Mathematics 9709/61 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The lengths, X centimetres, of a random sample of 7 leaves from a certain variety of tree are as follows.
5.2 4.8 5.5 6.1 4.8 3.9 4.4
(a) Calculate unbiased estimates of the population mean and variance of X . [3]
It is now given that the true value of the population variance of X is 0.55, and that X has a normal distribution.
(b) Find a 95% confidence interval for the population mean of X . [3]
Mark scheme
1(a) Σx = 34.7 = 4.9571 or 4.96 (3 sf) B1
7 7
2
(Σx = 175.15)
7 "175.15" M1
− "4.9571"2
6 7
0.523 (3 sf) A1
3
1(b) 0.523 M1
'4.96'± z ×
7
(FT their mean and standard deviation)
z = 1.96 B1
4.42 to 5.49 (3 sf) A1
3
In the past the yield of a certain crop, in tonnes per hectare, had mean 0.56 and standard deviation
0.08. Following the introduction of a new fertilizer, the farmer intends to test at the 2.5% significance level whether the mean yield has increased. He finds that the mean yield over 10 years is 0.61 tonnes per hectare.
(a) State two assumptions that are necessary for the test. [2]
(b) Carry out the test. [5]
Mark scheme
2(a) Assume standard deviation unchanged or standard deviation = 0.08 B1
Assume yields normally distributed B1
2
2(b) H0: Population mean yield (or μ) = 0.56 B1
H1: Population mean yield (or μ) > 0.56
0.61 − 0.56 M1
0.08
10
1.976 A1
Comp 1.96 M1
There is evidence that mean yield has increased A1
5
The masses, in kilograms, of large sacks of flour and small sacks of flour have the independent distributions N 40, 1.52 and N 12, 0.72 respectively.
(a) Find the probability that the total mass of 6 randomly chosen large sacks of flour is more than
245 kg. [4]
(b) Find the probability that the mass of a randomly chosen large sack of flour is less than 4 times the mass of a randomly chosen small sack of flour. [6]
Mark scheme
3(a) N(240, 6 × 1.52) or N(240, 13.5) M1
245 − "240" M1
(= 1.361)
"13.5"
1 – Φ(“1.361”) M1
0.0867 (3 sf) A1
4
3(b) Use of L – 4S or similar M1
E(L – 4S) = –8 B1
Var(L – 4S) = 1.52 + 16 × 0.72 or 10.09 B1
0 − ("− 8") M1
( = 2.519 )
"10.09"
Φ(“2.519”) M1
0.994 (3 sf) A1
6
A fair spinner has five sides numbered 1, 2, 3, 4, 5. The score on one spin is denoted by X .
(a) Show that Var X = 2. [1]
Fiona has another spinner, also with five sides numbered 1, 2, 3, 4, 5. She suspects that it is biased so that the expected score is less than 3. In order to test her suspicion, she plans to spin her spinner 40 times. If the mean score is less than 2.6 she will conclude that her spinner is biased in this way.
(b) Find the probability of a Type I error. [4]
(c) State what is meant by a Type II error in this context. [1]
Mark scheme
4(a) (12 + 22 + 32 + 42 + 52) ÷ 5 – 32 (= 2 AG) B1
1
4(b) N(3, 2) M1
2.6 − "3" M1
( = −1.789 )
2
40
Φ(“–1.789”) = 1 – Φ(“1.789”) M1
0.0367 to 0.0368 A1
4
4(c) Concluding that spinner is unbiased when it is biased B1
1
Each week a sports team plays one home match and one away match. In their home matches they score goals at a constant average rate of 2.1 goals per match. In their away matches they score goals at a constant average rate of 0.8 goals per match. You may assume that goals are scored at random times and independently of one another.
(a) A week is chosen at random.
(i) Find the probability that the team scores a total of 4 goals in their two matches. [2]
(ii) Find the probability that the team scores a total of 4 goals, with more goals scored in the home match than in the away match. [3]
(b) Use a suitable approximating distribution to find the probability that the team scores fewer than
25 goals in 10 randomly chosen weeks. [4]
(c) Justify the use of the approximating distribution used in part (b). [1]
Mark scheme
5(a)(i) 2.94 M1
e −2.9 ×
4!
0.162 (3 sf) A1
2
5(a)(ii) 2.14 −0.8 2.13 −0.8 B1
e −2.1 × ×e + e −2.1 × × e × 0.8
4! 3!
(B1 for either expression correct, M1 for P(4, 0) + P(3, 1)) M1
0.113 (3 sf) A1
3
5(b) N(29, 29) M1
24.5 − 29 M1
( = −0.83563)
29
1 – Φ(“0.836”) M1
0.202 (3sf) A1
4
5(c) 29 is large or 29 > 15 B1
1
The length of time, T minutes, that a passenger has to wait for a bus at a certain bus stop is modelled by the probability density function given by
T 3
20t − t2 0 ≤ t ≤ 20, f t = 4000
0 otherwise.
(a) Sketch the graph of y = f t. [1]
(b) Hence explain, without calculation, why E T = 10. [1]
(c) Find Var T . [3]
(d) It is given that P T < 10 + a = p, where 0 < a < 10.
Find P 10 − a < T < 10 + a in terms of p. [2]
(e) Find P 8 < T < 12. [3]
(f) Give one reason why this 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) ‘Tails down’ parabola only from x = 0 to 20 shown B1
1
6(b) Symmetrical B1
1
6(c) 3
20
3 t 4 t 5 20 M1
4000 0 ( )
20t 3 − t 4 dx = 20 −
4000 4 5 0
3 M1
Var(T) = × 160000 − 102
4000
20 A1
3
6(d) (p – 0.5) × 2 or 1 – 2(1 – p) M1
2p – 1 A1
2
6(e) 3
12 M1
4000 8( )
20t − t 2 dx
3 t 2 t 3 12 3 512 A1
20 − = 1440 − 576 − 640 +
4000 2 3 8 4000 3
37 A1
or 0.296
125
3
6(f) Does not allow times greater than 20 minutes B1
1
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Mathematics 9709/61 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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