Mathematics 9709/62 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A random variable X has the distribution Po (145) .
(a) Use a suitable approximating distribution to calculate P (X G 150) . [4]
(b) Justify the use of your approximating distribution in this case. [1]
Mark scheme
1(a) N(145, 145) B1 Stated or implied.
−145
± 150.5145 [= ±0.457] M1 Condone incorrect or omitted continuity correction.
Φ(‘0.457’) M1 For area consistent with their working.
= 0.676 (3sf) A1 SC: Unsupported answer of 0.676 scores B3.
Unsupported answer of 0.646 or 0.661 scores B2.
Unsupported answer of 0.6799 scores B1.
4
1(b) 145 > 15 B1 Explicit.
λ > 15 B0 if λ = 145 not stated.
Accept ⩾
Accept mean for λ.
1
Henri wants to choose a random sample from the 804 students at his college. He numbers the students from 1 to 804 and then uses random numbers generated by his calculator. The first 20 random digits produced by his calculator are as follows.
5 6 7 1 0 9 8 4 3 1 0 9 6 6 5 0 2 1 7 6
Henri’s first two student numbers are 567 and 109.
(a) Use Henri’s digits to find the numbers of the next two students in the sample. [2]
There were 30 students in Henri’s sample. He asked each of them how much time, X hours, they spent on social media each week, on average. He summarised the results as follows.
n = 30 Rx = 610 Rx 2 = 12 405
(b) Use this information to calculate an unbiased estimate of the mean of X and show that an unbiased estimate of the variance of X is less than 0.1 . [3]
(c) Henri’s friend claims that Henri has probably made a mistake in his calculation of Rx or Rx 2 .
Use your answer to part (b) to comment on this claim. [1]
Mark scheme
2(a) [567, 109], 665, 21 B2 B1 for each.
Allow 021.
If more than 2 answers given, count first two and
ISW.
2
2(b) Est(µ) = 610 or 613 B1 OE or 20.3.
30
Est(σ2) = 30
29
( 12405
30
− ( 610
30 (
) 2 ) or 291 12405 − 610
30
2
) M1 Use of correct formula.
= 0.0575 (3sf) A1 Accept 5 .
87
3
2(c) Variance is [unrealistically] small B1 FT Need both parts.
so Henri has [probably] made a mistake/claim is [probably] correct Need ‘small’ OE, not just < 0.1. FT their < 0.1
variance value (not –ve), e.g. 0.0556 (if omit 30
29
).
Accept ‘s.d. = 0.24 is small, so Henri has probably
made a mistake’.
Note: ‘mean is large/small’ scores B0, but ‘mean large
compared to variance so Henri prob made a mistake’
scores B1.
1
A student wishes to estimate the proportion, p, of students at her college who have exactly one brother.
She surveys a random sample of 50 students at her college and finds that 18 of them have exactly one brother. She calculates an approximate a% confidence interval for p and finds that the lower limit of the confidence interval is 0.244 correct to 3 significant figures.
Find a correct to the nearest integer. [4]
Mark scheme
3 18
18 ×(1− 18 ) M1 Use of correct equation.
50
−z× 50
50
50
= 0.244
z = 1.709 or 1.708 A1 Accept 1.71 if nothing better seen.
ɸ–1('1.709') = 0.956 ; 1 − 2(1 – ‘0.956’) [= 0.912] M1 Attempt area above or below their 1.709 and use
correct method to find α.
α = 91 A1 Allow α = 91%
0.91 or 91.2 score A0.
4
A random variable X has the distribution N(10, 12). Two independent values of X, denoted by X1 and
X2, are chosen at random.
(a) Write down the value of P (X 1 2 X2) . [1]
(b) Find P (X 1 2 2X2 - 3) . [5]
Mark scheme
4(a) 0.5 B1
1
4(b) E(X1 − 2X2 + 3) =10-20+3 [= −7] B1 Or equivalent using X1 − 2X2 =10 – 20 [= –10]
or E(2X2 − X1 − 3) = 20 − 10 – 3 [= 7] or 2X2 − X1 = 20 – 10 [= +10].
Var(X1 − 2X2 + 3) = 12 + 22×12 + 0 [= 60] B1
0 − ('− 7') M1 Or numerator 3–‘10’ or –3–(‘–10’), but not ‘–3 –10’
[= 0.904] (i.e. numerator must be ‘7’ or ‘–7’).
'60'
1 − Φ(‘0.904’) M1 For area consistent with their working.
= 0.183 A1
5
The number of goals scored by a sports team in the first half of any match has the distribution
X + Po (3.1) . The number of goals scored by the same team in the second half of any match has the distribution Y + Po (2.4) . You may assume that the distributions of X and Y are independent.
(a) Find P (X 1 4) . [2]
(b) Find the probability that, in a randomly chosen match, the team scores at least 5 goals. [3]
(c) Given that the team scores a total of 5 goals in a randomly chosen match, find the probability that they score exactly 3 goals in the first half. [4]
Mark scheme
5(a) 2 3
e−3.1(1 + 3.1 + 3.12! + 3.13! ) M1 Condone one end error. Any λ.
Accept fully correct Σ notation.
or e−3.1(1 + 3.1 + 4.805 + 4.965)
Expression must be seen.
or 0.0450 + 0.1397 + 0.2165 + 0.22368
= 0.625 (3sf) A1 Correct answer with no working scores SC B1.
2
5(b) [λ]= 5.5 B1 SOI
2 3 4
1 − e−5.5(1 + 5.5 + 5.52! + 5.53! + 5.54! ) M1 Condone one end error.
Any λ.
or 1 − e−5.5(1 + 5.5 + 15.125 + 27.7292 + 38.1276)
Accept fully correct Σ notation.
or 1 − e−5.5(0.004087 + 0.0224772 + 0.061812 + 0.113323 + 0.155819)
Expression must be seen.
= 0.642 or 0.643 (3sf) A1 Correct answer with no working scores SC B1 B1.
3
5(c) [P(X = 3) × P(Y = 2) = ] = e−3.1× 3.1
3
× e−2.4 × 2.42!
2
M1 Find P(3 in first half AND 2 in second half).
3!
Must see expression.
or 0.223676 × 0.261267 [= 0.05844]
[P(total 5) = ] e −5.5 × 5.55! or 0.17140
5
M1 Use of 5.5 to find P(5).
P(P(exactly 3 in 1st half given total 5) =
P(exactly 3 in 1st half and total 5) M1 Attempt at conditional probability;
P(total 5)
numerator = their 0.05844 and
denominator = P(total 5)
3 2 5
Note: ( 3.1
3!
× 2.42! )÷( 5.55! ) scores M1 M1 M1.
[= '0.05844'
'0.17140'
] = 0.341 (3sf) A1
4
The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustment is made to the machine and an inspector wishes to test whether the mean mass of cereal boxes filled by the machine has decreased.
After the adjustment is made, he chooses a random sample of 120 cereal boxes. The mean mass of these boxes is found to be 508 grams.
Assume that the standard deviation of the masses is 10 grams.
(a) Test at the 2.5% significance level whether the mean mass of cereal boxes filled by the machine has decreased. [5]
Later the inspector carries out a similar test at the 2.5% significance level, using the same hypotheses and another 120 randomly chosen cereal boxes.
(b) Given that the mean mass is now actually 506 grams, find the probability of a Type II error. [5]
Mark scheme
6(a) H0: Population mean mass = 510 g B1 Allow ‘μ’ but not just ‘mean’.
H1: Population mean mass < 510 g
508 − 510 M1 Standardising must have 120.
±
10 ÷ 120
= ± −2.191 or −2.190 A1
−2.191 < −1.96 or 2.191 > 1.96 M1 OE
Area comparison: 0.0143 or 0.0142 < 0.025 For valid comparison.
Inequality sign the wrong way round scores M1 A0.
[Reject H0] There is sufficient evidence to suggest that the [mean] mass has A1FT OE
decreased In context (must be ‘decreased’ OE, not ‘changed’);
not definite. No contradictions. Condone ‘there is
sufficient evidence to support the inspector’s claim’.
NB: Accept alternative method using critical value (=
508.21) and comparison with 508. Condone 509.79
compared with 510.
Two tail test scores maximum B0 M1 A1 M1 A0;
must have comparison with 0.0125 or 2.24/2.241.
5
6(b) cv − 510 M1 Standardising to find critical value (must use 510 and
= −1.96
10 ÷ 120 10÷ 120) . Accept ± 1.96.
cv = 508.21 A1 Accept 3 sf if nothing better seen.
Note: cv could be found in (a).
508.21 − 506 M1 Standardising with their 508.21 and 506 (must use
z=± [= 2.421]
10 ÷ 120 10÷ 120) .
P ( X > 508.21 | µ = 506) = 1 − Φ(‘2.421’) M1 For area consistent with their working.
= 0.0077 to 0.0080 (2sf) A1 510 − 506
Note: scores max M0 A0 M1 M1 A0.
10 ÷ 120
5
The probability density function, f, of a random variable X is given by
k (1 + cos x) f (x) = )
0 G x G r,
0 otherwise,
where k is a constant.
1
(a) Show that k = . [3]
r
(b) Verify that the median of X lies between 0.83 and 0.84 . [3]
(c) Find the exact value of E(X ). [4]
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) π M1 Attempt integrate f(x) with correct limits and equate to
k (1 + cos x)dx = 1
0
1.
π
k [ x + sin x ]0 = 1 A1 Correct integration.
[e.g. k(π + sin π − (0 + 0)) = 1], kπ = 1, k = π1 A1 AG
Some evidence of substitution of limits, i.e. at least
one interim step (e.g. kπ = 1) as minimum
requirement.
Convincingly obtained; no errors seen.
3
7(b) M1 Substitute correct limits into their integral.
π [
x + sin x ]0 π [
x + sin x ]0
1 0.83 1 0.84
OR1: integrate 0 to 0.83 and 0.84 to π.
or π1 (0.83 + sin 0.83) or π1 (0.84 + sin 0.84) OR2: use g(m) = m + sin m – (π/2) and find g(0.83)
and g(0.84).
OR3: use h(m) = m + sin m and find h(0.83) and
h(0.84).
Both attempted.
A1 OR1: 0.499 and 0.496.
= 0.499 (3 sf) = 0.504 (3 sf) OR2: g(0.83) = – 0.00286/7 and g(0.84) = 0.0138/9.
OR3: h(0.83) = 1.57 and h(0.84) = 1.58 or 1.59.
Both correct.
‘0.499’ < 0.5 < ‘0.504’ hence 0.83 < median < 0.84 A1FT FT their areas; dep 0.5 is between their areas OE.
Equivalent to OR1: 0.499 < 0.5 and 0.496 < 0.5, so
–0.000912 < 0 < 0.00441 hence 0.83 < median < 0.84 0.83 < m < 0.84.
OR2: g(0.83) > 0 g(0.84) < 0 OE, so
0.83 < m < 0.84.
OR3: h(0.83) < π2 h(0.84) > π2 , so 0.83 < m < 0.84.
Both statements needed.
Note: A score of M1 A0 A1FT is possible.
If 0 scored, SC: π1 (m + sin m) = 0.5 B1
and m = 0.831 to 0.832, so 0.83< m < 0.84 B1.
3
7(c) π M1* Attempt integrate xf(x).
1
π 0
( x + x cos x)dx Ignore limits.
π DM1 OE
1 x2
π
π
= + [ x sin x ]0 − ( sin xdx )
Attempt to integrate (using ‘parts’) with correct limits,
π 2 0 reaching an expression of the form
0
ax2 + uv – ∫v du.
OR using parts to integrate x(1+cosx) reaching an
expression of the form uv – ∫v du
i.e. π1 (x2 + x sin x – ∫(x + sin x) dx))
1 x2 A1 Integration fully correct.
= + x sin x + cos x
π 2
or e.g. π2 + π1 (0 − [ − cos x ]π0 )
or e.g. π2 + π1 ((−1 −1))
= π2 − π2 A1 OE
ISW after correct exact value seen.
SC1: Unsupported answer of π2 − π2 scores B3.
SC2: Unsupported answer of 0.934 scores B2.
4
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 9709/62 May/June 2024 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/62, May/June 2024 (9709_s24_ms_62), and the question paper (9709_s24_qp_62), to read or download, with each question's marking points written out underneath.
How many marks is 9709/62 out of?+
Mathematics 9709/62 for May/June 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/62 answers marked?+
Yes. Upload your answers to 9709/62/M/J/24 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.