Mathematics 9709/62 October/November 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Each of a random sample of 80 adults gave an estimate, h metres, of the height of a particular building.
The results were summarised as follows. n = 80 Σ h = 2048 Σ h2 = 52 760
(a) Calculate unbiased estimates of the population mean and variance. [3]
(b) Using this sample, the upper boundary of an !% confidence interval for the population mean is 26.0.
Find the value of !. [4]
Mark scheme
1(a) 2048 128 B1
Est μ = 25.6 or or
80 5
80 52760 2048 1 20482 M1 Substitution into a correct formula.
2
Est σ =
2
− or 52760 − Biased 4.14 scores M0.
79 80 80 79 80
1656 A1
= 4.19 (3 sf) or
395
3
1(b) '4.19' M1 Use of correct equation with their values.
‘25.6’ + z = 26.0
80
z = 1.748 or 1.747 A1 Accept 3sf.
FT Biased z = 1.758.
(ɸ(‘1.748’) = 0.960) ‘0.960’ – (1 – ‘0.960’) M1 Correct area using their values.
α = 92.0 or 91.9 A1 Allow 92 .
FT Biased 92.1.
A final answer of 0.92 or 0.919 scores A0.
4
In the past, the mean length of a particular variety of worm has been 10.3 cm, with standard deviation
2.6 cm. Following a change in the climate, it is thought that the mean length of this variety of worm has decreased. The lengths of a random sample of 100 worms of this variety are found and the mean of this sample is found to be 9.8 cm.
Assuming that the standard deviation remains at 2.6 cm, carry out a test at the 2% significance level of whether the mean length has decreased. [5]
Mark scheme
2 Ho: Population mean length = 10.3 cm B1 or μ = 10.3 (not just ‘mean’).
H1: Population mean length < 10.3 cm μ < 10.3
9.8 − 10.3 M1 If ± 1.923 (or 0.0272) seen allow M1 implied.
±
2.6 / 100
= –1.923 A1 Accept ± .
Accept 3sf.
–1.923 > –2.054 or –2.055 M1 OE For a valid comparison.
Or compare 1 – ϕ(‘1.923’’) with 0.02 e.g. 0.0272 > 0.02
Use of CV 9.8 > 9.766 scores M1 A1 for 9.766 and M1
for comparison.
[Not reject H0 ] No evidence that [mean] length has decreased A1 FT FT their z. No contradictions, not definite, in context.
5
1.6% of adults in a certain town ride a bicycle. A random sample of 200 adults from this town is selected.
(a) Use a suitable approximating distribution to find the probability that more than 3 of these adults ride a bicycle. [4]
(b) Justify your approximating distribution. [2]
Mark scheme
3(a) Use of Poisson. mean = 3.2 B1 B1
3.22 3.23 M1 Allow any λ .
1 − e−3.2 1 + 3.2 + + –3.2
or 1 – e (1 + 3.2 + 5.12 + 5.46133) Allow one end error.
2 3!
or 1 – (0.04076 + 0.1304 + 0.2087 + 0.2226)
= 0.397 or 0.398 A1 SC Use of binomial: B1 for answer 0.398 (3 sf).
0.397 or 0.398 with no working scores SC B1.
4
3(b) [Binomial with] [n =] 200 > 50 B1
[np =][200 × 0.016 =] 3.2 < 5 or [p =]0.016 < 0.1 B1 If B0 B0
SC n large (or n > 50),
and p small or p < 0.1 or np <5 : B1.
2
The number of faults in cloth made on a certain machine has a Poisson distribution with mean
2.4 per 10 m2. An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30 m2 of cloth is checked and the number of faults is found.
(a) State suitable null and alternative hypotheses for the test. [1]
(b) Find the probability of a Type I error. [3]
Exactly 3 faults are found in the randomly selected 30 m2 of cloth.
(c) Carry out the test at the 5% significance level. [2]
Later a similar test was carried out at the 5% significance level, using another randomly selected 30 m2 of cloth.
(d) Given that the number of faults actually has a Poisson distribution with mean 0.5 per 10 m2, find the probability of a Type II error. [2]
Mark scheme
4(a) H0: Population mean = 7.2 or 2.4 B1 or λ or μ = 7.2 or 2.4 (Not just ‘mean’).
H1: Population mean < 7.2 or 2.4 or λ or μ < 7.2 or 2.4
1
4(b) λ = 7.2 B1 SOI
7.22 M1 Both expressions needed, allow any λ
P ( X 2 ) = e−7.2 1 + 7.2 + –7.2
or e (1 + 7.2 + 25.92) If λ ≠ 7.2 allow P(X ⩽ n) for 2 consecutive values of n
2 with P(X ⩽ n) < 0.05 and P(X ⩽ n + 1) > 0.05.
or 0.0007465 + 0.0053754 + 0.01935 [= 0.0255]
7.23
P ( X 3) = '0.0255'+ e −7.2 × or ‘0.0255’ +e–7.2 (62.21)
3!
or ‘0.0255’ +0.04644 [= 0.0719]
P(Type I) = 0.02547 or 0.0255 (3 sf) B1
3
4(c) 3 > 2 or P(X ⩽ 3) > 0.05 or ‘0.0719’ > 0.05 M1 For a valid comparison
or 3 outside critical region. FT their CR in (b).
[Not reject H0] A1 FT No contradictions. In context, not definite.
No evidence that [mean] number of faults has decreased
2
4(d) 1 – e–1.5(1 + 1.5 + 1.52 / 2) or 1 – e–1.5(1 + 1.5 + 1.125) M1 Must see expression. FT their CR in (b).
or 1 – (0.2231 + 0.3347 + 0.2510)
= 0.191 (3 sf) A1
2
X is a random variable with distribution B 10, 0.2. A random sample of 160 values of X is taken.
(a) Find the approximate distribution of the sample mean, including the values of the parameters.
[3]
(b) Hence find the probability that the sample mean is less than 1.8. [3]
Mark scheme
5(a) For X, μ = 2 σ2 = 1.6
Mean = 2 B1
1.6 1 B1 Accept Var = 0.12 (accept sd=0.1 if clearly identified).
Variance = or or 0.01
160 100
Normal B1
3
5(b) 1.8 − 320
1
− '2' 1.8 − '2' M1 Allow with wrong continuity correction.
± or ± [= –2.03 or –2] M1 can be implied by correct final answer or for
'0.01' '0.01'
–2.03 / –2.0 or 0.9788 / 0.9772 seen.
or ± (287.5 – ’320’) / '256' or ± (288 – ‘320’) / '256' [= –2.03 or –2]
ɸ(‘–2.03’) = 1 – ɸ(‘2.03’) M1 Correct area consistent with their values.
M1 can be implied by correct final answer.
= 0.0212 or 0.0228 (3 sf) A1
3
The masses, in grams, of small and large bags of flour have the distributions N 510, 100 and
N 1015, 324 respectively. André selects 4 small bags of flour and 2 large bags of flour at random.
(a) Find the probability that the total mass of these 6 bags of flour is less than 4130 g. [5]
(b) Find the probability that the total mass of the 4 small bags is more than the total mass of the
2 large bags. [5]
Mark scheme
6(a) E(T) = 4 × 510 + 2 × 1015 [= 4070] B1
Var(T) = 4 × 100 + 2 × 324 [= 1048] B1 or ( 4 ×100 + 2 × 324 ) =
32.4 ( 3sf ) .
4130 − 4070 M1 Standardising with their values.
[= 1.853] Variance must be from a combination attempt.
'1048'
M1 can be implied by correct final answer.
Φ(‘1.853’) M1 For area consistent with their values.
M1 can be implied by correct final answer.
= 0.968 (3 sf) A1 As final answer.
5
6(b) E(D) = 4 × 510 – 2 × 1015 [= 10] B1
Var(D) = 4 × 100 + 2 × 324 [= 1048] ( 4 ×100 + 2 × 324 ) =
32.4 ( 3sf ) .
B1 Or
0 − '10' M1 Standardising with their values.
[= –0.309] Variance must be from a combination attempt.
'1048'
M1 can be implied by correct final answer.
1 – Φ(‘–0.309’) = Φ(‘0.309’) M1 For area consistent with their values.
M1 can be implied by correct final answer.
= 0.621 A1 As final answer.
5
The diagram shows the graph of the probability density function, f, of a random variable X which takes values between −3 and 2 only.
(a) Given that the graph is symmetrical about the line x = −0.5 and that P X < 0 = p, find P −1 < X < 0 in terms of p. [2]
(b) It is now given that the probability density function shown in the diagram is given by
T a − b x2 + x −3 ≤ x ≤ 2, f x =
0 otherwise, where a and b are positive constants.
(i) Show that 30a − 55b = 6. [3]
(ii) By substituting a suitable value of x into f x, find another equation relating a and b and hence determine the values of a and b. [3]
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) 1 – p or p – 0.5 M1 SOI, e.g. on diagram.
[P(–1 < X < 0) =] 2p – 1 A1 Clearly as final answer.
2
7(b)(i) 2 2 M1 OE Attempt integral, with correct limits and RHS.
−3
(a − b( x 2 + x))dx = 1 or
−3
(ax − b( x3 + x 2 ))dx = –0.5
x3 x 2 2 x2 x 4 x3 2 A1 OE Correct integration.
ax − b + (= 1) or a − b + (= –0.5)
3 2 −3 2 4 3 −3
2a – 8b/3 – 2b +3a – 9b +9b / 2 = 1 or 2a – 4b – 8b / 3 – 9a / 2 + 81b / A1 Correctly obtained. No errors seen.
4 – 9b = –0.5 leading to 30a – 55b = 6 AG
3
7(b)(ii) a – b(9 – 3) = 0 or a – b(4 + 2) = 0 [hence a – 6b = 0] *M1 Use f(–3) = 0 or f(2) = 0.
Further attempts at integration M0.
Attempt to solve 30a – 55b = 6 and their a – 6b = 0 DM1
36 6 A1
a= or 0.288 b = or 0.048
125 125
3
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Mathematics 9709/62 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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