Mathematics 9709/65 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.
The random variable X has the distribution Po(1.5). The sum of three independent values of X is denoted by S.
(a) Find P(S G 3) . [3]
P(S G 2) 125
(b) Show that the exact value of is . [3]
P(S G 1) 44
Mark scheme
1(a) λ = 4.5 B1 SOI.
e-4.5(1 + 4.5 + 4.52! + 4.53! )= e −4.5 (1 + 4.5 + 10.125 + 15.1875 ) M1 Allow any λ; allow one end error; expression or terms
2 3
must be seen.
= 0.01111 + 0.04999 + 0.11248 + 0.16872
= 0.342 (3 sf) A1 SC Unsupported answer of 0.342 scores B1.
3
1(b) P(S ⩽ 1) = e–4.5(1 + 4.5) or 5.5e–4.5 M1 For either expression.
2
or P(S ⩽ 2) = e–4.5(1 + 4.5 + 4.52 ) or 125 e-4.5 Allow any λ , allow one end error.
8
P(S ⩽ 2) ÷ P(S ⩽ 1) [= 125
8
÷ 5.5] M1 Both expressions.
No end error, λ = 4.5 .
Division attempted.
Allow in decimal form for this mark.
P(S ≤ 2) 125 A1 AG.
P( S ≤1)
= 44
Must be exact, not decimal approximation.
No decimals seen (condone exact decimals).
4.52
Convincingly obtained, 1 + 4.5 and 1 + 4.5 +
2
evaluated.
3
A random sample of 200 values of a random variable X gives the following results.
n = 200 / (x - 2) = 60 / ( x - 2 ) 2 = 20
(a) Find a 95% confidence interval for the population mean of X. [6]
(b) State which part of your solution to part (a) makes use of the Central Limit Theorem. [1]
Mark scheme
2(a) 60 B1 Can be given at an early stage.
Unbiased estimate of mean of x = 2 + = 2.3
200
200 20 M1 Biased estimate scores M0.
Unbiased estimate of variance of ( x − 2 ) = − '0.3'2 or
Accept use of Σ ( x − x ) correctly substituted.
2
199 200
1 602
20 −
199 200
2 A1 Condone 0.01 , from unbiased estimate.
0.01005... = 0.0101 (3 s.f.) or
199
'0.01005' M1 FT their mean and variance.
CI for x is ‘2.3’ ± z× 200 Must be a z value (note: if only one side of the interval
still allow M1).
z = 1.96 B1 Seen.
CI for x is 2.29 to 2.31 (3 sf) A1 Must be an interval.
SC Use of biased variance (0.01) can score A1FT max
4/6.
6
2(b) Use of z = ‘1.96’, or Value of z used [comes from normal distribution]. B1 Must refer to z (i.e. ‘value of z ’ or ‘ 1.96 ’).
Not "X not normally distributed".
Not "Assume that X is normally distributed".
1
Maroulla’s calculator can generate random numbers between 0.000 and 0.999 inclusive, correct to
3 significant figures. She plans to use her calculator to choose a sample of members from the 851 members in her health club. She numbers the members from 1 to 851. Then she uses her calculator to generate some random numbers. She multiplies each random number by 851 and rounds up to the next whole number to give the number of a member in the sample. This is called a ‘member number’.
(a) Maroulla’s first random number is 0.401.
Find the member number that is produced by this random number. [1]
(b) Find all possible random numbers, correct to 3 decimal places, that would produce the following member numbers.
(i) A member number of 680. [1]
(ii) A member number of 850. [1]
(c) Explain briefly how your answers to part (b) show that Maroulla’s method does not produce a random sample. [1]
Mark scheme
3(a) 342 B1 Only.
1
3(b)(i) 0.798, 0.799 B1 Only.
1
3(b)(ii) 0.998 B1 Only.
1
3(c) 680 and 850 are not equally likely to be chosen for the sample. B1 Or a sample including 680 is more likely to be selected
than a sample including 850.
Must relate to part (b).
Allow FT if number of solutions in (i) and (ii) are
different.
1
The numbers of cars and trucks arriving per minute at a fuel station are modelled by independent variables with distributions Po(0.8) and Po(0.5) respectively.
(a) Find the probability that at least 4 cars and at least 2 trucks arrive at the fuel station during a randomly chosen 5-minute period. [4]
(b) Use a suitable approximating distribution to find the probability that a total of fewer than 145 cars and trucks arrive at the fuel station during a randomly chosen 2-hour period. [5]
Mark scheme
4(a) 32 M1 Any λ.
1 – e-4 (1 + 4 + 42! + 43! ) 1 − e−4 1 + 4 + 8 +
2 3
3 Allow one end error.
Must see full expression or terms.
= 1 − ( 0.018316 + 0.073263 + 0.14653 + 0.19537 ) [= 0.56653]
1 – e–2.5(1 + 2.5) = 1 − ( 0.08208 + 0.20521) [= 0.71270] M1 Any λ.
Allow one end error.
Must see full expression or terms.
‘0.56653’ × ‘0.71270’ M1 Multiply their values.
= 0.404 (3 sf) A1 SC1 Unsupported answer of 0.404 scores B3.
SC2 Unsupported 0.567 and 0.713 scores B1, leading
to M1A1 as per scheme.
4
4(b) λ = 156 B1
N(156, 156) B1FT Seen or implied.
144.5 − '156' M1 For standardising with their values.
[= –0.921] Allow with wrong or no cc.
'156'
Φ(‘–0.921’) = 1 – Φ('0.921') M1 Finding the area consistent with their value.
= 0.179 or 0.178 (3 sf) A1
5
Candidates for a certain diploma take two tests. Their marks for the first test and the second test are modelled by the independent variables with distributions N(38.1, 3.8 2) and N(64.0, 6.1 2) respectively.
The final mark, F, for each candidate is found by doubling the mark in the first test and adding the result to the mark in the second test.
Find the probability that the mean, F , of the final marks of a random sample of 25 candidates is greater than 143. [5]
Mark scheme
5 E(F) = 140.2 B1
Var(F) = 4 × 3.82 + 6.12 [= 94.97] M1 Must be a combination of 3.8 and 6.1.
143− '140.2 '
(= 1.437) or
143+ 501 − '140.2'
[= 1.447] M1 For standardising with their values.
' 94.97 ' ' 94.97 '
25 25
Must have 25 .
Allow incorrect continuity correction.
1 – Φ('1.437') or 1 – Φ('1.447') M1 Finding the area consistent with their value.
Allow attempted continuity correction
= 0.0754 (3 sf) or = 0.0739 or 0.0740 (3 sf) A1 Allow 0.074.
5
It is claimed that 28% of voters in a certain town support the Forward Now political party. A researcher suspects that the true figure is less than 28%. She interviews a random sample of 30 voters from the town and she finds that 4 voters in the sample say that they support the Forward Now party. She plans to carry out a hypothesis test at the 10% significance level.
(a) Use a binomial distribution to carry out the test. [5]
(b) State, with a reason, whether it is possible that a Type I error was made in carrying out the test.
[1]
Later the researcher carries out a similar test at the 10% significance level, using a new random sample of 30 voters from the town.
(c) Find the probability of a Type I error. [2]
Mark scheme
6(a) H0: p = 0.28 H1: p < 0.28 B1
P(X < 4) = 0.7230 + 30×0.7229×0.28 + 30C2×0.7228×0.282 + 30C3×0.7227×0.283 M1 For attempting to find P ( X ≤ 4 ) using B ( 30,0.28) ,
+ 30C4×0.7226×0.284 no end errors.
= 0.0000525 + 0.0006122 + 0.0034524 + 0.012531 + 0.03893
Expression or terms must be seen.
= 0.0495 (3 sf) A1 SC 0.0495, no working: B1.
‘0.0495’ < 0.1 M1 Valid comparison of their 0.0495 (must be a tail
probability) with 0.1 .
[Reject H 0 ] There is sufficient evidence [at 10% level] to suggest that the A1FT FT their 0.0495 .
percentage [who support Forward Now] is less than 28%. No contradictions, in context, not definite.
5
6(b) Yes, because H0 was rejected. B1FT OE.
FT their conclusion from 6(a).
1
6(c) P(X ⩽ 5) = 0.116 (3 sf) B1
P(Type I) = 0.0495 B1
2
A firm makes a certain type of battery-powered toy. The battery life is denoted by X hours and the population mean of X is supposed to be 12. The Quality Control department wished to test whether the population mean of X is actually less than 12. They tested a random sample of 50 of these toys and found that the sample mean, X , was 11.4.
(a) State suitable null and alternative hypotheses for the test. [1]
You may assume that the standard deviation of the battery life is 2.3 hours.
(b) Show that the value X = 11.4 leads to rejection of the null hypothesis at the 5% significance level.
[2]
(c) It is given that the value X = 11.4 leads to rejection of the null hypothesis at the a% significance
level.
Find the set of possible values of a . [2]
Mark scheme
7(a) H0: Population mean (or μ) = 12 B1 Accept population mean, but not just mean.
H1: Population mean (or μ) < 12
1
7(b) 11.4 −12 M1 For standardising.
2.3÷ 50 [= –1.845] Must have 50 .
–1.845 < –1.645 or 1.845 > 1.645 or 0.0325 < 0.05 A1 OE.
[Hence H 0 is rejected] Correct comparison seen.
No contradictions.
2
7(c) 1 – Φ('1.845') [= 0.0325 3 sf] M1 Or P(X < 11.4) = 0.0325 or 3.25% (3 sf).
May be implied.
α > 3.25 (3 sf) A1 Or α ⩾ 3.25 (3 sf).
Allow α > 3 with correct working.
0.0325 scores M1A0.
Condone 3.25%.
2
f (x)
a x
O
The diagram shows the graph of the probability density function f of a random variable X. Between x = 0 and x = a the graph consists of a straight line through O with gradient k, where k and a are
positive constants. Elsewhere f (x) = 0 .
It is given that the median of X is 2.
(a) Find the value of k. [2]
(b) Find the value of E(X ). [4]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
8(a) 2 M1 Using the median =0.5.
1
2
× 2×k 2 = 12 or
0
kxdx = 12 Must attempt integration for ‘or’ method.
Correct limits and = 0.5 for ‘or’ method.
For a method involving k and a , M0 until a is
evaluated.
k = 12 A1
2
8(b) a M1 FT their k.
1
2
× a × ' 12 ' a = 1 or
0
' 12 ' xdx = 1
a=2 A1FT FT their k.
Can be seen in 8(a).
'2' M1 Attempt to integrate x × (their k)x, limits 0 to their a.
'2'
' 12 ' x 2 dx [= x6 ]
3 a
kx dx with integration attempted.
2
0 Allow
0
0
= 43 or 1.33 (3 sf) A1
4
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Mathematics 9709/65 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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