Mathematics 9709/65 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A student notes the length, t minutes, of certain lectures. The results for a random sample of 80 lectures are summarised as follows.
n = 80 / t = 6430 / t 2 = 519 740
(a) Calculate a 96% confidence interval for the population mean length of the lectures. [6]
(b) The method used in part (a) is valid because the sample size was large. Explain why the method would not be valid if the sample size were small. [1]
* 0000800000004 * DFD
4
Mark scheme
1(a) 6430 643 B1 Accept 80.4 if nothing better seen.
Est ( μ ) = or or 80.375
80 8
80 519740 6430 1
2
64302 M1 Biased [36.61] scores M0A0.
( )
Est σ 2 =
79 80
− =
80 79
519740 −
80
11715 A1
= 37.073 [37.1 3 sf ] or
316
z = 2.054 or 2.055 B1
6430 '37.073' M1 Use of correct formula, any z .
± z×
80 80
= 79.[ 0] to 81.9 or 81.8 [3 sf] A1 Must be an interval.
Use of biased [79.0 to 81.8] can score B1M0A0B1M1A1.
6
1(b) CLT would not apply B1 Or distribution of the sample means is not [necessarily]
normal.
1
A researcher is investigating whether the proportion of families who do not own a car in his town is different from the proportion of the population in the whole country, which is 10.1%. He takes a large random sample of families in his town and finds the proportion of families that do not own a car.
(a) Explain why a two-tailed test is appropriate in this context. [1]
(b) State suitable null and alternative hypotheses for the test. [1]
The researcher calculates the value of the test statistic z and finds that z = 1.82. He carries out the test at the 5% significance level.
(c) State the conclusion of the test, explaining your answer. [2]
* 0000800000005 * DFD
5
Mark scheme
2(a) Researcher is looking for evidence of ‘different from’ or B1 OE.
researcher is not looking for ‘less than’ or ‘more than’
1
2(b) H0 : p = 0.101 H1 : p ≠ 0.101 B1
[where p is proportion [in his town] not owning a car]
1
2(c) 1.82 < 1.96 M1 Valid comparison, accept 0.034 > 0.025 OE.
[Do not reject H0 ] A1 In context, not definite, no contradictions.
Insufficient evidence that [town] proportion is different from 10.1% If p is used, it must be defined.
SC: if in 2(b) H1 : p > 0.101 [or p < 0.101 ], then
1.82 > 1.645 or 0.0344 < 0.05 scores M1A0.
2
A certain website receives an average of n hits per hour. In the past the value of n was 14.4. After making some improvements, the owner of the website wishes to test whether the value of n has increased. He chooses a 10-minute period at random and finds that there were 6 hits during this period.
You may assume that the number of hits the website receives in any given time period follows a Poisson distribution.
(a) Carry out the test at the 2.5% significance level. [6]
(b) Explain whether it is possible that a Type I error or a Type II error or both may have been made in carrying out the test. [2]
* 0000800000006 * DFD
6
Mark scheme
3(a) H0 : λ = 2.4 or μ = 14.4 B1 OE.
H1 : λ > 2.4 or μ > 14.4
[λ =] 2.4 soi B1
2.42 2.43 2.44 2.45 M1 Any λ .
1 − e−2.4 1 + 2.4 + + + + = Allow one end error.
2! 3! 4! 5!
This expression must be seen, accept fully correct sigma
1 − e−2.4 (1 + 2.4 + 2.88 + 2.304 + 1.3824 + 0.6636) = notation.
1 − ( 0.09072 + 0.2177 + 0.2613 + 0.2090 + 0.1254 + 0.060196 )
= 0.0357 (3 sf) A1 SC unjustified answer of 0.0357 scores B1M0B1.
'0.0357' > 0.025 M1 Valid comparison.
[Do not reject H0 ] A1FT In context, not definite, no contradictions.
Insufficient evidence that μ has increased Condone sufficient evidence that μ has not increased OE.
6
3(b) H0 not rejected, so Type II error may have been made. B1FT OE.
H0 not rejected, so Type I error cannot have been made. B1FT OE.
‘H0 not rejected’ is necessary, but only needs to be stated
once
2
A sports fan produces a magazine each month.
(a) On average 1 in 540 characters in the magazine is incorrect.
(i) Use an appropriate approximating distribution to find the probability that, in a magazine containing 2430 characters, there are at least 4 incorrect characters. [3]
(ii) Justify your approximating distribution. [1]
* 0000800000007 * DFD
7
(b) On average the number of copies, X, of the magazine sold per month is 123.4.
(i) State one condition for X to have a Poisson distribution. [1]
You are now given that X has a Poisson distribution.
(ii) Use an appropriate approximating distribution to find the probability that in a randomly chosen month, more than 130 copies of the magazine are sold. [5]
* 0000800000008 * DFD
8
Mark scheme
4(a)(i) 2430 B1 SOI.
[λ =] or [ λ =] 4.5
540
4.52 4.53 M1 Any λ .
1 − e−4.5 1 + 4.5 + + = Allow one end error.
2! 3!
This expression must be seen, accept fully correct sigma
1 − e−4.5 (1 + 4.5 + 10.125 + 15.1875) = notation.
1 − ( 0.01111 + 0.04999 + 0.11248 + 0.16872 )
= 0.658 (3sf) A1 SC unjustified answer of 0.658 scores B1M0B1.
1
SC use of B 2430, and probability = 0.658 scores
540
B2.
Note: use of normal could score B1 only.
3
4(a)(ii) [ n =] 2430 > 50 B1 Explicit, both needed.
Note: n large and p small is insufficient.
and [ np = ] 4.5 < 5 or [ p = ] 0.00185 < 0.01
1
4b(i) Mean of X is constant B1 Any of these stated in context OE.
Sales occur singly or randomly or independently of each other
1
4(b)(ii) N (123.4,123.4 ) B1 SOI.
B1 for N (123.4,)
B1 B1 for Var( X ) = 123.4
130.5 − '123.4' M1 For standardising with their values
[ = 0.639] Condone with no cc or incorrect cc.
'123.4'
1 − Φ ( '0.639') M1 Correct area consistent with their working.
= 0.261 (3 sf) A1
5
D ~ N ( 2.10 − 1.51, 0.12 + 0.09 ) = B1 B1 for N ( 2.10 − 1.51,) OE.
N ( 0.59, 0.21)
The masses, in kg, of large and small bags of tomatoes have the distributions L ~ N(2.10, 0.12) and
S ~ N(1.51, 0.09) respectively.
(a) Find P(L 2 S + 0.5) . [5]
* 0000800000009 * DFD
9
The price of tomatoes is $4.30 per kg.
(b) A large bag of tomatoes and a small bag of tomatoes are chosen at random. Find the probability that the total price of the tomatoes in the two bags is less than $16. [5]
* 0000800000010 * DFD
10
Mark scheme
5(a)
B1 B1 for Var( D) = 0.12 + 0.09 SOI.
0.5 − '0.59' M1 For standardising with their values.
[ = −0.196]
'0.21'
( ) (
1 − Φ ' − 0.196' = Φ '0.196' ) M1 Correct area consistent with their working.
= 0.578 (3 sf) A1
5
5(b) (
T ~ N 4.3 × ( 2.10 + 1.51) , 4.32 × ( 0.12 + 0.09) ) B1 B1 for N ( 4.3 × ( 2.10 − 1.51) , ) .
N (15.532, 3.8829)
=
M1 M1 for Var( T ) = 4.32 × ( 0.12 + 0.09) SOI.
16 − '15.523' M1 For standardising with their values.
[ = 0.242]
'3.8829'
(
Φ '0.242' ) M1 Correct area consistent with their working.
= 0.596 (3 sf) A1 16
− '3.61'
Accept alternative method: 4.30 .
'0.21'
5
f (x) a
x
O b
The diagram shows the graph of the probability density function, f, of a random variable X. The graph is a straight line from (0, a) to (b, 0) where a and b are constants. Elsewhere f (x) = 0 .
(a) Find an expression for b in terms of a. [2]
(b) Given that E (X ) = 49 find the value of a. [5]
* 0000800000011 * DFD
11
(c) Using the value of a found in part (b) find the value of k such that P(X 1 k) = 34 . [4]
* 0000800000012 * DFD
12
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
6(a) 1 M1 Attempt area of triangle = 1 .
ab = 1
2 b a
0 − b x + a dx = 1 and attempt to integrate with correct
limits to reach an equation in a and b .
2 A1
b=
a
2
6(b) a2 a B1FT FT their b .
f ( x) = a − x or f ( x) = a − x SOI.
2 b
2 M1 Attempt to integrate xf ( x ) , ignore limits.
a b
a2 x2 2 2
0
ax −
2
dx or
0
b
x − 2 x 2 dx
b
FT their linear f ( x ) .
2
b A1FT FT their b .
ax 2 a 2 x3 a 2 x 2 2 x3 1 Correct integration and limits.
− = or − 2 = b
2 6 0 3a b 3b 0 3 Could be in terms of a or a and b .
2 4 1 4 2 M1 4
= or b = and b = ' ' Equate their integral to to obtain an equation in a .
3a 9 3 9 a 9
3 A1
a=
2
5
6(c) k
3 9
k
3 9 M1 Attempt to integrate f ( x ) , with limits 0 to k .
0
' ' − 0.5 ' 'x dx =
2 4 0
− x dx
2 8 FT their linear f ( x ) and their a .
k
3 9 2 3
2 x − 16 x = 4
0
3 9 3 M1 Attempting to find a quadratic equation in k in any form.
k − k2 = 3k 2 − 8k + 4 = 0
2 16 4
2 A1
( 3k − 2 )( k − 2 ) = 0 k = 2 , k =
3
2 A1 4
k= only with explanation k b= so reject k = 2 .
3 3
4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/65, October/November 2025 (9709_w25_ms_65), and the question paper (9709_w25_qp_65), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/65 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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