Mathematics 9709/63 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The number of characters in emails sent by a particular company is modelled by the distribution
N 1250, 4802 .
Find the probability that the mean number of characters in a random sample of 100 emails sent by the company is more than 1300. [3]
Mark scheme
1 1300 + 1 − 1250 M1 Allow with incorrect or omitted continuity correction
200 or 1300 − 1250 [= 1.042] Must have 10
480 480 Accept totals method
10 10
1 – Φ(‘1.042’) M1 For area consistent with their values
0.149 ( 3 s.f.) A1
3
Anton believes that 10% of students at his college are left-handed. Aliya believes that this is an under- estimate. She plans to carry out a hypothesis test of the null hypothesis p = 0.1 against the alternative hypothesis p > 0.1, where p is the actual proportion of students at the college that are left-handed.
She chooses a random sample of 20 students from the college. She will reject the null hypothesis if at least 5 of these students are left-handed.
(a) Explain what is meant by a Type I error in this context. [1]
(b) Find the probability of a Type I error in the test. [3]
(c) Given that the true value of p is 0.3, find the probability of a Type II error in the test. [2]
Mark scheme
2(a) Conclude more than 10% of the students are left handed when this is not true B1 OE. Must be in context (accept use of p).
Need the context of one tail test.
1
2(b) 1 – (0.920 + 20 × 0.919 × 0.1 + 20C2 × 0.918 × 0.12 + 20C3 × 0.917 × 0.13 + 20C4 × M2 M2: fully correct
0.916 × 0.14) M1: attempt 1 – P(X = 0, 1, 2, 3, 4);
allow 1 – P(X = 0,1,2,3,4,5) or 1 – P(X = 0,1,2,3)
need 1 – …
the method mark cannot be implied
0.0432 (3 s.f.) A1 If M0 awarded allow SC B2 for 0.0432
3
2(c) 0.720 + 20 × 0.719 × 0.3 + 20C2 × 0.718 × 0.32 + 20C3 × 0.717 × 0.33 + 20C4 × 0.716 M1 Attempt to find P (≤ 4) using B(20,0.3)
× 0.34
Allow one end error
The method mark cannot be implied
0.238 or 0.237 (3 s.f.) A1 If M0 awarded allow SC B1for 0.238 or 0.237
2
Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries.
(a) Give a reason for using a sample rather than the whole population in carrying out this test. [1]
A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours2.
(b) Test, at the 2% significance level, whether type B batteries have a shorter mean life than type A batteries. [5]
(c) Calculate a 94% confidence interval for the population mean life of type B batteries. [3]
Mark scheme
3(a) Batteries unusable after testing or Population too big B1
or too costly or too time consuming to use the whole population oe
1
3(b) H0: μ = 150 B1 Or population mean = 150; not just ‘mean’ = 150
H1: μ < 150
147 − 150 M1 Allow with continuity correction
225 ÷ 120 Need 120
–2.191 A1 Condone – 2.19
–2.191 < –2.054 [or –2.055] M1 OE. For valid comparison with 2.054 or 2.055
Or 0.0143 (or 0.0142) < 0.02
For two tail test allow comp –2.326 OE if H1: μ ≠ 150
(can score B0M1A1M1A0 max 3/5 )
[Reject Ho] There is evidence that the (mean) life of type B is less than type A1 FT In context, not definite with no contradictions
A (or less than 150) Accept critical value method 147.19 M1A1
147 < 147.19 M1 conclusion A1
Or 150 > 149.81
5
3(c) 147 ± z × 15 M1 Expression of correct form must be a z value
120
z = 1.881 [or 1.882] B1
144 to 150 (3 s.f.) A1 Must be an interval
Incorrect z value can only score M1B0A0
3
Each box of Seeds & Raisins contains S grams of seeds and R grams of raisins. The weight of a box, when empty, is B grams. S, R and B are independent random variables, where S ∼ N 300, 45,
R ∼ N 200, 25 and B ∼ N 15, 4. A full box of Seeds & Raisins is chosen at random.
(a) Find the probability that the total weight of the box and its contents is more than 500 grams. [5]
(b) Find the probability that the weight of seeds in the box is less than 1.4 times the weight of raisins in the box. [5]
Mark scheme
4(a) T~N(515, 74) B1 B1 for N(515, ..) give at early stage
B1 B1 for Var = 45 + 25 + 4 = 74 give at early stage
500 − '515' [= –1.744] M1 Standardise with their values. No standard
'74' deviation/variance mix
Need combination for variance. Allow continuity
correction.
Φ(’1.744’) M1 Area consistent with their working
0.959 or 0.96[0] (3 s.f.) A1
5
4(b) E(S – 1.4R) = 300-1.4x200 =20 B1 Give at early stage
Var(S – 1.4R) = 45 + 1.42×25 = 94 B1 Give at early stage
SC: if B0B0 awarded allow SC B1 for 14 and 105.84
0 − (20) M1 Standardise with their values. No standard
[= –2.063] deviation/variance mix. Need combination for
'94'
variance.
1 – Φ(‘2.063’) M1 Area consistent with their working
0.0196 (3 s.f.) A1
5
The number of clients who arrive at an information desk has a Poisson distribution with mean 2.2 per
5-minute period.
(a) Find the probability that, in a randomly chosen 15-minute period, exactly 6 clients arrive at the desk. [3]
(b) If more than 4 clients arrive during a 5-minute period, they cannot all be served.
Find the probability that, during a randomly chosen 5-minute period, not all the clients who arrive at the desk can be served. [2]
(c) Use a suitable approximating distribution to find the probability that, during a randomly chosen
1-hour period, fewer than 20 clients arrive at the desk. [4]
Mark scheme
5(a) λ = 6.6 B1
6 M1 Any λ
e-6.6 × 6.6
6!
0.156 (3 s.f.) A1 If M0 awarded SC B1 for 0.156
3
5(b) 2 3 4 M1 Allow one end error. Need 1 – …
1 – e-2.2(1 + 2.2 + 2.2 + 2.2 + 2.2 )
2 3! 4! Any λ
0.0725 (3 s.f.) A1 If M0 awarded SC B1 for 0.0725
2
5(c) N(26.4, 26.4) B1 Give at early stage 2.2 × 12
19.5 − '26.4' [= –1.343] M1 Standardising with their values. Allow wrong or no
'26.4' continuity correction
ɸ('–1.343') = 1 – ɸ('1.343') M1 Area consistent with their working
0.0897 or 0.0896 (3 s.f.) A1
4
A random sample of 5 values of a variable X is given below.
2 3 3 5 a
(a) Find an expression, in terms of a, for the mean of these values. [1]
It is given that an unbiased estimate of the population variance of X , using these values, is 4. It is also given that a is positive.
(b) Find and simplify a quadratic equation in terms of a and hence find the value of a. [3]
Mark scheme
6(a) 13 + a B1 2+ 3+ 3+ 5+ a
5 Accept .
5
Do not ignore subsequent working
1
6(b) M1 Use of correct formula using their value from (a), in
(13 + a )
( )
2
5 47 + a 2 − 13 + a
2
1
= 4 or 4 47 + a − =4
2
terms of a, and equate to 4
4 5 5 5
2a2 – 13a – 7 = 0 A1 Any correct three-term quadratic equation rearranged
to a form ready to solve
a=7 A1 Condone the other value of a ( − 1 )
2
3
The random variables X and W have probability density functions f and g defined as follows:
T p a2 − x2 0 ≤ x ≤ a, f x =
0 otherwise,
T q a2 − w2 −a ≤ w ≤ a, g w =
0 otherwise, where a, p and q are constants.
(a) (i) Write down the value of P X ≥ 0. [1]
(ii) Write down the value of P W ≥ 0. [1]
(iii) Write down an expression for q in terms of p only. [1]
(b) Given that E X = 3, find the value of a. [6]
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)(i) 1 B1 no ambiguity
1
7(a)(ii) 1 B1 No ambiguity
2
1
7(a)(iii) [q = ] 1 p B1 Accept 2q = p
2
1
7(b) a M1 Attempt to integrate f(x) and equated to 1
p (a 2 − x 2 )dx = 1
0
2 a3 p = 1 A1 OE, simplified
3
a a M1 Attempt to integrate xf(x), with multiplier p or 3
2 3
" 23a3 " (a x − x )dx = 3 or " 23a3 " ( a 2 x − x3 )dx = 3 2a 3
0 0 or their p, and equate to 3
4 A1 May be implied by next line
p× a =3
4
4 M1 OE. Substitute from one equation into the other.
" 3 3 "× a = 3
2a 4 FT their equations
a=8 A1
6
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/63, May/June 2022 (9709_s22_ms_63), and the question paper (9709_s22_qp_63), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/63 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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