Mathematics 9709/63 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.
A random sample of 100 values of a variable X is taken. These values are summarised below.
n = 100 Σ x = 1556 Σ x2 = 29 004
Calculate unbiased estimates of the population mean and variance of X . [3]
Each day at the gym, Sarah completes three runs. The distances, in metres, that she completes in the three runs have the independent distributions W ∼ N 1520, 450, X ∼ N 2250, 720 and
Y ∼ N 3860, 1050.
Find the probability that, on a particular day, Y is less than the total of W and X . [5]
Mark scheme
2 Y – W – X ~ N(90, …) B1
Var(Y – W – X) = 1050 + 450 + 720 or 2220 B1
0 − 90 M1
(= –1.910)
'2220 '
Φ(‘–1.910’) = 1 – Φ(‘1.910’) M1
0.0281 A1
5
The number of customers who visit a particular shop between 9.00 am and 10.00 am has the distribution
Po ,. In the past the value of , was 5.2. Following some new advertising, the manager wishes to test whether the value of , has increased. He chooses a random sample of 20 days and finds that the total number of customers who visited the shop between 9.00 am and 10.00 am on those days is 125.
Use an approximating distribution to test at the 2.5% significance level whether the value of , has increased. [6]
Mark scheme
3 H0: λ = 104 (or 5.2) B1
H1: λ > 104 (or 5.2)
N(104, 104) stated or implied B1
124.5 − 104 M1
104
2.010 A1
2.010 > 1.96 M1
There is evidence that λ has increased A1
6
The random variable A has the distribution Po 1.5. A1 and A2 are independent values of A.
(a) Find P A1 + A2 < 2. [3]
(b) Given that A1 + A2 < 2, find P A1 = 1. [4]
(c) Give a reason why A1 − A2 cannot have a Poisson distribution. [1]
Sunita has a six-sided die with faces marked 1, 2, 3, 4, 5, 6. The probability that the die shows a six on any throw is p. Sunita throws the die 500 times and finds that it shows a six 70 times.
(a) Calculate an approximate 99% confidence interval for p. [4]
(b) Sunita believes that the die is fair. Use your answer to part (a) to comment on her belief. [1]
(c) Sunita uses the result of her 500 throws to calculate an !% confidence interval for p. This interval has width 0.04.
Find the value of !. [5]
Mark scheme
5(a) 70 B1
p= or 0.14
500
z = 2.576 B1
"0.14" (1 − "0.14") M1
"0.14" ± z ×
500
0.100 to 0.180 A1
4
5(b) 0.1666... is within confidence interval B1
Belief supported or justified
1
5(c) "0.14"(1 − "0.14") M1
z× = 0.02
500
z = 1.289 A1
Φ(‘1.289’) = 0.9013 M1
α = ‘0.9013’ – (1 – ‘0.9013’) M1
80.3% (3 sf) A1
5
The length, X centimetres, of worms of a certain type is modelled by the probability density function
T 6
10 − x x − 5 5 ≤ x ≤ 10, f x = 125
0 otherwise.
(a) State the value of E X . [1]
(b) Find Var X . [3]
(c) Two worms of this type are chosen at random.
Find the probability that exactly one of them has length less than 6 cm. [5]
A market researcher is investigating the length of time that customers spend at an information desk.
He plans to choose a sample of 50 customers on a particular day.
(a) He considers choosing the first 50 customers who visit the information desk.
Explain why this method is unsuitable. [1]
The actual lengths of time, in minutes, that customers spend at the information desk may be assumed to have mean - and variance 4.8. The researcher knows that in the past the value of - was 6.0. He wishes to test, at the 2% significance level, whether this is still true. He chooses a random sample of
50 customers and notes how long they each spend at the information desk.
(b) State the probability of making a Type I error and explain what is meant by a Type I error in this context. [2]
(c) Given that the mean time spent at the information desk by the 50 customers is 6.8 minutes, carry out the test. [5]
(d) Give a reason why it was necessary to use the Central Limit theorem in your answer to part (c).
[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
7(a) Later customers might spend times different from first ones B1
1
7(b) 0.02 B1
Concluding that μ ≠ 6.0, when actually μ = 6.0 B1
2
7(c) H0: μ = 6.0 B1
H1: μ ≠ 6.0
6.8 − 6.0 M1
4.8
50
2.582 A1
comp 2.326 M1
Evidence that μ ≠ 6.0 A1
5
7(d) Population distribution unknown B1
1
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 9709/63 May/June 2020 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/63, May/June 2020 (9709_s20_ms_63), and the question paper (9709_s20_qp_63), to read or download, with each question's marking points written out underneath.
How many marks is 9709/63 out of?+
Mathematics 9709/63 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/63 answers marked?+
Yes. Upload your answers to 9709/63/M/J/20 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.