Mathematics 9709/61 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 random variables X and Y have independent distributions X ~ Po(3) and Y ~ Po(2) respectively.
(a) Find P( 2 1 X 1 5). [2]
(b) Find P (X + Y 2 2) . [3]
* 0000800000003 * DFD
3
(c) The total of 100 random values of X and 150 random values of Y is denoted by T.
Use a suitable approximating distribution to find P(T 1 560 ). [4]
* 0000800000004 * DFD
4
Mark scheme
1(a) 3 4 M1 Allow one end error.
e−3 ( 3 + 3 ) = e−3 (4.5 + 3.375) = 0.22404 + 0.16803 Allow incorrect λ.
3! 4!
= 0.392 (3 sf) A1
2
1(b) λ=5 B1 SOI. Need to be stating or using Poisson.
2 M1 Allow one end error and any λ.
1 − e−5 (1 + 5 + 5 ) = 1 – e−5 (1+5+12.5)
2!
= 1 – (0.0067379 + 0.0336897 + 0.0842243 )
= 0.875 (3 sf) A1 Alternative combinations method.
B1 for one correct product, M1 for correct structure
( incl 1 – … ) and probs of full set of totals = 1 or
probs of full set of totals = 2.
A1 for 0.875.
3
1(c) T ~ N(600, 600) B1 SOI.
559.5−600 [= −1.653] M1 For standardising. Must have sqrt.
600 Allow with no or incorrect continuity correction.
Φ(‘−1.653’) = 1 − Φ(‘1.653’) M1 For correct area consistent with their working.
= 0.0492 (3 sf) or 0.0491 A1
4
The mean mass of packets of Trueleaf tea is supposed to be 500 grams. An inspector wishes to test whether this value is correct. He weighs 60 randomly chosen packets and notes the mass, x grams, of each packet. The results are summarised as follows.
n = 60 / x = 29 970 / x 2 = 14 970 300
Test, at the 5% significance level, whether the population mean mass is 500 grams. [8]
* 0000800000005 * DFD
5
Mark scheme
2 est (μ) = 499.5 or 29970/60 B1
est (σ2 ) = 60 ( 14970300 – ‘499.5’2 ) or 1/59(14970300 – (29970)2 /60 ) M1 Biased var = 4.75 M0.
59 60
= 4.83 or 285/59 A1
H0 : Pop mean (or μ) = 500 B1
H1 : Pop mean (or μ) ≠ 500 Both. Not just ‘mean’.
'499.5'−500 M1 For standardising with their values.
' 4.83' Must have ÷√60 . Ignore cc s.
60
= –1.762 A1 Allow −1.778 (from biased variance)
accept 3 sf if nothing better.
‘1.762’ < 1.96 or −’1.762’ > −1.96 M1 For valid comparison.
0.0390 > 0.025 Allow 1.778 < 1.96 0.0377 > 0.025.
There is insufficient evidence that [mean ] mass is not 500g A1FT In context. Not definite. No contradictions.
FT from incorrect z.
µ only accepted if defined.
Accept CV method 499.5 > 499.44.
Using biased var or using incorrect hypothesis can
score a maximum of 6/8.
8
The data produced by a certain data entry firm always include a small number of incorrect characters that occur at random. The proportion of incorrect characters is denoted by p, and experience has shown that p = 0.0001. A particular data set from the firm contains 14 500 characters, of which X characters are incorrect.
(a) Use a suitable approximating distribution to find P(X 1 4) . [3]
The firm’s management wishes to decrease the value of p by giving their employees some training.
Their aim is that, for a data set containing 14 500 characters, the value of P(X = 0) for the new value of p should be double the value of P(X = 0) when p = 0.0001.
(b) Use a suitable approximating distribution to find the new value of p. [3]
* 0000800000006 * DFD
6
Mark scheme
3(a) λ = 1.45 B1 Need indication of Poisson.
2 3 M1 Allow one term extra or omitted or incorrect.
e−1.45 (1 + 1.45 + 1.45 + 1.45 ) = e−1.45 (1+1.45+1.05125+0.5081041) Allow incorrect lambda.
2! 3!
= 0.94[0] or 0.941 A1 SC use of Binomial(14500, 0.0001) to find
prob (< 4) = 0.940485 = 0.940 scores B2.
Unjustified 0.94[0] or 0.941scores B1B1.
3
3(b) e−14500 p seen or 2e−1.45 [= 0.46914 ] seen B1FT FT their λ from 3(a).
e−14500 p = 2e−1.45 [ −14500p = ln 2 + (−1.45)] M1 Forming an equation in p and attempting to solve.
FT their e−14500 p .
Factor of 2 on the wrong side of the equation can
still score M1.
p = 0.0000522 (3 sf) A1 OE.
Use of binomial reaching 0.0000522 scores SCB2.
3
The masses of a certain species of animal are known to be normally distributed with standard deviation v kg. A researcher obtains the masses of a random sample of n animals of this species and uses these masses to find two confidence intervals ( a % and 90%) for the population mean. The width of the a % confidence interval is 1.414 # the width of the 90% confidence interval.
(a) Find a . [3]
(b) Find the probability that the 90% confidence interval contains the population mean given that the a % confidence interval contains the population mean. [1]
* 0000800000007 * DFD
7
Mark scheme
4(a) [2 ×] z × σ = 1.414 × [2 ×] 1.645 × σ [ z = 2.326 ] M1 Form an equation in z (accept z = 1.414 × 1.645
n n OE).
Factor of 1.414 on the wrong side of the equation
can still score M1.
Condone 1.282 instead of 1.645 for M1.
2Φ(‘2.326’) –1 M1 OE.
α = 98 (3 sf) A1 Allow 98%.
3
4(b) 90 = 45 or 0.918 (3 sf) B1FT FT their α.
'98' 49 no ft if alpha < 90 in 4(a).
1
It is known that 20% of households in a certain country contain more than 4 people. Laxmi believes that, in her town, the percentage is lower than 20%. She chooses a random sample of 40 households in her town and notes the number which contain more than 4 people. She then carries out a test at the
2.5% significance level using a binomial distribution.
(a) Find the probability of a Type I error. [4]
(b) State the rejection region for the test. [1]
Laxmi finds that exactly 2 households in her sample contain more than 4 people.
(c) Explain why it is impossible for Laxmi to make a Type II error. [1]
* 0000800000008 * DFD
8
Mark scheme
5(a) P(X < 2) = 0.840 + 40×0.839 ×0.2 + 40 C2 ×0.838 ×0.22 M1 Attempt P(X ⩽ 2) or P(X ⩽ 3) using B(40, 0.2).
= 0.0001329 + 0.0013292 + 0.0064799 Need to see expressions.
= 0.00794 ( < 0.025 ) A1 SC unjustified X ⩽ 2 = 0.00794 scores M0B1.
P(X < 3) = [0.00794 + 40 C3 ×0.837 ×0.23 ] B1FT Correct term P(X = 3) added to P(X ⩽ 2) and one
= 0.0079421 + 0.0205199 = 0.0285 > 0.025 relevant comparison seen. Need to see expressions.
P(Type I) = 0.00794 (3 sf) with both relevant comparisons seen B1 Unjustified final answer 0.00794 scores SCB1B1.
4
5(b) Rejection region is X ⩽ 2 B1 OE.
1
5(c) H0 will be rejected B1
Or result lies in the rejection region
1
The masses, in kilograms, of large and small bags of potatoes have the independent distributions
N(2.5, 0.05) and N(0.8, 0.02) respectively.
(a) Find the probability that the total mass of a randomly chosen large bag of potatoes and a randomly chosen small bag of potatoes is more than 3.55 kg. [5]
* 0000800000009 * DFD
9
(b) Find the probability that the mass of a randomly chosen large bag of potatoes is less than 3 times the mass of a randomly chosen small bag of potatoes. [5]
* 0000800000010 * DFD
10
Mark scheme
6(a) N(2.5 + 0.8 , 0.05+ 0.02) = N(3.3, 0.07) B1B1 SOI B1 for N(3.3, …) B1 for 0.07 give at early stage
OE.
3.55− '3.3' [= 0.945] M1 For standardising with their values.
'0.07 '
1 – ɸ('0.945') M1 For area consistent with their values.
= 0.172 (3 sf) A1
5
6(b) E(D) = 2.5 – 3 x 0.8 = 0.1 B1 Give at early stage.
OE.
Var(D) = 0.05 + 32 ×0.02 = 0.23 M1 Give at early stage.
0 − ('0.1') M1 For standardising with their values.
[= −0.2085 or 0.209 or 0.208] Must have used the vars for denom.
'0.23'
Φ(‘−0.2085’) = 1 – Φ(‘0.2085’) M1 For area consistent with their values.
= 0.417 (3 sf) or 0.418 A1
5
The time, in minutes, taken by students to complete a test is modelled by the random variable X with probability density function
- 34 (x - 3) (x - 5) 3 G x G 5, f (x) = *
0 otherwise.
(a) Find the probability that a randomly chosen student takes longer than 4.5 minutes to complete the test. [4]
* 0000800000011 * DFD
11
(b) Write down the median of X. [1]
(c) Without performing an integration, use your answer to part (a) to find P( 3.5 1 X 1 4.5). [2]
* 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
7(a) [f(x) = − 3 (x2 − 8x + 15)]
4
5 M1 Attempt to integrate their f(x).
− 3 ( x 2 − 8 x + 15) dx
4
4.5 Condone missing -3/4.
3 5 A1 Correct integration and correct limits 4.5 and 5.
= − 3 x − 4 x 2 + 15 x OE e.g. 1– integration from 3 to 4.5
4 3 4.5 condone missing –3/4.
3 M1 Attempt substitute correct limits in correct integral
− 3 125 − 100 + 75 − ( 4.5 − 4 × 4.52 + 15 × 4.5) must have the –3/4.
4 3 3 OE.
May be implied by correct answer.
= 5 or 0.15625 or 0.156 (3 sf) A1
32
4
7(b) Median = 4 B1
1
7(c) 1 − 2 × ‘ 5 ’ or 2(0.5 − ‘ 5 ’) M1 Must see working. FT their 5 .
32 32 32
11 or 0.6875 or 0.687 or 0.688 (3 sf) A1FT FT their 5 .
16 32
SC not using 7(a) and integrating from 3.5 to 4.5
and getting 11/16 OE scores B1 only.
2
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/61 October/November 2025 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/61, October/November 2025 (9709_w25_ms_61), and the question paper (9709_w25_qp_61), to read or download, with each question's marking points written out underneath.
How many marks is 9709/61 out of?+
Mathematics 9709/61 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/61 answers marked?+
Yes. Upload your answers to 9709/61/O/N/25 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.