Mathematics 9709/63 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.
At a certain shop, customers arrive independently and randomly at a constant average rate of 23.4 per hour.
(a) Find the probability that, in a randomly chosen 1-minute period, at least 2 customers arrive. [3]
(b) The random variable X denotes the number of customers who arrive in a randomly chosen 1-hour period.
(i) State a suitable approximating distribution for X, giving the value(s) of any parameter(s). [2]
(ii) Use your approximating distribution to find P (20 1 X 1 30) . [3]
Mark scheme
1(a) 23.4 B1 First B1 only scored if see 0.39.
λ= or 0.39
60
1 − e−0.39(1 + 0.39) = 1 – ( 0.67706 + 0.26405 ) M1 Any λ. Allow one end error.
= 0.0589 (3 sf) A1 SC unsupported answer score B1 instead of M1A1.
3
1(b)(i) N(23.4, 23.4) B1B1 B1 for N(23.4, ….) B1 for Var(X) = 23.4.
Note: marks for (i) cannot be recovered from (ii).
2
1(b)(ii) 20.5 − 23.4 29.5 − 23.4 M1 Attempt to standardise both with their values.
[= –0.5995] [= 1.2610] Must have square roots.
23.4 23.4
Allow no (or incorrect) continuity correction(s).
Φ(‘1.2610’) − Φ(‘−0.5995’) ( 0.8964 – 0.2743 ) M1 For attempt at area between 20 and 30 consistent
with their working.
= 0.622 (3 sf) A1 SC no working seen can score B2 for 0.622.
3
The lengths of pencils made at a factory are normally distributed. The standard deviation of the lengths is v cm , and the mean is supposed to be 10 cm. An inspector thinks that the mean is actually greater than 10 cm. He takes a random sample of 50 pencils produced at the factory and finds that the mean of these 50 lengths is 10.03 cm. He then carries out a hypothesis test.
(a) He finds that the value of the test statistic z is 1.995 correct to 3 decimal places.
(i) Calculate the value of v. [3]
(ii) Carry out the hypothesis test at the 2.5% significance level. [3]
(b) Explain whether it was necessary to use the Central Limit Theorem in carrying out the test. [1]
Mark scheme
2(a)(i) 10.03 − 10 M1 For standardising.
= 1.995 Must have square root 50.
σ
50
M1 Equating to 1.995 (not phi 1.995).
= 0.106 (3 sf) A1
3
2(a)(ii) H0: µ = 10 H1: µ > 10 B1 Accept if seen in part (i) but not in (ii).
Accept population mean but not just mean.
1.995 > 1.96 M1 Valid comparison.
0.977 > 0.975 or 0.023 < 0.025.
Accept 2.00 or 2.001 > 1.96.
[Reject H0 ] There is sufficient evidence [at 2.5% level] to suggest that mean A1 No contradictions. In context. Not definite.
length is greater than 10 cm. Accept ‘….. that mean length has increased’
Or There is sufft evidence to support the inspectors thought /claim. Note 2 tail test scores B0M1A0.
3
2(b) No, because population distribution [of lengths] is normal B1
1
A machine dispenses coffee into cups. The volume, V cm 3 , of coffee in a cup was measured for a random sample of 150 cups. The results were summarised as follows.
/ v = 46 350 / v 2 = 14 410 800
(a) (i) Calculate unbiased estimates of the population mean, n, and population variance, v 2. [3]
(ii) Calculate a 95% confidence interval for n. [3]
Another random sample of n cups of coffee is taken, where 100 1 n 1 120 . A 95% confidence interval for n is calculated using this sample. You may assume that, for large samples, unbiased estimates of v 2 are very similar.
(b) Without calculation, state whether this confidence interval would be wider or narrower than the confidence interval found in part (a)(ii). Give a reason for your answer. [2]
Mark scheme
3(a)(i) 46350 B1
Est(μ) = or 309
150
150 14410800 46350 150
2
M1 Use of correct formula.
( )
Est σ 2
=
149 150
− = 149 × 591
150
1
( 14410800 –
463502
).
149 150
OE.
= 595 (3 sf) or 88650/149 A1
3
3(a)(ii) z = 1.96 B1 Seen.
'595' M1 Any z.
309 ± z × Use of correct formula.
150
= 305 to 313 (3 sf) A1 Use of biased (591) gives same interval but scores
B1M1A0.
Unsupported correct answer scores max B1B1.
3
3(b) Either Smaller sample means more uncertainty about the value of µ. OE.
1 1 B1* OE.
or n < 120 means that >
n 150
Therefore Wider B1dep Dep on previous B1.
2
Emma needs to choose one person at random from three people, P, Q and R. She plans to throw two fair coins and note the number, n, of heads. If n is 0, she will choose P. If n is 1, she will choose Q. If n is 2, she will choose R.
(a) By considering probabilities, show that the choice made by this method is not random. [2]
Later, Emma has to choose two people at random from three people.
(b) Describe how Emma could use a single throw of a fair six-sided dice to make this random choice.
[2]
Mark scheme
4(a) The probabilities are not equal B1 OE.
SOI.
1 1 B1 Numerical justification or there is one way to get 0
e.g. P ( n = 0 ) = and P ( n = 1) = or P ( n = 1) = 2 × P ( n = 0 ) or similar but two ways to get 1, or similar B1B1.
4 2
2
4(b) Choose PQ or QR or RP soi or reject R or P or Q soi B1 Identifying the three correct cases to choose or reject
and correctly identifying choosing or rejecting.
eg 1 or 2 [ choose ]PQ. 3 or 4 [choose] QR. 5 or 6 [choose ]RP OE.
or 1 or 2 [reject] R. 3 or 4[ reject] P. 5 or 6 [reject] Q B1 OE.
Using the score on the dice to give each of the three
cases equal probability.
More than one throw of the dice scores B0B0.
2
In Urberia, the masses, in kilograms, of men have the distribution N(70.3, 5.9 2 ). A certain footbridge in
Urberia can take a maximum safe load of 1500 kg. When n men stand on the bridge, the probability that the bridge is unsafe is less than 0.01.
Stating a necessary assumption, find the maximum value of n. [4]
Mark scheme
5 Assume the men on the bridge are a random sample B1 OE.
Allow: assume that all the men on the bridge are
independent of one another.
1500 − 70.3n M1 Must have square root n.
= 2.326 Allow wrong z but must be a z value.
5.9 n
70.3n + 13.7234 n − 1500 = 0 A1 Correct quadratic equation or inequality in square
Or 4942.09 n2 - 211088.33 n + 2250000 = 0 root n.
OE.
[ n = 4.5226 (3 sf)] Max value of n is 20 A1 CWO.
4
A random variable X has probability density function given by
ax 0 G x G b, f (x) = *
0 otherwise, where a and b are constants.
2
(a) Show that a = . [3]
b2
(b) Show that P (X 1 E (X )) = 49 . [6]
Mark scheme
6(a) 1 b M1 Attempt triangle area or integrate f(x) (ignore limits)
2
× b × ab = 1 or
a xdx =1
0 and = 1.
x2
b A1 Correct integration.
a =1 ½ × b × ab = 1.
2 0
ab 2 2 A1 Convincingly obtained. No errors seen.
=1 a = 2
2 b
3
6(b) 2
b M1 Attempt to integrate xf ( x ) . Ignore limits.
E ( X ) = 2 x 2 dx
b 0
b A1 Correct integration and limits.
2 x3
=
b2 3 0
2 A1 OE.
= b E.g. ab3 /3
3
2
b M1 OE.
Integrate f ( x ) with limits 0 and their E(X)
3
2
P ( X < E ( X ) ) = 2 xdx
b 0 2
Condone a instead of
b2
1 2b 2
Or using areas, prob = × × ab
2 3 3
2
b A1FT Correct integration and limits, FT their E ( X ) .
2 x2 3
= 2 2
b 2 0 Need .
b2
1 2b 2 2b
Or prob = × ' '× × 2 .
2 3 3 b
4 2 A1 Must see correct expression after substituting limits.
b Convincingly obtained. No errors seen.
2 9 4
= 2× = Or expression correctly simplified.
b 2 9
6
In the past, one quarter of job applicants at a certain firm had first-class degrees. A change is made in the job description and a director of the firm believes that, on average, the proportion of job applicants with first class degrees has decreased.
In the month following the change, there were 35 job applicants, and r of these had first-class degrees.
The firm carried out a hypothesis test at the 4% significance level to test the director’s belief.
(a) Use a binomial distribution to find the largest value of r that would provide sufficient evidence that the director’s belief is correct. [6]
In another month, the director carries out a similar test at the 4% significance level using the 35 job applicants from that month.
(b) Explain the meaning of a Type I error in this context, and state the probability of a Type I error.
[2]
(c) Given that the proportion of job applicants with first class degrees this year is actually 0.05, find
the probability of a Type II error. [2]
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) B(35, 0.25) M1 SOI.
0.7535 + 35×0.7534×0.25 + 35C2×0.7533×0.252 + 35C3×0.7532×0.253 M1 Attempt P(r ⩽ 3)
= 0.000042378 + 0.0004944 + 0.00280168 + 0.01027283 Expressions or terms.
= 0.0136 (3 sf) [ < 0.04 ] A1 SC unsupported answer 0.0136 B1.
‘0.0136’ + 35C4×0.7531×0.254 = 0.0136 + 0.0273942 M1 Attempt P(r ⩽ 4)
Expressions or terms
= 0.041[0] (2 sf) [ > 0.04 ] A1 SC unsupported answer 0.041 B1.
Maximum value of r is 3 A1 Dep on above two correct probabilities seen and at
least one comparison with 0.04.
6
7(b) Concluding proportion has decreased when in fact it hasn’t B1 Or concluding that the director’s belief is correct
when in fact it isn’t.
0.0136 B1 FT their P(r ⩽ 3), dep < 0.04 OE.
2
7(c) 1 − (0.9535 + 35×0.9534×0.05 + 35C2×0.9533×0.052 + 35C3×0.9532×0.053) M1 Attempt 1 − P(r ⩽ 3) with B(35, 0.05).
= 1 – ( 0.16608 + 0.305943 + 0.2737385 + 0.158480 ) Expressions or terms.
= 0.0958 (3 sf) A1 SC unsupported answer 0.0958 B1.
2
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Mathematics 9709/63 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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