Mathematics 9709/61 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
It is known that 1% of houses in a certain area have a wind turbine. A random sample of 400 houses in this area is chosen for a survey on domestic heating. The number of houses in the sample that have a wind turbine is denoted by X.
Use a suitable approximating distribution to find P (X G 3) . [3]
3
Mark scheme
1 (λ ) = 4 B1 SOI.
4 2 43 M1 Allow any λ, allow one end error.
e −4 1 + 4 + + = e −4 (1 + 4 + 8 + 10.6666)
2! 3!
= 0.018316 + 0.073263 + 0.146525 + 0.19537
= 0.433 A1 SC1 Unsupported answer of 0.433 scores B1B1.
SC2 Use of Normal mean = 4 B1.
SC3 Use Binomial 0.432 or 0.433 scores B1.
3
The random variable X has the distribution B b8, l. A random sample of 100 values of X is chosen, and
4 the sample mean, X , is found.
(a) Find P (X 2 6.2) . You are not expected to use a continuity correction. [6]
(b) State why the Central Limit Theorem was needed in the calculation in part (a). [1]
Mark scheme
2(a) N (6, ….) B1 Normal used with a mean of 6.
3 6 M1 Var of 1.5 (or sd found).
σ2 = = 1.5 (or σ = 1.5 or or 1.22 (3sf)
2 2
6.2 - 6 M1 For standardising. Must have 100 . Ignore
1.5 incorrect continuity correction (correct cc if used is
100 1/2n).
= 1.633 A1 Or accept 1.674 with correct cc.
( )
P X > 6.2 = 1 − Φ(‘1.633’) M1
= 0.0512 or 0.0513 (3sf) A1 Or accept 0.0471 with correct cc (scores full marks).
6
2(b) Population (X) is not normally distributed / Population (X) is Binomial B1 OE.
Accept underlying distribution.
1
The time, T minutes, for a certain daily bus journey is normally distributed. The bus company claims that the mean of T is 45. A passenger believes that the mean of T is actually greater than 45. She notes the times taken for this journey on a random sample of 60 days. The results are summarised below.
n = 60 / t = 2750 / t 2 = 127000
(a) Calculate unbiased estimates of the population mean and variance. [3]
(b) Test the passenger’s belief at the 5% significance level. [5]
Mark scheme
3(a) 2750 275 B1
Est(μ) = or or 45.8 (3 sf)
60 6
60 127000 275
2 M1 1 27502
Est( σ 2 ) = − Or 127000 −
59 60 6 59 60
Note: σ (4.03) can score M1 for correct expression
Use of biased (15.97) M0.
2875 A1
=16.2 (3 sf) or
177
3
3(b) Ho: Population mean time = 45 H1: Population mean time > 45 B1 Allow ‘μ’ but not just ‘mean’.
'275' M1 Standardise using their values from 3(a).
- 45
6 Must have 60 (ignore cc).
'16.24294'
60
= 1.602 to 1.595 A1 Accept 3sf 1.6(0) if nothing better seen.
(or area = 0.0546 to 0.0553).
FT Biased in 3(a) scores A1 for 1.608 to 1.615.
‘1.602’ < 1.645 M1 Or compare areas.
i.e. 0.0546 to 0.0553 > 0.05.
[Accept H0] A1FT OE.
There is insufficient evidence [at 5% level] to reject the company’s claim FT their z-calc.
OR There is insufficient evidence to accept the passenger’s belief No contradictions, In context, Not definite, e.g. not
OR There is insufficient evidence that the mean time is more than 45 minutes ‘Mean time is more than 45 mins’.
Note: accept cv method (45.856 > 45.83 or 44.98 <
45).
5
At an entertainment centre, the cost for using a particular video game is $0.40 per minute. The number of minutes for which people use the video game has mean 15 and variance 9.
(a) Find the mean and variance of the amount people pay for using the video game. [3]
Each day, 35 people independently use the video game.
(b) Find the mean and variance of the total amount paid by 35 people. [3]
Mark scheme
4(a) Mean = 6 [$] B1 Ignore units in both parts of question 4.
Variance = 0.402 × 9 M1
= 1.44 [$2] A1
3
4(b) Mean = 35 × ‘6’ = 210 [$] B1FT FT their mean in 4(a)
Variance = 35 × ‘1.44’ M1 FT their variance of cost for one person.
= 50.4 [$2] A1
3
(a) The random variables W and X have the independent distributions Po(1.2) and Po(2.3) respectively.
(i) Find P (3 G W + X G 5) . [2]
The random variable S is the sum of 100 independent values of W and 200 independent values of X.
(ii) Use a suitable approximation to find P (S 2 600) . [6]
(b) The random variable Y has the distribution Po (m) , where m 2 0 .
5
It is given that P (Y = 3) + P (Y = 4) = P (Y = 5) .
2
Find the value of m. [3]
Mark scheme
5(a)(i) 3.53 3.54 3.55 M1 Any λ, allow one end error.
e-3.5 + + -3.5
= e (7.1458 + 6.2526 + 4.37682) = Must see expression or terms.
3! 4! 5!
0.21579 + 0.18881 + 0.13217
= 0.537 A1 SCB1 for unsupported answer of 0.537.
2
5(a)(ii) Use of N( 1.2 ×100 + 2.3 × 200, …) = N(580, ….) B1 SOI give at early stage.
Var(S) = 1.2 × 100 + 2.3 × 200 = 580 M1 SOI give at early stage.
600.5 − '580' M1 For standardising with their values.
'580' Allow with omitted or incorrect continuity
correction.
= 0.851 A1
1 – Φ(‘0.581’) M1 For area consistent with their working
= 0.197 (3sf) A1
6
5(b) 5 λ3 λ 4 λ5 5 λ3 λ4 λ5 B1 Accept with or without each term multiplied by e-λ.
× + = Or e− λ +e − λ = e − λ
2 3! 4! 5! 2 3! 4! 5!
λ 2 − 5λ − 50 = 0 M1 Legitimately obtain quadratic in λ.
5 ± 25 + 4 × 50 A1
( λ − 10)( λ + 5) = 0 or λ =
2
λ = 10 Alone.
3
A manufacturer of cell phones claims that 25% of students own a Pumpkin phone. Jeyeraj thinks that the proportion of students at his large college who own a Pumpkin phone is less than 25%. He plans to test the manufacturer’s claim. He chooses a random sample of 30 students at his college. If the number of students who own a Pumpkin phone is less than 5, Jeyeraj will reject the manufacturer’s claim.
(a) State suitable hypotheses for the test. [1]
(b) Given that the true proportion of students at the college who own a Pumpkin phone is 10%, use a binomial distribution to find the probability of a Type II error. [3]
At Florence’s college, in a random sample of 40 students, it was found that 5 own a Pumpkin phone.
(c) Calculate an approximate 95% confidence interval for the proportion of students at Florence’s college who own a Pumpkin phone. [3]
Mark scheme
6(a) H0: Proportion (at college) owning Pumpkin phone = 0.25 B1 Allow p = 0.25, p < 0.25. Allow 25%.
H1: Proportion (at college) owning Pumpkin phone < 0.25
1
6(b) B(30, 0.1) and P(X ⩾ 5) attempted M1 May be implied.
1− (0.930 + 30×0.929×0.1 + 30C2×0.928×0.12 + 30C3×0.927×0.13 + 30C4×0.926×0.14) M1 For expression or terms. No end errors.
= 1 – (0.042391 + 0.141304 + 0.22766 + 0.236088 + 0.177066)
= 0.175 (3 sf) accept 0.176 A1 SC Unsupported working and correct answer scores
M1B1.
3
6(c) 1 7 M1 Any z must be a z.
× Only one side calculated can score M1.
1 8 8
±z
8 40
z = 1.96 B1 Seen.
0.0225 to 0.227 (3 sf) A1 Must be an interval.
3
X is a random variable with probability density function given by
(1 + cos rx) 0 G x G 1, f (x) = )
0 otherwise.
1 1 1
(a) Show that P b X 1 l = + . [3]
2 2 r
1 2
(b) Show that E (X ) = - . [5]
2 r2
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) 1 M1 Attempt to integrate f(x). Ignore limits.
( 1+ cosπ x ) dx
2
0
1 A1
sin π x 2 Correct integration and limits.
= x +
π 0
π A1 AG.
sin Must see intermediate step.
1
= + 2 = 1+1
2 π 2 π Convincingly obtained. No errors seen.
3
7(b) 1 M1* Attempt to integrate xf(x). Ignore limits.
( x + x cos π x ) dx
0
x2
1 M1* Attempt to integrate by parts to reach expression
sin π x sin π x
= + x× - dx with at least two terms correct (must be more than
2 π π 0 two terms).
May be implied by next line.
x2
1 A1
sin π x cos π x
= + x× +
2 π π 2 0
1 sinπ cosπ cos0 DM1 Substitute correct limits, dep M1M1.
= + + 2 - 2
2 π π π
1 2 A1 AG.
= − Legitimately obtained. No errors seen
2 π2
5
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/61, May/June 2025 (9709_s25_ms_61), and the question paper (9709_s25_qp_61), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/61 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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