Mathematics 9709/62 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) A random variable X has the distribution Po 25.
Use the normal approximation to the Poisson distribution to find P X > 30. [4]
(b) A random variable Y has the distribution B 100, p where p < 0.05.
Use the Poisson approximation to the binomial distribution to write down an expression, in terms of p, for P Y < 3. [2]
Mark scheme
1(a) X ~ N(25, 25) B1 soi
30.5 − 25 M1 Standardising with their values.
[= 1.1] Allow with missing or incorrect continuity correction.
25
1 – ɸ('1.1') M1 For area consistent with their working.
= 0.136 (3 sf) A1
4
1(b)
e −100 p 1 + 100 p +
(100 p )2 M1 For Po(100p) expression.
Accept un-simplified terms (e.g. p0/0! For M1).
2!
Allow one end error (e.g. for correct with extra term
e −100 p
×
(100 p )
3
oe), or brackets omitted.
3!
(100 p )2 (100 p )2 A1 Must have brackets.
e −100 p 1 + 100 p + or e–100p + e–100p × 100p + e −100 p × Allow with or without ! sign (but not 0! or p0).
2! 2!
ISW once a fully correct answer seen.
or e–100p(1 + 100p + 5000p2) oe
2
The length, in minutes, of mathematics lectures at a certain college has mean - and standard deviation 8.3.
(a) The total length of a random sample of 85 lectures was 4590 minutes.
Calculate a 95% confidence interval for -. [3]
The length, in minutes, of history lectures at the college has mean m and standard deviation s.
(b) Using a random sample of 100 history lectures, a 95% confidence interval for m was found to have width 2.8 minutes.
Find the value of s. [2]
Mark scheme
2(a) 4590 8.3 M1 For expression of correct form. Any z (but not ϕ(z)).
± z×
85 85
z = 1.96 B1
52.2 to 55.8 (3 sf) A1 Must be an interval.
3
2(b) s s M1 Equation of correct form (any z).
1.96 × = 1.4 or 2 × 1.96 × = 2.8 Allow factor of 2 error (i.e. first equation = 2.8).
100 100
50 A1
s = 7.14 (3 sf) or
7
2
A researcher read a magazine article which stated that boys aged 1 to 3 prefer green to orange. It claimed that, when offered a green cube and an orange cube to play with, a boy is more likely to choose the green one.
The researcher disagrees with this claim. She believes that boys of this age are equally likely to choose either colour. In order to test her belief, the researcher carried out a hypothesis test at the 5% significance level. She offered a green cube and an orange cube to each of 10 randomly chosen boys aged 1 to 3, and recorded the number, X , of boys who chose the green cube.
Out of the 10 boys, 8 boys chose the green cube.
(a) (i) Assuming that the researcher’s belief that either colour cube is equally likely to be chosen is valid, a student correctly calculates that P X = 8 = 0.0439, correct to 3 significant figures.
He says that, because this value is less than 0.05, the null hypothesis should be rejected.
Explain why this statement is incorrect. [1]
(ii) Carry out the test on the researcher’s claim that either colour cube is equally likely to be chosen. [5]
(b) Another researcher claims that a Type I error was made in carrying out the test.
Explain why this cannot be true. [1]
A similar test, at the 5% significance level, was carried out later using 10 other randomly chosen boys aged 1 to 3.
(c) Find the probability of a Type I error. [2]
Mark scheme
3(a)(i) Need to find P(X ⩾ 8) B1 oe (e.g. invalid because it should be a tail probability
compared with 0.05).
1
3(a)(ii) H0: P(green) = 0.5 B1 Allow p = 0.5.
H1: P(green) > 0.5 Allow p > 0.5.
P(X ⩾ 8) = 0.0439 + 10C9 × (0.5) × (0.5)9 + 0.510 M1 Attempt 0.0439 + P(X = 9) + P(X = 10).
Must see Binomial expressions B(10,0.5).
= 0.0547 or 0.0546 (3 sf) A1 SC B1 0.0547 or 0.0546 with no working.
0.0547 > 0.05 M1 Valid comparison of tail probability with 0.05.
[Do not reject H0] ‘There is insufficient evidence [at the 5% level] to accept A1FT In context, not definite. No contradictions.
the hypothesis that boys prefer green.’ Allow ‘There is insufficient evidence to reject the
Or ‘There is sufficient evidence to support the researcher’s claim.’ hypothesis that boys like green and orange equally’.
Not definite, e.g. not ‘They don't prefer green" or
‘Researchers claim true’ ‘Magazine’s claim untrue’.
Any mention of ‘claim’ must be clear which claim it is.
5
3(b) H0 was not rejected B1 Mark independently.
1
3(c) 10
C9×(0.5)×(0.5)9 + 0.510 or "0.0547" – 0.0439 M1 Finding P(9,10) using B(10,0.5).
Could be seen in (a)(ii).
P(Type I error) = 11/1024 or 0.0107 (3 sf) or 0.0108 A1
2
The height H , in metres, of mature trees of a certain variety is normally distributed with standard deviation 0.67. In order to test whether the population mean of H is greater than 4.23, the heights of a random sample of 200 trees are measured.
(a) Write down suitable null and alternative hypotheses for the test. [1]
The sample mean height, h metres, of the 200 trees is found and the test is carried out. The result of the test is to reject the null hypothesis at the 5% significance level.
(b) Find the set of possible values of h. [3]
(c) Ajit said, ‘In (b) we had to assume that H is normally distributed, so it was necessary to use the
Central Limit Theorem.’
Explain whether you agree with Ajit. [1]
Mark scheme
4(a) H0: population mean [of H ] = 4.23 B1 Allow μ = 4.23 or population mean of h = 4.23
H1: population mean [of H ] > 4.23 but NOT h = 4.23 or H = 4.23 or h = 4.23 or H = 4.23.
1
4(b) h − 4.23 M1 For standardising and forming an equation. Must have
= 1.645
0.67 200 . Allow ± 1.645 or ± 1.96.
200 Accept ‘>’ and ‘<’.
Allow H or any letter instead of h .
h = 4.31 (3 sf) A1 May be implied by h > 4.31. Allow h < 4.31 for this A1
only, condone 4.15 also seen.
h > 4.31 or h ⩾ 4.31 (3 sf) A1 Condone any letter instead of h .
3
4(c) Incorrect, because the population of H is given as normally distributed B1 Allow h instead of H or just ‘The population is normal.’
[with known variance]. Must use ‘population’ or ‘underlying distribution’.
1
The random variable X has probability density function, f, given by t
1 a < x < b, f x = x2
0 otherwise, where a and b are positive constants.
(a) It is given that E X = ln 2.
Show that b = 2a. [3]
(b) Show that a = 12 . [3]
(c) Find the median of X . [3]
Mark scheme
5(a) b
x M1 Attempt to integrate xf(x) from a to b.
a
x2
dx
= [ln x]a or ln b – ln a A1
b
b A1 Must see both statements. No errors seen.
ln = ln 2 or ln 2a = ln b oe
a
Hence b = 2a (AG)
3
5(b) b
1 M1 Attempt to integrate f(x) and equate to 1. Ignore limits.
a
x2
dx = 1
1 b 1 1 1 1 A1 Integrate with correct limits and substitute b = 2a.
[ − = 1 or − = 1 − =1
x a a b a 2a
1 A1 Obtain convincingly (at least one step from previous
e.g. = 1 or 2 + (−1) = 2a answer), no errors seen (ignore a = 0).
2a
1
Hence a = a = (AG)
2
3
5(c) m
1 1
1
1 1 M1 Attempt integrate f(x), with correct limits, and equate to
0.5
x 2
dx =
2
or
m
x 2
dx =
2
1
.
2
m 1
1 1 1 1
− x = 2 or − x = 2
0.5 m
1 1 1 1 A1 oe. Correct equation after substituting limits.
2− = or −1 =
m 2 m 2
2 A1
m= or 0.667 (3 sf)
3
3
A factory makes loaves of bread in batches. One batch of loaves contains X kilograms of dried yeast and Y kilograms of flour, where X and Y have the independent distributions N 0.7, 0.022 and
N 100.0, 3.02 respectively.
Dried yeast costs $13.50 per kilogram and flour costs $0.90 per kilogram. For making one batch of bread the total of all other costs is $55. The factory sells each batch of bread for $200.
Find the probability that the profit made on one randomly chosen batch of bread is greater than $40.
[7]
Mark scheme
6 Cost of dried yeast and flour: $D and $F B1 One of these soi – can be given at early stage.
E(D) = 13.5 × 0.7 = 9.45 E(F) = 0.9 × 100 = 90
Var(D) = 0.022 × 13.502 =0.0729 Var(F) = 3.02 × 0.902 = 7.29 B1 One of these soi – can be given at early stage.
Total cost: T ~ N(99.45, (0.022 × 13.502 + 3.02 × 0.902)) M1 Attempt to combine their D and their F with or without
55 and 200 (but variance must not include 55 or 200).
Or N(99.45 +55, (0.022 × 13.502 + 3.02 × 0.902)).
Or N(200-55 -99.45, (0.022 × 13.502 + 3.02 × 0.902)).
Or N(99.45 +c, (0.022 × 13.502 + 3.02 × 0.902)).
N(99.45, 7.3629) accept 99.4 or 99.5 A1 Or N(154.45,7.3629) or N(45.55, 7.3629).
Accept 3sf (accept 45.5 or 45.6).
[P(profit > $40) = P(T < 105)] M1 160–154.45/ 7.3629 or 40–45.55/ 7.3629 .
105 − '99.45' No mixed methods.
[= 2.045]
'7.3629'
[P(T < 105) = P(z < 2.045) =] M1 For area consistent with their working.
ɸ(2.045)
= 0.9795 or 0.9796 or 0.98(0) or 0.979 (3 sf) A1
7
A random variable X has the distribution Po 2.4.
(a) Find P 2 ≤ X < 4. [2]
(b) Two independent values of X are chosen.
Find the probability that both of these values are greater than 1. [3]
(c) It is given that P X = r < P X = r + 1.
(i) Find the set of possible values of r. [3]
(ii) Hence find the value of r for which P X = r is greatest. [1]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) 2.42 2.43 −2.4 M1 2.42 2.43 2.44
( e−2.4 + = e ( 2.88 + 2.304 ) = 0.2613 + 0.2090 Allow M1 for e−2.4 + + .
2! 3! 2! 3! 4!
Expression must be seen.
= 0.47(0) A1 SC B1 0.47(0) with no working.
2
7(b) 1 – e–2.4(1 + 2.4) [= 0.691558] M1 Allow one end error. Allow any λ.
(1 – e–2.4(1 + 2.4))2 M1 Squaring their probability (λ ≠ 4.8).
= 0.478 (3 sf) A1 SC B2 0.478 with no working.
3
7(c)(i) 2.4r 2.4r +1 M1 For both expressions seen.
e−2.4 × < e−2.4 ×
r! ( r + 1)!
r + 1 < 2.4 A1* Or r < 1.4 (must have correct inequality).
Set is r = 0, 1 DA1
3 For trial and error solutions.
M1 for substituting one value into correct expression
(can be implied by correct values).
A1* for [P(0) = 0.0907], P(1) = 0.218, P(2) = 0.261, P(3)
= 0.209 (accept 2sf accuracy).
DA1 Set is r = 0,1.
7(c)(ii) r=2 B1
1
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Mathematics 9709/62 for October/November 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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