Mathematics 9709/62 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.
The number, X, of used computers donated to a charity has a constant average rate of 2.4 computers per week.
(a) State a necessary condition for X to have a Poisson distribution. [1]
Now assume that X has a Poisson distribution.
(b) Calculate the probability that the number of computers donated during a 4-week period is more than 6 and less than 9. [3]
(c) Use a suitable approximating distribution to calculate the probability that more than 50 computers are donated during a 20-week period. [4]
* 0000800000003 * DFD
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Mark scheme
1(a) Computers are donated independently or singly or randomly B1 Context i.e. ‘computers’ or ‘donations’ must be
mentioned. ‘Events/trials/X/It is independent’ B0.
ISW after a correct statement.
1
1(b) λ = 9.6 B1
7
9.68 M1 Allow one end error and incorrect λ.
e−9.6 ( 9.6
7! + 8! ) = e
-9.6 (1490.97 + 1789.16) = 0.10098 + 0.12118
Accept fully correct sigma notation.
= 0.222 (3 sf) A1 As final answer.
SC Unjustified 0.222 scores B1M0B1.
3
1(c) N(48, 48) B1 SOI.
50.5− 48
[= 0.361] M1 For standardising with their N (…, …)
48 Ignore missing or incorrect cc for M1.
1 − Φ(‘0.361’) M1 For area consistent with their working.
= 0.359 (3 sf) A1
4
The random variable X has a normal distribution with mean 10 and standard deviation 3. The independent random variable Y has a Poisson distribution with mean 4.
(a) Find the standard deviation of X + Y . [3]
(b) Find the standard deviation of 5X - Y . [3]
* 0000800000004 * DFD
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Mark scheme
2(a) Var(Y) = 4 B1 SOI.
Accept sd = 2 for B1 if var = 4 not seen.
Var(X + Y) = 32 + ‘4’ M1 SOI.
FT their Var(Y).
sd of X + Y = 13 = 3.61 A1
3
2(b) Var(5X ) = 52 × 32 B1 May be implied.
Var(5X − Y) = Var(5X) + Var(Y) = “52 × 32 ” + “4” M1 ft their Var (5X), and their Var(Y) from 2(a).
sd of 5X − Y = 229 or 15.1 (3 sf) A1
3
The times, in minutes, taken by students to complete a test have mean n and standard deviation v. The times taken by a random sample of 100 students are noted and are used to calculate a 95% confidence interval for n.
(a) Given that the end points of the 95% confidence interval are 31.02 and 33.98, correct to
4 significant figures, calculate the value of v. [3]
(b) The calculation of the confidence interval required the use of the Central Limit theorem.
Explain why it is valid to use the Central Limit theorem in this case. [1]
(c) A researcher calculates a number, r, of 95% confidence interval for n.
Find the largest value of r such that the probability that all r confidence intervals contain the true value of n is greater than 0.5. [4]
* 0000800000005 * DFD
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Mark scheme
3(a) Sample mean = 32.5 soi or Width = 2.96 B1 SOI.
Could be implied.
32.5 + 1.96 × σ
= 33.98 or 2.96 = 2 × 1.96 × σ M1 OE.
100 100
Factor of 2 error can still score M1.
Accept z=1.96 or 1.645 for M1.
σ = 7.55 (3 sf) A1 (Final answer of 15.1 scores 2/3).
3
3(b) Because the sample size is large. B1 Accept n is large. Accept n > 30 condone > 50
‘Number of students is large’ B0 need ‘sample’ or
100.
1
3(c) 0.95r > 0.5 Allow ‘=’ M1 OR 0.95r evaluated for any positive integer r M1.
rln 0.95 > ln 0.5 Allow ‘=’ Correctly take logs and use log rule A1 OE.
0.9513 = 0.51… A1.
r < 13.5…. Allow ‘=’ A1 0.9514 = 0.48… A1.
Largest value of r is 13 A1 (Condone incorrect inequality signs throughout).
4
An inspector believes that 18% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%. The inspector examines a random sample of 40 cups and finds that 3 of them contain flaws.
(a) Stating a necessary assumption, use a binomial distribution to test the factory owner’s claim at the
5% significance level. [6]
(b) Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a). [1]
* 0000800000006 * DFD
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Mark scheme
4(a) Assume flaws in cups are independent OR prob of containing a flaw is the same B1 OE in context (e.g. prob… is consistent) but p =
for all cups 0.18 is B0.
H0 : P(contains flaw) = 0.18 B1 Both. Allow ‘p’
H1 : P(contains flaw) < 0.18
[P(X ⩽ 3) =] 0.8240 + 40×0.8239 ×0.18 + 40C2 ×0.8238 ×0.182 + 40 C3 ×0.8237 ×0.183 M1
=0.0003569 + 0.0031338 + 0.013414 + 0.037298
0.0542 (3 sf) A1 Unsupported 0.0542 scores M0B1.
0.0542>0.05 M1 Valid comparison – must be a tail probability
Or if CR found (⩽2) then comparison with 3 M1.
[Accept Ho ] Insufficient evidence to accept factory owners claim. A1FT In context, not definite, no contradictions
[Insufficient evidence that percentage less than 18%] no use of ‘p’ unless defined.
e.g. not ‘%age is not less than 0.18’ or ‘%age is
18%’
2TT scores max B1B0M1A1M1(with 0.025) is A0.
Normal approx’n: (First B1B1* if earned) then:
CV = 3.5− 40×0.18 = ±1.523 SCB1 (need cc).
40×0.18×0.82)
1.523< 1.645 or 0.0639>0.05 No evidence less than
18% SCB1.
*hypotheses µ=7.2 and µ<7.2 can score if Normal
approx. used.
6
4(b) np = 7.2 which is too large, or > 5. B1 Context required (7.2 or 40 or 0.18).
Or n = 40 which is not greater than 50. One correct condition required ISW.
Or p = 0.18 which is not small or > 0.1.
1
A random variable X has probability density function given by
k (2x 2 - x 3) 0 G x G 2, f (x) = )
0 otherwise.
3
(a) Show that k = . [3]
4
* 0000800000007 * DFD
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The median of X is denoted by m.
(b) (i) Write down the value of P(X G m) . [1]
(ii) Hence find P(E(X ) G X G m) . [5]
* 0000800000008 * DFD
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Mark scheme
5(a) 2 M1 Attempt to integrate f(x) & = 1, ignore limits.
k (2 x 2 − x3 ) dx = 1
0
2 A1 Correct integration and limits.
k 23x − x4
3 4
=1
0
k × 43 = 1 k = 34 A1 AG.
Correctly obtained, at least one step shown and no
errors seen.
3
5(b)(i) 0.5 B1 Could be in a correct expression.
1
5(b)(ii) 2 M1 Attempt to integrate xf(x), ignore limits condone k
E(X) = 34 (2 x3 − x 4 ) dx
0
or missing k.
2 A1
= 34 24x − x5
4 5
= 65
0
'1.2' M1 Attempt to integrate f(x), limits from 0 to their E(X)
P(X < E(X)) = 34
0
(2 x 2 − x3 ) dx (OR integrate f(x) limits from their E(X) to 2).
'1.2' A1 OR 0.5248 or 656/1250 or 328/625.
= 34 23x − x4
3 4
594
= 1250 or 297/625 or 0.4752
0
P(E(X) ⩽ X ⩽ m) = 0.5 − ‘0.4752’ OR 0.5248 – 0.5.
= 0.0248 or 31/1250 A1 CWO.
SC solutions attempting to find the value of m
(1.22854) 1.229 Must be correct.
Can score M1A1 then B1 for integrate correctly
from 1.2 to 1.229 (accept 3sf here for 1.229) and
0.0248 to 0.0252 obtained B1.
(Incorrect values of m could score first M1A1 only).
5
The weekly profit, in dollars, made by a certain firm has a normal distribution. In the past, the weekly profit had the distribution N(736, 262). Following a change in management, the mean weekly profit for
35 randomly chosen weeks is $725.
(a) Stating a necessary assumption, test at the 2% significance level whether the mean weekly profit has decreased. [6]
* 0000800000009 * DFD
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The mean weekly profit for another random sample of 35 weeks is found and a similar test is carried out at the 2% significance level.
(b) State the probability of a Type I error. [1]
(c) Given that the mean weekly profit is now in fact $718, find the probability of a Type II error. [5]
* 0000800000010 * DFD
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Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
DFD
11
* 0000800000012 * DFD
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Mark scheme
6(a) Assume sd is still 26 or is unchanged B1
H0 : Population mean = 736 B1 Both. Allow ‘µ’, but not just ‘mean’.
H1 : Population mean < 736
±
725− 736 M1 For standardising (ignore cc attempts).
26
35
= −2.503 A1 (± ) Accept 3sf if nothing better seen.
−2.503 < −2.054(or 5 or 6) or 0.0062 < 0.02 M1 Valid comparison.
[Reject Ho] There is sufficient evidence that (mean) weekly profit has decreased A1 In context, not definite, no contradictions.
No use of µ unless defined.
e.g. not ‘Mean weekly profit has decreased.’.
2TT scores max B1B0M1A1M1 (comparison 2.326
or 0.01) A0.
CV method: 726.973=727 M1 A1 > 725 and concl
M1A1.
Condone: 734.027 =734 M1 A1 < 736 and concl
M1A1.
6
6(b) 0.02 B1
1
6(c) a − 736
26
= −2.054 or -2.055 M1 For standardising to find CV Must be – 2.054 OE.
35
(2TT in part a accept –2.326 in expression for M1).
a = 726.973 A1 CWO Accept 727 or better (working may be in
6(a)).
'726.973' − 718
26
(= 2.042) M1 For standardising with their cv and 718.
35
= 1 – ɸ('2.042') M1 For area consistent with their working.
= 0.0203 to 0.0207 A1 CWO.
Note: (736–718)/(26/√35) or (725–718)/(26/√35) or
similar scores M0A0M1 and possibly M1. Max 2/5.
5
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Mathematics 9709/62 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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