Mathematics 9709/62 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.
(a) A javelin thrower noted the lengths of a random sample of 50 of her throws. The sample mean was 72.3 m and an unbiased estimate of the population variance was 64.3 m2.
Find a 92% confidence interval for the population mean length of throws by this athlete. [3]
(b) A discus thrower wishes to calculate a 92% confidence interval for the population mean length of his throws. He bases his calculation on his first 50 throws in a week.
Comment on this method. [1]
Mark scheme
1(a) M1 Expression of correct form (allow only one side for M1).
72.3 ± z 64.3 Must be a z value
50
z = 1.751 B1 Accept 1.75 if nothing better seen
CI is 70.3 to 74.3 metres (3 s.f.) A1 Allow without units Must be an interval
3
1(b) Not random sample B1 Need ‘random’ or ‘not representative/biased because…’
OE
1
In the past, the mean height of plants of a particular species has been 2.3 m. A random sample of
60 plants of this species was treated with fertiliser and the mean height of these 60 plants was found to be 2.4 m. Assume that the standard deviation of the heights of plants treated with fertiliser is 0.4 m.
Carry out a test at the 2.5% significance level of whether the mean height of plants treated with fertiliser is greater than 2.3 m. [5]
Mark scheme
2 H0: Pop mean height = 2.3 B1 Not just ‘mean’
H1: Pop mean height > 2.3 Allow μ
2.4 − 2.3 M1 For standardising, must have 60
0.4
60
1.936 or 1.937 or 1.94 A1
‘1.936’ < 1.96 M1 Valid comparison with 1.96
Or 2.64% > 2.5% OE
Accept 1.936 < 2.24 or 2.64% > 1.25% OE if H1µ ≠ 2.3
[Do not reject H0] A1 FT FT their z
No evidence that (mean) height (with fertiliser) is more than without In context, not definite. E.g. not ‘Mean height is not
greater’ with no contradictions
No FT for 2 tail test (max B0 M1 A1 M1 A0 3/5)
Accept critical values method 2.401 (M1 A1) 2.4 < 2.401
(M1) Condone 2.299 (M1 A1) < 2.3 (M1)
A1 conclusion
5
It is known that 1.8% of children in a certain country have not been vaccinated against measles. A random sample of 200 children in this country is chosen.
(a) Use a suitable approximating distribution to find the probability that there are fewer than 3 children in the sample who have not been vaccinated against measles. [4]
(b) Justify your approximating distribution. [2]
Mark scheme
3(a) Poisson B1 SOI
Mean = 3.6 B1 Can be awarded for N(3.6, …)
2 M1 Allow any λ
e-3.6(1 + 3.6 + 3.6 )
2 Allow one end error
Expression must be seen
0.303 (3 s.f.) A1 If M0 awarded allow SC B1 for 0.303
SC Use of binomial: B1 for answer 0.300 (3 sf)
4
3(b) [Binomial with] 200 > 50 B1
[200 × 0.018 =] 3.6 < 5 or [p =] 0.018 < 0.1 B1 If B0 B0 then SC n large, p small: B1 or n large np < 5:
B1 or n > 50 and either np < 5 or p < 0.1: B1
2
The number of cars arriving at a certain road junction on a weekday morning has a Poisson distribution with mean 4.6 per minute. Traffic lights are installed at the junction and a council officer wishes to test at the 2% significance level whether there are now fewer cars arriving. He notes the number of cars arriving during a randomly chosen 2-minute period.
(a) State suitable null and alternative hypotheses for the test. [1]
(b) Find the critical region for the test. [4]
The officer notes that, during a randomly chosen 2-minute period on a weekday morning, exactly
5 cars arrive at the junction.
(c) Carry out the test. [2]
(d) State, with a reason, whether it is possible that a Type I error has been made in carrying out the test in part (c). [1]
The number of cars arriving at another junction on a weekday morning also has a Poisson distribution with mean 4.6 per minute.
(e) Use a suitable approximating distribution to find the probability that more than 300 cars will arrive at this junction in an hour. [3]
Mark scheme
4(a) H0: Pop mean = 4.6 [or 9.2] B1 or λ = 4.6 or μ (Not just ‘mean’)
H1: Pop mean < 4.6 [or 9.2] or λ < 4.6
1
4(b) Use of Poisson with λ = 9.2 B1 SOI
2 3 M1 At least one of these attempted correct λ
P(X ⩽ 3) = e–9.2(1 + 9.2 + 9.2 + 9.2 ) = 0.0184 or 0.018 [< 0.02]
2 3! (with Poisson expression seen not implied)
4
P(X ⩽ 4) = 0.0184 + e–9.2 × 9.2 = 0.0486 or 0.049 [> 0.02] *A1 Both correct
4!
SC Use of λ = 4.6 scores B1 for P(X = 0) = 0.01[0][1]
and P(X ⩽ 1) = 0.056[3]only
CR is X ⩽ 3 DA1 From CWO and at least one comparison seen
SC If M0 awarded allow *B1 for both 0.018 and 0.049 or
better and DB1 for correct critical region from CWO and
at least one comparison seen.
4
4(c) 5 is not in critical region OR P(X ⩽ 5) =0.104 > 0.02 M1 For a comparison (i.e. 5 > 3) OE
so [not reject H0] no evidence that number of cars arriving is now fewer A1 FT In context, not definite No contradictions
e.g. not ‘No. of cars arriving is not fewer’
ft their critical region if used (but must be from Poisson
and integers)
2
4(d) No, because H0 was not rejected B1 FT OE, FT their (c)
1
4(e) N(276, 276) B1 SOI
300.5 − 276 [= 1.475] M1 Standardising with their values
276 Allow with wrong or no continuity correction
1 – ɸ(‘1.475’) = 0.0701 (3 s.f.) A1 SC Use of Poisson: B1 for answer 0.0727 (3 sf)
3
A random variable X has probability density function given by
T
3 4x − x2 2 ≤ x ≤ 4, f x = 16
0 otherwise.
(a) Show that E X = 11
4
. [3]
(b) Find Var X . [3]
(c) Given that the median of X is m, find P m < X < 3. [3]
Mark scheme
5(a) 4
3 (4 x 2 − x3 )dx M1 Attempt to integrate xf(x) Ignore limits
16 2 (must see a power increase for attempted integration)
3 4 x3 x 4 4 M1 Attempt integrate xf(x) with correct limits
= −
16 3 4 2 (must see a power increase for attempted integration)
= 3 ( 256 − 64 − ( 32 − 4) ) = 11 (AG) A1 Correct substitution of correct limits (at least 2 terms
16 3 3 4 seen) and answer seen. No errors seen i.e. NO recovery of
errors and no non-exact decimals (e.g. 21.33) seen
3
5(b) 4
3 (4 x3 − x 4 )dx *M1 Attempt to integrate x2 f(x) with correct limits
16 2 (integration must be seen not implied. Must see a power
increase for attempted integration)
DM1 their
x f ( x ) dx – ( 114 ) , with x f ( x ) dx evaluated, not
5 4
= 3 x 4 − x [= 39 or 7.8]
2 2 2
16 5 2 5
necessarily simplified
Var(X) = ‘ 39 ’ – ( 11 )2
5 4
= 19 or 0.2375 (or 0.238 (3 sf)) A1 39 19
80 SC If M0 then score B1 for and B1 for
5 80
3
5(c) 3
3 (4 x − x 2 )dx M1 Attempt to integrate correct integral and limits
16 2 must see a power increase for attempted integration
Oe (Integrate 3 to 4)
OR ALTERNATIVE METHOD integrate from m to 4 or
2 to m and equate to 0.5 to obtain cubic (m3 -6m2 +
24 = 0 oe)
(NB Integrating from m to 3 and equating to 0.5M0)
( ( )) f ( x ) dx – 12 oe (1/2 – 5/16)
3 3
3 8 11
= 3 2 x2 − x = 16 18 − 9 − 8 − 3 = 16
Their
16 3 2
M1 OR
‘ 11 ’ – 1 ALTERNATIVE METHOD m obtained from cubic (m =
16 2
2.69459) and attempt to integrate f(x) from ‘their m’ (2 <
m < 4) to 3 must see a power increase for attempted
integration and limits substituted
3 or 0.1875 A1 Condone 0.187 or 0.188
16
3
The masses, in kilograms, of large and small sacks of grain have the distributions N 53, 11 and
N 14, 3 respectively.
(a) Find the probability that the mass of a randomly chosen large sack is greater than four times the mass of a randomly chosen small sack. [5]
(b) A lift can safely carry a maximum mass of 1000 kg.
Find the probability that the lift can safely carry 12 randomly chosen large sacks and 25 randomly chosen small sacks. [5]
Mark scheme
6(a) E(D) = 53 – (4 × 14) = –3 B1 OE Give at early stage
Var(D) = 11 + 42 × 3 [= 59] B1 or
(11 + 42 ´3) (= 7.68 (3 s.f.))
Give at early stage
0 − (−3) M1 For standardising with their values (var must be from a
[= 0.391] combination attempt)
59
Ignore continuity correction attempts
1 – Φ(‘0.391’) M1 For area consistent with their values
0.348 (3 s.f.) A1 As final answer
5
6(b) E(T) = 12 × 53 + 25 × 14 [= 986] B1 Give at early stage (N.B. accept E(T – 1000)= –14)
Var(T) = 12 × 11 + 25 × 3 [= 207] B1 Or (12´11 + 25´ 3) (= 14.4 (3sf))
Give at early stage
1000 − 986 [= 0.973] M1 For standardising with their values (var must be from a
207 combination attempt)
Ignore continuity correction attempts
Φ(‘0.973’) M1 For area consistent with their values
0.835 (3 sf) A1 As final answer
5
X is a random variable with distribution Po 2.90. A random sample of 100 values of X is taken.
Find the probability that the sample mean is less than 2.88. [5]
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 X ~ N(2.9, 2.9 ) OR Totals method N(290, 290) B1 B1 for N(2.9, …) OR N(290, …)
100
B1 B1 for Var = 2.9 OR for var = 290 SOI
100
2.88 − 2.90 [= –0.1174] OR 288 − 290 M1 Standardising with their values Allow without –ve sign
2.9 290 AND/OR with incorrect continuity correction
100 No mixed methods
1– Φ(‘0.1174’) M1 For area consistent with their values
0.453 (3 sf) A1 As final answer
Alternative method for question 7
X ~ N(2.9, 2.9 ) OR Totals method N(290, 290) B1 B1 for N(2.9, …) OR N(290, …)
100
B1 B1 for Var = 2.9 OR Var = 290 stated or implied
100
(2.88 − 1 ) − 2.90 M1 Standardising with their values Allow without –ve sign
200 [= –0.1468] OR (287.5 – 290)/√290 AND/OR with incorrect continuity correction
2.9 No mixed methods
100
1 – ɸ(‘0.1468’) M1 For area consistent with their values
0.442 (3 sf) A1 As final answer
5
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Mathematics 9709/62 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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