Mathematics 9709/61 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The heights of a certain species of deer are known to have standard deviation 0.35 m. A zoologist takes a random sample of 150 of these deer and finds that the mean height of the deer in the sample is 1.42 m.
(a) Calculate a 96% confidence interval for the population mean height. [3]
(b) Bubay says that 96% of deer of this species are likely to have heights that are within this confidence
interval.
Explain briefly whether Bubay is correct. [1]
Mark scheme
1(a) z = 2.054 or 2.055 B1 Accept 3 sf if nothing better seen (2.05 or 2.06).
0.35 M1 Must be a z value.
1.42 ± z 150
1.36 to 1.48 [m] (3 sf) A1 Correct working only. Must be an interval.
3
1(b) No. CI is about mean, not individual values. B1 Or similar. Need both.
1
The masses, in kilograms, of small and large bags of wheat have the independent distributions
N(16.0, 0.4) and N(51.0, 0.9) respectively.
Find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag. [5]
Mark scheme
2 E(S1 + S2 + S3 – L) = 16 x 3 – 51 [= −3 ] B1 Oe, using L – (S1 + S2 + S3 ).
Var(S1 + S2 + S3 – L) = 3 × 0.4 + 0.9 [= 2.1] M1
0−( −3) M1 For standardising with their values.
'2.1' [= 2.070]
1 − Φ(‘2.070’) M1 For area consistent with their values.
= 0.0192 (3 sf) A1
5
The times, T minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by / t = 230 and / t 2 = 930 .
(a) Calculate unbiased estimates of the population mean and variance of T. [3]
You should now assume that your estimates from part (a) are the true values of the population mean and variance of T.
(b) The times taken by another random sample of 75 students were noted, and the sample mean, T , was found.
Find the value of a such that P ( T 2 a) = 0.234 . [3]
Mark scheme
3(a) B1
75 [= 3.0666… or 3.07 (3 sf)] [ 0r 46/15 ]
t = 230
s2 = 74 ( 75 − ( 230
75 930
75
)2 ) or 1/74(930 – 2302/75 ) M1 Use of correct formula.
= 3.0360… or 3.04 (3 sf) or = 337/111 A1
3
3(b) [ Φ−1(1 − 0.234) ] = 0.726 B1
a −'3.0667' M1 Ft their 0.726 but must be a z value.
± '3.04'/75 = ± ‘0.726’ Note using 0.766 is M0.
Must have sqrt 75.
a = 3.21 (3 sf) A1 CWO
3
A random variable X has probability density function f defined by
a 18
2- 3 2 G x G 3, f (x) = * x x
0 otherwise,
where a is a constant.
(a) Show that a = 27
2 . [3]
(b) Show that E (X ) = 27 3
2 ln 2 - 3. [3]
Mark scheme
4(a) 3 M1 Attempt integrate f(x), ignore limits and ‘= 1’.
( − )dx = 1
2
a
x2
18
x3
3 A1 OE Correct integration and limits.
− ax + 92 = 1
x 2
− a3 + 1 + a2 − 94 = 1 a = 272 (AG) A1 Must see correct substitution of limits.
Correct working no errors seen.
3
4(b) 3 M1 Attempt to integrate xf(x), ignore limits.
2
( 227x − 18
x2
)dx
3 3 A1 Correct integration and limits.
272 ln x + 18x or 272 ln 2 x + 18x OE e.g. using ln 2x.
2 2
= 272 ln 3 + 6 − 272 ln 2 − 9 = 272 ln 32 − 3 AG A1 Must see correct substitution of limits.
Correct working no errors seen.
3
The lengths, in centimetres, of worms of a certain kind are normally distributed with mean n and standard deviation 2.3 . An article in a magazine states that the value of n is 12.7 . A scientist wishes to test whether this value is correct. He measures the lengths, x cm, of a random sample of 50 worms of this kind and finds that / x = 597.1 . He plans to carry out a test, at the 1% significance level, of whether the true value of n is different from 12.7 .
(a) State, with a reason, whether he should use a one-tailed or a two-tailed test. [1]
(b) Carry out the test. [5]
Mark scheme
5(a) Two-tailed because looking for difference B1
1
5(b) H0: μ = 12.7 H1: μ ≠ 12.7 B1 No ft from part (a).
597.1 −12.7 M1
50
2.3
50
= −2.330 A1 or 0.00989 or 0.0099.
Accept – 2.336 or – 2.337 or 0.0097 if area comparison used.
‘−2.330’ > −2.576 or ‘2.330’ < 2.576 M1 Accept 2.574 to 2.579.
or ‘0.00989’ > 0.005 Or use of CV.
or `0.0097` > 0.005 12.7- 2.576 x ( 2.3 / sqrt 50 ) = 11.862 M1A1.
11.942 > 11.862 M1A1.
[Not reject H0] There is insufficient evidence to suggest that µ is A1 FT OE ft their zcalc.
not 12.7 In context, not definite, e.g. not ‘µ =12.7’.
No contradictions.
SC use of 1 tailed test can score B0M1A1M1 for comparison with
0.01 A0 max 3/5.
5
The numbers of customers arriving at service desks A and B during a 10-minute period have the independent distributions Po(1.8) and Po(2.1) respectively.
(a) Find the probability that during a randomly chosen 15-minute period more than 2 customers will arrive at desk A. [2]
(b) Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4. [3]
(c) An inspector waits at desk B. She wants to wait long enough to be 90% certain of seeing at least one customer arrive at the desk.
Find the minimum time for which she should wait, giving your answer correct to the nearest minute. [4]
Mark scheme
6(a) [λ = 2.7] 1 − e−2.7(1 + 2.7 + 2.72 ) or 1 − e−2.7(1 + 2.7 +3.645)
2
M1 Any λ. Allow one end error.
Must see expression.
or 1- ( 0.06721 + 0.1815 + 0.2450 )
= 0.506 (3 sf) A1 SC unsupported answer 0.506 scores B1.
2
6(b) λ = 1.95 B1
e−1.95(1 + 1.95 + 1.95
2
+ 1.95 ) or e−1.95(1 + 1.95 + 1.90125 +1.2358)
3
M1 Any λ. Allow one end error.
2 3!
Must see expression.
or 0.1423 + 0.2774+ 0.2705 + 0.1758
= 0.866 A1 SC unsupported answer 0.866 scores B1B1.
3
6(c) 1 – e-2.1x ⩾ 0.90 or 1 – e−λ ⩾ 0.90 M1 OE Condone use of ‘=’ throughout.
[e−2.1x < 0.1] or e−λ < 0.1 M1 Rearrange and attempt take logs of relevant form.
−2.1x < ln0.1 or −λ < ln0.1 [ λ > 2.3026, 2.3026/2.1 ]
1.096 or 10.96 accept 1.097 or 10.97 *A1 Seen.
She must wait for at least 11 minutes A1 dep
SC Use of trial and improvement.
Use of 1–e-λ any numerical λ (not 2.1) ie one trial M1.
Use of enough trials to give an answer of 0.90 (2sf) M1.
λ=2.30 i.e. 3sf accuracy AND 1.09… or 10.9 … A1.
Then 11 A1 dep.
4
The number of accidents per year on a certain road has the distribution Po(m). In the past the value of m was 3.3 . Recently, a new speed limit was imposed and the council wishes to test whether the value of m has decreased. The council notes the total number, X, of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the 5% significance level.
(a) Calculate the probability of a Type I error. [4]
(b) Given that X = 2, carry out the test. [3]
(c) The council decides to carry out another similar test at the 5% significance level using the same hypotheses and two different randomly chosen years.
Given that the true value of m is 0.6, calculate the probability of a Type II error. [3]
(d) Using m = 0.6 and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years. [4]
Additional page
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Mark scheme
7(a) λ = 6.6 B1
P(X < 2) = e−6.6(1 + 6.6 + 6.62 ) [= 0.0400] [ < 0.05 ]
2
M1 Expression must be seen. No end errors.
Allow use of 3.3 here.
or e−6.6(1 + 6.6 + 21.78 )
or 0.001360 + 0.008978 + 0.02963
P(X < 3) = e−6.6(1 + 6.6 + 6.62 + 6.6 ) or 0.0400 + e−6.6× 6.6
2 3 3
= B1 Condone unsupported 0.105.
3! 3!
0.105 [ > 0.05 ]
P(Type I error) = 0.0400 (3 sf) A1 Allow 0.040 or 0.04 AWRT
SC unsupported ans of 0.0400 can score max B1B1B1.
4
7(b) H0: λ = 6.6, H1: λ < 6.6 B1 May be seen in part (a) and award B1 mark here.
Accept µ or λ.
Accept 3.3 or 6.6.
[P(X < 2) = 0.0400] ` 0.04 ` < 0.05 M1 For comparing their P(X < 2) any λ with 0.05.
[Reject H0] There is evidence to suggest that mean number of A1
accidents has decreased In context, not definite.
No contradictions.
CWO.
3
7(c) P(X > 2) attempted, with any λ M1
P(X > 2) = 1 − e−1.2(1 + 1.2 + 1.22 )
2
M1 Expression must be seen.
Correct λ.
or = 1 − e−1.2(1 + 1.2 +0.72)
No end errors.
or = 1 – ( 0.3012 + 0.3614 + 0.2169 )
0.121 (3 sf) or 0.120 A1 SC unsupported answer scores B2.
3
7(d) N(18, 18) seen or implied B1
10.5−18 M1 Allow with no or incorrect continuity correction.
18 [= −1.768] Their 18.
P(X > ‘−1.768’) = Φ(‘1.768’) M1 ft their standardised value.
Area consistent with their values.
= 0.961 or 0.962 (3 sf) A1
4
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Mathematics 9709/61 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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