Mathematics 9709/61 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
It is known that the height H , in metres, of trees of a certain kind has the distribution N 12.5, 10.24.
A scientist takes a random sample of 25 trees of this kind and finds the sample mean, H , of the heights.
(a) State the distribution of H , giving the values of any parameters. [2]
(b) Find P 12 < H < 13. [3]
Mark scheme
1(a) N(12.5, ….) B1
Variance = 0.4096 B1 10.24
Accept 0.410 (3sf), condone
25
2
1(b) 13 − '12.5' [= 0.781] M1 For standardising with their values. Accept standardising
'0.4096 ' with 12.
ɸ(‘0.781’) – (1 – Φ(‘0.781’)) M1 For attempting to find their central area.
0.565 (3sf) A1
3
The number of enquiries received per day at a customer service desk has a Poisson distribution with mean 45.2. If more than 60 enquiries are received in a day, the customer service desk cannot deal with them all.
Use a suitable approximating distribution to find the probability that, on a randomly chosen day, the customer service desk cannot deal with all the enquiries that are received. [4]
Mark scheme
2 N(45.2, 45.2) B1 SOI
60.5 − 45.2 [= 2.276] M1 Allow with wrong or no continuity correction.
45.2
1 –ɸ(‘2.276’) M1
0.0114 A1
4
A random sample of 75 students at a large college was selected for a survey. 15 of these students said that they owned a car. From this result an approximate !% confidence interval for the proportion of all students at the college who own a car was calculated. The width of this interval was found to be 0.162.
Calculate the value of ! correct to 2 significant figures. [5]
Mark scheme
3 15 B1 SOI
est(p) = 0.2 accept
75
0.2 × 0.8 = 0.162 M1 Expression of the correct form.
2×z× Condone missing 2x.
75
75 = 1.754 A1 Correct z. Condone 3sf accuracy.
z = 0.081 ×
0.2 × 0.8
Φ(‘1.754’) = 0.96[03] M1 OE. Using their z to find alpha.
'0.96' – (1 – '0.96')
α = 92 A1 Following correct working.
5
A random variable X has probability density function given by
T 1
9 − x2 0 ≤ x ≤ 3, f x = 18
0 otherwise.
(a) Find P X < 1.2. [3]
(b) Find E X . [3]
The median of X is m.
(c) Show that m3 − 27m + 27 = 0. [3]
Mark scheme
4(a) 1.2 M1 Attempt to integrate f(x), ignore limits.
1 (9 − x 2 )dx
18
Must see an increase of power.
0
1 9 x − x 3 1.2 A1 Correct integration and correct limits.
18 3 0
71 or 0.568 A1 SC unsupported answer scores B2 only.
125
3
4(b) 3 M1 Attempt to integrate xf(x), ignore limits.
1
18
0
(9 x − x3 )dx Must see an increase of power.
A1 Correct integration and correct limits.
1 9 x − x4 3
2
18 2 4 0
9 A1 SC unsupported answer scores B2 only.
or 1.125
8
3
4(c) M1 Attempt to integrate f(x) with correct limits and = 0.5.
1 9 x − x3 m = 0.5
18 3 0
OE. Accept limits m to 3.
Allow x instead of m.
A1 Any correct cubic equation in m or x.
1 9m − m3 – 0.5 = 0
18 3
m3 – 27m + 27 = 0 A1 AG. Correctly obtain this equation. No errors seen.
3
(a) The proportion of people having a particular medical condition is 1 in 100 000. A random sample of 2500 people is obtained. The number of people in the sample having the condition is denoted by X .
(i) State, with a justification, a suitable approximating distribution for X , giving the values of any parameters. [2]
(ii) Use the approximating distribution to calculate P X > 0. [2]
(b) The percentage of people having a different medical condition is thought to be 30%. A researcher suspects that the true percentage is less than 30%. In a medical trial a random sample of 28 people was selected and 4 people were found to have this condition.
Use a binomial distribution to test the researcher’s suspicion at the 2% significance level. [5]
Mark scheme
5(a)(i) Po(0.025) B1 For Poisson and correct parameter.
B1 Must show 2500 and 0.025.
n = 2500 > 50, np = 0.025 < 5 1
Accept p = < 0.1 in place of np = 0.025 < 5.
100000
2
5(a)(ii) M1 Allow any λ.
1 – e–0.025
FT their (a)(i) if normal; must have continuity correction.
0.0247 (3sf) A1 Must be from Poisson.
Unsupported correct answer scores B1 instead of M1 A1.
2
5(b) H0: p = 0.3 B1
H1: p < 0.3
0.728 + 28 × 0.727 × 0.3 + 28C2 × 0.726 × 0.32 + 28C3 × 0.725 × 0.33 + 28C4 × M1 Use of B(28, 0.3 ). Addition of terms must be intended.
0.724 × 0.34 Allow one term wrong or omitted or extra.
0.0474 A1 Unsupported correct answer scores B1 instead of M1 A1.
0.0474 > 0.02 [Not reject H0 ] M1 Valid comparison.
No evidence that suspicion is true. A1 ft Not definite e.g. not ‘Suspicion is not true’, in context, no
contradictions.
SC use of N(8.4, 5.88) leading to 0.054 > 0.2 OE can
score B1 only for comparison and correct conclusion.
Correct hypotheses with p will also score B1.
5
The random variable T denotes the time, in seconds, for 100 m races run by Tania. T is normally distributed with mean - and variance 32 . A random sample of 40 races run by Tania gave the following results. n = 40 Σ t = 560 Σ t2 = 7850
(a) Calculate unbiased estimates of - and 32 . [3]
The random variable S denotes the time, in seconds, for 100 m races run by Suki. S has the independent distribution N 14.2, 0.3.
(b) Using your answers to part (a), find the probability that, in a randomly chosen 100 m race, Suki’s time will be at least 0.1 s more than Tania’s time. [5]
Mark scheme
6(a) 560 B1
est μ = 14 accept
40
M1
7850 5602
est σ = 40
2
− 142 or 1 7850 −
39 40 39 40
0.25641 or 0.256 (3sf) A1 10
Accept
39
Without 40 i.e. biased: est σ2 = 0.25 M0 A0.
39
3
6(b) E(S – T) = 14.2 – ‘14’ [= 0.2 ] B1 FT FT their 14.
B1 FT 217
Accept
Var(S – T) = 0.3 + ` 0.256` [= 0.55641] 390
FT their 0.256 including FT biased.
Var(S – T) = 0.55.
0.1 − '0.2' [= –0.134] M1 Standardising with their values (note biased gives
'0.55641' –0.135). FT their E & Var.
P(S – T > 0.1) = 1 – Φ(‘–0.134’) = Φ(‘0.134’) M1 Finding correct area consistent with their values.
0.553 (3sf) A1 Use of biased gives 0.554 (3sf) can score the A1.
Similar scheme for P(T – S) < –0.1.
Similar scheme for S – T – 0.1 > 0.
And T – S + 0.1 < 0.
5
The masses, in grams, of apples from a certain farm have mean - and standard deviation 5.2. The farmer says that the value of - is 64.6. A quality control inspector claims that the value of - is actually less than 64.6. In order to test his claim he chooses a random sample of 100 apples from the farm.
(a) The mean mass of the 100 apples is found to be 63.5 g.
Carry out the test at the 2.5% significance level. [5]
(b) Later another test of the same hypotheses at the 2.5% significance level, with another random sample of 100 apples from the same farm, is carried out.
Given that the value of - is in fact 62.7, calculate the probability of a Type II error. [5]
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) H0: μ = 64.6 B1
Allow population mean, not just ‘mean’.
H1: μ < 64.6
[ ± ] 63.5 − 64.6 M1
Standardising. Must have 100 .
5.2 ÷ 100
[ ± ] –2.115 A1 Accept -2.12 (3sf)
‘2.115’ > 1.96 or ‘–2.115’ < –1.96 [do not accept H0] M1 Valid comparison (0.0172 < 0.025 for area comparison).
There is evidence that μ < 64.6 A1 FT Not definite, e.g. not ‘μ < 64.6’. in context. No
contradictions.
Accept critical value method leading to 63.5 < 63.58 or
64.6 > 64.52.
5
7(b) m − 64.6 = –1.96 M1 5.2
5.2 ÷ 100 Finding the critical value using N 64.6, and a
100
z value.
m = 63.5808 A1
63.5808 − 62.7 [= 1.694] M1 5.2
5.2 ÷ 100 Standardising using N 62.7, and a critical value.
100
1 – Φ(‘1.694’) M1 For area consistent with their values.
0.0451 A1 Accept answers that round to 0.045.
5
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/61 October/November 2021 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/61, October/November 2021 (9709_w21_ms_61), and the question paper (9709_w21_qp_61), to read or download, with each question's marking points written out underneath.
How many marks is 9709/61 out of?+
Mathematics 9709/61 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/61 answers marked?+
Yes. Upload your answers to 9709/61/O/N/21 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.