18.36} - 100$ Combining columns A and B: $\frac{(29 - 31.36)^{2}}{31.36} + \frac{(20 - 17.64)^{2}}{17.64} + \dots + \frac{(4 - 5.04)^{2}}{5.04}$ $[= 0.17760 + 0.31574 + 0.07358 + 0.13081 + 0.12071 + 0.21460]$ OR $\frac{29^{2}}{31.36} + \frac{20^{2}}{17.64} + \dots + \frac{4^{2}}{5.04} - 100$ No rows / columns combined: $\frac{(7 - 12.25)^{2}}{12.25} + \frac{(22 - 19.11)^{2}}{19.11} + \dots + \frac{(4 - 5.04)^{2}}{5.04}$ $[= 2.250 + 0.43705 + 0.31574 + 2.4392 + 0.81531 + 0.13081 + 0.07143 + 0.05341 + 0.21460]$
A1
Combining rows 2 and 3: Test statistic = 5.888 OR Combining columns A and B: Test statistic = 1.033
M1
Combining rows 2 and 3: ‘5.888’ < 5.991, accept H0 / do not reject H0 / not significant. OR Combining columns A and B: ‘1.033’ < 5.991, accept H0 / do not reject H0 / not significant. OR No cells combined: ‘6.73’ < 9.488, accept H0 / do not reject H0 / not significant.
A1
Insufficient evidence to suggest reliability (of service) is not independent of (bus) company. OR Insufficient evidence to suggest reliability (of service) is associated with (bus) company. OR Insufficient evidence to suggest that reliability (of service) is dependent on (bus) company.

7 (12.25)	22 (19.11)	20 (17.64)
14 (9.25)	11 (14.43)	12 (13.32)
4 (3.50)	6 (5.46)	4 (5.04)

7 (12.25)	22 (19.11)	20 (17.64)
18 (12.75)	17 (19.89)	16 (18.36)

29 (31.36)	20 (17.64)
25 (23.68)	12 (13.32)
10 (8.96)	4 (5.04)

2b · 2 marks

B1
Company A (appears less reliable).
1 mark
The observed frequencies are higher than the (combined) expected frequencies for company A for late and cancelled, but not for company B and company C.

3 · 6 marks

M1
$\sum (x - \overset{ˉ}{x})^{2} = 29144 - \frac{1}{7} (445^{2}) [= 854.7]$ $\sum (y - \overset{ˉ}{y})^{2} = 11622 - \frac{1}{9} (310^{2}) [= 944.2]$
M1
$s_{p}^{2} = \frac{‘854.7’ + ‘944.2’}{7 + 9 - 2}$
A1
128.495
B1
B1 for 2.145 seen.
M1
$(\frac{445}{7} - \frac{310}{9}) \pm 2.145 \sqrt{‘128.5’ (\frac{1}{7} + \frac{1}{9})}$
A1
$[16.9, 41.4]$

4a · 1 mark

B1
$F (3) = \frac{1}{12} \times 3^{2} - \frac{1}{108} \times 3^{3} = 0.5$ [Since $F (3) = 0.5$ , the median of $X$ is 3.]

4b · 3 marks

B1
$- \frac{1}{32} \times 3^{2} + \frac{7}{16} \times 3 + a = 0.5, a = - \frac{17}{32}$
M1
$F (b) = 1 \Rightarrow - \frac{1}{32} b^{2} + \frac{7}{16} b - \frac{17}{32} = 1$ OR $F (b) - F (3) = \frac{1}{2} \Rightarrow {[- \frac{1}{32} x^{2} + \frac{7}{16} x (+ a)]}_{3}^{b} = 0.5 \Rightarrow - \frac{1}{32} b^{2} + \frac{7}{16} b - \frac{33}{32} = \frac{1}{2}$
A1
$b = 7$

4c · 2 marks

B1
Both parts differentiated correctly, ignore domain.
1 mark
Fully defined including correct domain, may be in terms of b.

4d · 3 marks

1 mark
$\int_{0}^{3} \frac{1}{x} (- \frac{1}{36} x^{2} + \frac{1}{6} x) d x + \int_{3}^{7} \frac{1}{x} (- \frac{1}{16} x + \frac{7}{16}) d x$
M1
${[\frac{1}{6} x - \frac{1}{72} x^{2}]}_{0}^{3} + {[- \frac{1}{16} x + \frac{7}{16} \ln x]}_{3}^{7}$
A1
$\frac{1}{8} + \frac{7}{16} \ln (\frac{7}{3}) [= 0.496]$

5a · 7 marks

B1
H0: $μ_{d} = 5$ , H1: $μ_{d} > 5$ OR H0: $μ_{A} - μ_{B} = 5$ , H1: $μ_{A} - μ_{B} > 5$
M1
| d | 6 | 24 | 12 | 8 | 5 | 26 | 13 | 2 |
M1
$s_{d}^{2} = \frac{1}{7} [‘1694’ - \frac{{‘96’}^{2}}{8}] = \frac{542}{7} = 77.43$
M1
$[t =] \frac{\frac{‘96’}{8} - 5}{\sqrt{\frac{‘77.43’}{8}}}$
A1
2.25
M1
‘2.25’ > 1.415, reject H0 / significant.
A1
Sufficient evidence to support Margaret’s claim. OR Sufficient evidence to suggest that the course will improve a student’s score by more than 5 marks (on average).

5b · 1 mark

B1
Population of differences are normally distributed.

6a · 5 marks

B1
$k (a + 1)^{2} = 1$
M1
$k a^{2} - k = \frac{1}{3}$
1 mark
$\frac{a^{2} - 1}{(a + 1)^{2}} = \frac{1}{3}$
A1
$a = 2$
B1
$k = \frac{1}{9}$

6b · 2 marks

1 mark
EITHER: $G_{X}^{''} (t) = \frac{2}{9 t^{3}}, G_{X}^{''} (1) = \frac{2}{9} [= 2 k]$ OR: $E (X^{2}) = \frac{5}{9} [= 5 k]$
1 mark
$Var (X) = \frac{2}{9} + \frac{1}{3} - (\frac{1}{3})^{2} = \frac{4}{9}$ OR $Var (X) = \frac{5}{9} - (\frac{1}{3})^{2} = \frac{4}{9}$

6c · 3 marks

1 mark
EITHER: $G_{Y} (t) = \frac{1}{729 t^{3}} (2 t + 1)^{6} [= \frac{k^{3}}{t^{3}} (2 t + 1)^{6}]$ OR: Relevant values for $(X_{1}, X_{2}, X_{3})$ are $(- 1, 0, 1)$ , 6 possible orderings, and $(0, 0, 0)$ .
M1
Coefficient of $t^{0}$ term: $\frac{1}{729} (\begin{array}{c} 6 \\ 3 \end{array}) (2)^{3} [= k^{3} (\begin{array}{c} 6 \\ 3 \end{array}) (2)^{3}]$ OR $6 \times \frac{1}{9} \times \frac{4}{9} \times \frac{4}{9} + (\frac{4}{9})^{3} [= 6 \times k \times (4 k)^{2} + (4 k)^{3}]$
A1
$P (Y = 0) = \frac{160}{729}$ OR $\frac{160}{729}$

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