Mathematics 9709/62 February/March 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics, February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The booklets produced by a certain publisher contain, on average, 1 incorrect letter per 30 000 letters, and these errors occur randomly. A randomly chosen booklet from this publisher contains
12 500 letters.
Use a suitable approximating distribution to find the probability that this booklet contains at least
2 errors. [3]
Mark scheme
1 (λ =) 125 = 0.417 or better B1
−5
1 – e 12 (1 + 125 ) M1 1 – P(X = 0 or 1), by Poisson, using any λ, allow
1 – P(X = 0 or 1 or 2) for M1
= 0.0661 or 0.0662 (3 sf) A1 Final answer
SC use of Binomial (from 0.06607…) B1 only
3
Lengths of a certain species of lizard are known to be normally distributed with standard deviation
3.2 cm. A naturalist measures the lengths of a random sample of 100 lizards of this species and obtains an !% confidence interval for the population mean. He finds that the total width of this interval is
1.25 cm.
Find !. [5]
Mark scheme
2 2 × z × 3.2
10
= 1.25 M1 OE Allow without ‘2 ×’
z = 1.953 A1 SOI
ɸ(‘their 1.953’) (= 0.9746) M1
= 1 – 2(1 – ‘0.9746’) M1 OE
= 0.9492
α = 94.9 or 95 A1 CWO
5
In the past, the mean time taken by Freda for a particular daily journey was 39.2 minutes. Following the introduction of a one-way system, Freda wishes to test whether the mean time for the journey has decreased. She notes the times, t minutes, for 40 randomly chosen journeys and summarises the results as follows. n = 40 Σt = 1504 Σt2 = 57 760
(a) Calculate unbiased estimates of the population mean and variance of the new journey time. [3]
(b) Test, at the 5% significance level, whether the population mean time has decreased. [5]
Mark scheme
3(a) 1504 188 B1
est (μ) = 37.6 or or
40 5
40 57760 2016 M1 Correct substitution in any correct formula
est (σ2) = 39 2
40 − 37.6 = 31.0154 = 65 1
57760 − 1504
2
39 40
= 31.(0) (3 sf) A1 2016
Accept or 31 651
65
3
3(b) H0: Pop mean (or μ) = 39.2 B1 Both. Not just ‘mean’
H1: Pop mean (or μ) < 39.2
'37.6'−39.2 M1 Allow use of biased variance (30.2), must have √40
'31.0154'
40
= –1.817 A1 SC FT use of biased = –1.840 for A1
‘1.817’ > 1.645 OE M1 Valid comparison ‘their 1.817’ with 1.645 or valid area
comparison 0.0346 < 0.05 OE
There is evidence that mean time has decreased A1FT FT their 1.817; in context, not definite, no contradictions
SC For 2 tail test: H1: μ ≠ 39.2 and comp 1.96, max
B0M1A1M1A0 (no FT for final mark)
5
The number of accidents on a certain road has a Poisson distribution with mean 0.4 per 50-day period.
(a) Find the probability that there will be fewer than 3 accidents during a year (365 days). [3]
(b) The probability that there will be no accidents during a period of n days is greater than 0.95.
Find the largest possible value of n. [4]
Mark scheme
4(a) λ (= 0.4 × 365 ÷ 50) = 2.92 B1
2
e–2.92(1 + 2.92 + 2.92 ) M1 Any λ. Allow one end error
2
= 0.441 (3 sf) A1
3
4(b) e–λ > 0.95 M1 Allow ‘=’ throughout
–λ > ln 0.95 or λ < 0.051293 OE M1 Attempt ln both sides
‘0.051293’ × 50 ÷ 0.4 (= 6.411) M1
Largest n is 6 (3 sf) A1 SC Trial and Improvement
Allow n = 6 or n ⩽ 6 (NOT n < 6 or n ⩾ 6 as final answer) 0.4
M1 for e–λ > 0.95 SOI; M1 for λ = n × ; M1 for use of both
50
n = 6 giving 0.9531 and n = 7 giving 0.9455; A1 n = 6
4
Bottles of Lanta contain approximately 300 ml of juice. The volume of juice, in millilitres, in a bottle is 300 + X , where X is a random variable with probability density function given by
T 3
100 − x2 −10 ≤ x ≤ 10, f x = 4000
0 otherwise.
(a) Find the probability that a randomly chosen bottle of Lanta contains more than 305 ml of juice.
[3]
(b) Given that 25% of bottles of Lanta contain more than 300 + p ml of juice, show that
p3 − 300p + 1000 = 0. [4]
(c) Given that p = 3.47, and that 50% of bottles of Lanta contain between 300 − q and 300 + q ml of juice, find q. Justify your answer. [2]
Mark scheme
5(a) 10 M1 Attempt integration of f(x), ignore limits. Condone omission
(100 − x )dx
3 2
4000 5 3
of
3 3 10 4000
= 4000 [100x - x3 ]
5
3
= 4000 (1000 – 1000
3
– 500 + 125
3
) M1 Correct limits 5 and 10. OE SOI
5 A1 For fully correct working seen including substitution of limits
= 0.156 (3 sf) or
32
3
5(b) 10 M1 1 3
(100 − x )dx =
3 2 1
4000 4 Attempt integration of f(x) with any limits and = or =
p 4 4
3
seen. Condone omission of
4000
3 3 10 1 A1 Correct integration with correct limits seen (or implied for
4000
[100x – x3 ] = 4 limits p and 10) and = 14 OE
p
3
Condone omission of
4000
3
(1000 – 1000
3
– 100p + p3 ) = 14 M1 Attempt substitution correct limits in their integration of f(x).
4000 3
Accept limits 0 to p if clearly seen, accept limits –10 and p.
Substitution must be seen.
e.g. 2000 – 100p + p3 = 1000
3
A1 AG
3 3
No errors seen
p3 – 300p + 1000 = 0
4
5(c) Curve is symmetrical about x = 0 B1 May be implied by sketch. No contradictions
or integrate f(x) between –q and +q and equate to 0.5 leading
to q3 -300q + 1000 = 0 oe
q = 3.47 B1
2
The volumes, in millilitres, of large and small cups of tea are modelled by the distributions N 200, 30 and N 110, 20 respectively.
(a) Find the probability that the total volume of a randomly chosen large cup of tea and a randomly chosen small cup of tea is less than 300 ml. [4]
(b) Find the probability that the volume of a randomly chosen large cup of tea is more than twice the volume of a randomly chosen small cup of tea. [6]
Mark scheme
6(a) N(310, 50) B1 SOI
300−'310' (= –1.414) M1 Standardise using their values
'50'
Φ(‘–1.414’) = 1 – ɸ(‘1.414’) M1 Area consistent with their values
= 0.0786 or 0.0787 (3 sf) A1 As final answer
4
6(b) P(L – 2S > 0 ) M1 OE SOI
E(X) = 200-2x110 or = – 20 B1 OE seen
Var = 30 + 22 × 20 or = 110 B1 Seen
N(–20, 110) M1 Standardising with their values. Mean and variance must come
0 − ('−20' ) from a combination attempt.
(= 1.907)
'110'
1 – Φ(‘1.907’) M1 Correct area consistent with their working
= 0.0283 (3 sf) A1 Final answer
6
A national survey shows that 95% of year 12 students use social media. Arvin suspects that the percentage of year 12 students at his college who use social media is less than the national percentage.
He chooses a random sample of 20 students at his college and notes the number who use social media.
He then carries out a test at the 2% significance level.
(a) Find the rejection region for the test. [4]
(b) Find the probability of a Type I error. [1]
(c) Jimmy believes that the true percentage at Arvin’s college is 70%. Assuming that Jimmy is correct, find the probability of a Type II error. [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) P(X ⩽ n) (n ⩽ 20) attempted, using B(20, 0.95) M1 OE
P(X ⩽ 17) or P(X ⩽ 16) attempted, using B(20, 0.95) M1 OE
(P(X ⩽ 17)) = 0.0755 and (P(X ⩽ 16)) = 0.0159 A1 OE (0.925 and 0.984) both correct
Rej region is X ⩽ 16 or X < 17 A1 Dependent on M1M1 and previous answers correct to at least
0.075/0.076 and 0.016 or 0.92/0.93 and 0.98
Correct unsupported answers of 0.0755 and 0.0159 OE scores
M1 M1 A0
4
7(b) 0.0159 B1 FT their rejection region, from Binomial in a,
if P(X in rejection region) < 0.025
1
7(c) Use of B(20, 0.7) M1
P(X > 16 | p = 0.7) M1 Correct method using B(20, 0.7)
= 0.107 A1
3
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Mathematics 9709/62 for February/March 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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