Mathematics 9709/71 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 7, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
At an internet café, the charge for using a computer is 5 cents per minute. The number of minutes for which people use a computer has mean 23 and standard deviation 8.
(i) Find, in cents, the mean and standard deviation of the amount people pay when using a computer.
[2]
(ii) Each day, 15 people use computers independently. Find, in cents, the mean and standard deviation of the total amount paid by 15 people. [3]
Mark scheme
1(i) Mean = 115 B1
SD = 40 B1
2
1(ii) Mean = 15 × ‘115’ = 1725 B1ft
15 × ‘40’2 (= 24000) M1 or SD = √15 × ‘40’. ft their (i)
SD = √24000 A1 Accept √24000 SC: Allow correct answers in dollars
SD = 155 (cents) (3 sf)
3
The time, in minutes, that John takes to travel to work has a normal distribution. Last year the mean and standard deviation were 26.5 and 4.8 respectively. This year John uses a different route and he finds that the mean time for his first 150 journeys is 27.5 minutes.
(i) Stating a necessary assumption, test at the 1% significance level whether the mean time for his journey to work has increased. [6]
(ii) State, with a reason, whether it was necessary to use the Central Limit theorem in your answer to part (i). [1]
Mark scheme
2(i) Assume sd still 4.8 or is unchanged B1 or Assume the 150 times can be treated as a random sample / are independent
H0: Pop mean = 26.5 B1 Allow ‘µ ’ but not just ‘mean’
H1: Pop mean > 26.5
27.5 − 26.5 M1 Standardise, with √
4.8 Accept CV method
150
= 2.552 A1
Comp with z-value M1 or comp 1 – Φ(‘2.552’) with 0.01
‘2.552’ > 2.326 1 – 0.9946 = 0.0054 < 0.01
There is evidence time has increased A1ft oe No contradictions
(2 tail test scores max. B1 B0 M1 A1 M1 (for comparison with 2.576) A0 no ft)
6
2(ii) No because pop is normal so distr of X is B1 Condone just ‘No because pop is normal’
normal
1
Sumitra has a six-sided die. She suspects that it is biased so that it shows a six less often than it would if it were fair. She decides to test the die by throwing it 30 times and noting the number of throws on which it shows a six.
(i) It shows a six on exactly 2 throws. Use a binomial distribution to carry out the test at the 5% significance level. [5]
(ii) Later, Sumitra repeats the test at the 5% significance level by throwing the die 30 times again.
Find the probability of a Type I error in this second test. [2]
Mark scheme
3(i) H0: P(6) = 16 B1
H1: P(6) < 16
( 56 )30 + 30( 16 ) × ( 56 )29 + 30C2( 16 )2 × ( 56 )28 M1 Allow one term incorrect, omitted or extra
= 0.103 A1
‘0.103’ > 0.05 M1
No evidence (at 5% level) that die biased A1ft oe No contradictions
5
3(ii) ( 56 )30 + 30( 16 ) × ( 56 )29 M1
P(Type I) = 0.0295 A1
2
(a) f x
x
0 a
The diagram shows the graph of the probability density function, f, of a random variable X , where a is a constant greater than 0.5. The graph between x = 0 and x = a is a straight line parallel to the x-axis.
(i) Find P X < 0.5 in terms of a. [2]
(ii) Find E X in terms of a. [1]
1 a2 .
(iii) Show that Var X = 12 [2]
(b) A random variable T has probability density function given by t
3
2 ≤ t ≤ 4, g t = 2 t − 12
0 otherwise.
Find the value of b such that P T ≤ b = 34 . [4]
Mark scheme
4(a)(i) 0.5 × 1/a = ( 0.5 ) M1 Or attempt to integrate f(x) (=1/a) between 0 and 0.5
a
1
= 2a oe A1 Accept 0.5/a for A1
2
4(a)(ii) a B1
2
1
4(a)(iii) a M1 Integ their x²f(x) from 0 to a and sub their mean2
∫ 0
x2
a
dx – (‘ a2 ’)2
Var(X) = a3 – a4
2 2
A1 Must see this line oe
2
a
(Var(X) = 12 AG)
2
4(b) b M1 Attempt integ g(t) ignore limits
∫
2
3
2( t −1)2
dt
b A1 Correct integral
− 2(t3−1)
2
− 32 ( ( b−1 1) − 1) = 34 M1 Attempt subst correct limits in their integ and = 3
4
(1 – ( b−1 1) = 12 )
b=3 A1
4
(a) The random variable X has the distribution Po 2.3.
(i) Find P 2 ≤ X ≤ 4. [2]
(ii) Find the probability that the sum of two independent values of X is greater than 2. [3]
(iii) The random variable S is the sum of 50 independent values of X . Use a suitable approximating distribution to find P S ≤ 110. [4]
(b) The random variable Y has the distribution Po ,. Given that P Y = 3 = P Y = 5, find ,. [3]
Mark scheme
5(a)(i) 2
e–2.3( 2.32 + 2.3
3
+ 2.34!
4
M1 Allow one end error
3!
= 0.585 A1
2
5(a)(ii) (λ) = 4.6 B1
1 – e–4.6(1 + 4.6 + 4.62 )
2
M1 any λ, Allow one end error
= 0.837 (3 sf) A1
3
5(a)(iii) S ~ N(115, 115) B1 May be implied
110.5 −115 M1 Allow with wrong or no cc OR no √
115 (= –0.420)
1 – Φ(‘0.420’) (= 1 – 0.663) M1
= 0.337 A1 Accept alternative method using N(2.3, 2.3) no mixed methods.
4
5(b) 3
e–λ × λ3! = e–λ × λ5!
5
M1
5
λ3 = 4λ×5 or λ2 = 20 oe A1 any correct simplification without e–λ or !
λ = √20 or 2√5 or 4.47 (3 sf) A1
3
Ramesh plans to carry out a survey in order to find out what adults in his town think about local sports facilities. He chooses a random sample from the adult members of a tennis club and gives each of them a questionnaire.
(i) Give a reason why this will not result in Ramesh having a random sample of adults who live in the town. [1]
(ii) Describe briefly a valid method that Ramesh could use to choose a random sample of adults in the town. [2]
Ramesh now uses a valid method to choose a random sample of 350 adults from the town. He finds that 47 adults think that the local sports facilities are good.
(iii) Calculate an approximate 90% confidence interval for the proportion of all adults in the town who think that the local sports facilities are good. [4]
(iv) Ramesh calculates a confidence interval whose width is 1.25 times the width of this 90% confidence interval. Ramesh’s new interval is an x% confidence interval. Find the value of x.
[3]
Additional Page
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Mark scheme
6(i) Biased towards people who like tennis B1 or other sensible
Excludes people who don't like tennis
1
6(ii) Obtain a list of all people in the town B1
Use random numbers B1 or, e.g. pick numbers from a hat or other sensible
2
6(iii) 47 (1− 47 ) M1
350 350
Var(p) = 350 (= 0.000332152)
z = 1.645 B1
47 (1− 47 ) M1 Must be a z value
47
350
±z 350
350
350
0.104 to 0.164 (3 sf) A1 Must be an interval
4
6(iv) 1.25 × 1.645 (= 2.056) M1 47 (1− 47 )
350 350
or 1.25 × their width ÷ 2 ÷ their 350 (Complete method)
Φ(‘2.056’) (= 0.980) M1 Attempt Φ(their z)
x = 96 (2 sf) A1 Allow 0.96 (2 sf) CWO
3
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Mathematics 9709/71 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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