Mathematics 9709/63 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The number of goals scored by a team in a match is independent of other matches, and is denoted by the random variable X , which has a Poisson distribution with mean 1.36. A supporter offers to make a donation of $5 to the team for each goal that they score in the next 10 matches.
Find the expectation and standard deviation of the amount that the supporter will pay. [5]
In the past, the time, in hours, for a particular train journey has had mean 1.40 and standard deviation
0.12. Following the introduction of some new signals, it is required to test whether the mean journey time has decreased.
(a) State what is meant by a Type II error in this context. [1]
(b) The mean time for a random sample of 50 journeys is found to be 1.36 hours.
Assuming that the standard deviation of journey times is still 0.12 hours, test at the 2.5% significance level whether the population mean journey time has decreased. [5]
(c) State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (b). [2]
Mark scheme
2(a) Conclude (mean) (journey) time has not decreased when in fact it has. B1 OE in context
1
2(b) H0: Pop mean (or μ) = 1.4 B1 May be seen in (a)
H1: Pop mean (or μ) < 1.4
1.36 − 1.4 M1 68 − 70
0.12 Accept totals method
50 × 0.12
50 No mixed methods or no standard deviation/variance
mixes
–2.357 or – 2.36 A1 Correct z or correct area if used
–2.357 < –1.96 or 0.0092 < 0.025 or 0.9908 > 0.975 M1 valid comparison
Or CV method 1.36 < 1.367
There is evidence that (mean) (journey) times have decreased A1 FT in context not definite no contradictions
NB use of two tail test scores max B0M1A1M1A0 no ft
for two tail test
5
2(c) H0 was rejected OE *B1 FT FT H0 was accepted OE
Type I DB1 FT FT Type II
2
The local council claims that the average number of accidents per year on a particular road is 0.8.
Jane claims that the true average is greater than 0.8. She looks at the records for a random sample of
3 recent years and finds that the total number of accidents during those 3 years was 5.
(a) Assume that the number of accidents per year follows a Poisson distribution.
(i) State null and alternative hypotheses for a test of Jane’s claim. [1]
(ii) Test at the 5% significance level whether Jane’s claim is justified. [4]
(b) Jane finds that the number of accidents per year has been gradually increasing over recent years.
State how this might affect the validity of the test carried out in part (a)(ii). [1]
Mark scheme
3(a)(i) H0: λ = 2.4 B1 Accept λ or µ
H1: λ > 2.4 Accept 2.4 or 0.8 (per year)
1
3(a)(ii) 2 3 4 M1 Any λ; allow one end error
1 – e–2.4(1 + 2.4 + 2.4 + 2.4 + 2.4 )
2 3! 4!
0.0959 (3 sf) A1 SC unsupported answer 0.0959 scores B1 only not
M1A1
0.0959 > 0.05 M1 Valid comparison
Use of 0.9041 < 0.95 can recover either M1A1 or B1
There is evidence that Jane’s claim not justified A1 FT OE. In context, not definite, e.g. not ‘Jane is wrong’, no
or There is insufficient evidence to support Jane’s claim contradictions.
Condone omission of Jane.
4
3(b) Mean not constant so Poisson model not valid B1
1
The masses, m kilograms, of flour in a random sample of 90 sacks of flour are summarised as follows.
n = 90 Σ m = 4509 Σ m2 = 225 950
(a) Find unbiased estimates of the population mean and variance. [3]
(b) Calculate a 98% confidence interval for the population mean. [3]
(c) Explain why it was necessary to use the Central Limit theorem in answering part (b). [1]
(d) Find the probability that the confidence interval found in part (b) is wholly above the true value of the population mean. [2]
Mark scheme
4(a) 4509 [= 50.1] B1
90
M1 Attempted.
90 ( 225950 − '50.1'2 ) or 1 225950 − 4509
2
89 90 Use of biased = 0.5455 scores M0A0
89 90
491 or 0.552 (3 sf) A1
890
3
4(b) M1 Expression of the correct form, allow any z-value but
' 491 ' must be a z-value
‘50.1’ ± z 890
90
z = 2.326 B1 Accept 2.326 to 2.329
49.9 to 50.3 (3 sf) A1 FT from biased variance. Must be an interval.
3
4(c) Population of masses is unknown B1 Accept population of masses is not normal
1
4(d) 1 – 0.98 M1 0.02 seen
0.02 ÷ 2 = 0.01 A1 As final answer
2
Most plants of a certain type have three leaves. However, it is known that, on average, 1 in 10 000 of these plants have four leaves, and plants with four leaves are called ‘lucky’. The number of lucky plants in a random sample of 25 000 plants is denoted by X .
(a) State, with a justification, an approximating distribution for X , giving the values of any parameters.
[2]
Use your approximating distribution to answer parts (b) and (c).
(b) Find P X ≤ 3. [2]
(c) Given that P X = k = 2P X = k + 1, find k. [2]
The number of lucky plants in a random sample of n plants, where n is large, is denoted by Y .
(d) Given that P Y ≥ 1 = 0.963, correct to 3 significant figures, use a suitable approximating distribution to find the value of n. [3]
Mark scheme
5(a) Po(2.5) B1 Accept Poisson with mean = 2.5 not just np = 2.5
n = 25 000 > 50 and np (or λ) = 2.5 which is < 5 B1 Must see 2.5 (or 0.0001) and 25000 OE, not just np < 5
or n = 25 000 > 50 and p = 0.0001 < 0.1 (or p < 0.1) and n > 50
2
5(b) 2 3 M1 Any λ, accept one end error.
e–2.5(1 + 2.5 + 2.5 + 2.5 )
2 3! FT binomial from part (a) scores M1 only for equivalent
binomial expressions
FT normal from part (a) must use correct continuity
correction and tables scores M1 only for complete
method
0.758 (3 sf) A1 Unsupported answer of 0.758 scores B1 instead of
M1A1
2
5(c) k k +1 M1 Any λ
e−2.5 × 2.5 = 2e −2.5 × 2.5
(k )! (k + 1)! FT binomial from (a) scores M1 only for equivalent
binomial expression
FT from (a) normal for equivalent expressions
continuity correction must be included
k=4 A1 No errors seen
SC k = 4 unsupported scores B1 only, but see full
Poisson expressions for P(4) and P(5) and 0.134 scores
M1A1
2
5(d) 1 - e-λ = 0.963 M1 Accept their attempt at λ
λ = –ln0.037 (= 3.2968 or 3.30 or 3.3) M1 Correct use of lns
n = 33 000 (3 sf) A1 Allow n = 32 950 to 33 050 (must be an integer)
SC use of binomial leading to 32 967 scores
B1 for (0.9999)n = 0.037
B1 for 33 000 to 3sf (32 967)
3
Alethia models the length of time, in minutes, by which her train is late on any day by the random variable X with probability density function given by
T 3 x − 202 0 ≤ x ≤ 20, f x = 8000
0 otherwise.
(a) Find the probability that the train is more than 10 minutes late on each of two randomly chosen days. [4]
(b) Find E X . [4]
(c) The median of X is denoted by m.
Show that m satisfies the equation m − 203 = −4000, and hence find m correct to 3 significant figures. [4]
(d) State one way in which Alethia’s model may be unrealistic. [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
6(a) 20
2
3 ( x − 20) dx M1 Attempt integration of f(x), ignore limits.
P(X > 10) = 8000
10
20 M1 Substitute correct limits 10 to 20
( x − 20)3
20
3 3 x 3 40 x 2 or 1 – … limits 0 to 10 in their integral
= × or − + 400 x
8000 3 8000 3 2 10
10
= 1 0 − ( −10)3
8000
1 or 0.125 A1 SC Unsupported answer of 1 scores B1 only
8 8
( 1 )2 = 1 or 0.0156 (3 sf) B1 FT FT their P(X > 10) dependent on first M1 gained
8 64
4
6(b) 20
3 ( x − 40 x + 400 x)dx M1 Attempt integration of xf(x). Ignore limits.
8000
3 2
0
3 x4
20 A1 Correct integral
− 403x + 4002 x
3 2
8000 4
0 (by expanding or by parts)
3 x ( x − 20 )3 1 ( x − 20 )
4
or × −
8000 3 8000 4
3 160000 − 40 × 8000 + 200 × 400 M1 Subst correct limits in their (4th degree) integral
8000 4 3
5 A1
4
6(c) m
3 ( x − 20) dx = 0.5 M1 Attempt to integrate f(x) and equate to 0.5. Ignore limits.
8000
2
0
m M1 Attempt integral and substitute limits 0 and m
( x − 20)3
m
3 3 x3 40 x 2 or m and 20 and = 0.5
× = 0.5 or − + 400 x = 0.5
8000 3 8000 3 2 0
0
1 ( m − 20)3 − ( −20)3 = 0.5
8000
(m – 20)3 = –4000 A1 AG. Found convincingly.
(m = 20 + 3 −4000 ) B1
m = 4.13 (3 sf)
4
6(d) Doesn't allow for trains > 20 mins late or Doesn't allow for trains being early B1 or any relevant comment e.g. trains on Sun may be
different to trains on Mon
1
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Mathematics 9709/63 for May/June 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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