Mathematics 9709/63 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.
The random variable X has the distribution B (4000, 0.001) .
(a) Use a suitable approximating distribution to find P (2 G X 1 5) . [3]
(b) Justify your approximating distribution in this case. [1]
Mark scheme
1(a) [λ =] 4 B1
2 3 4
e−4 ( 42! + 43! + 44! ) M1 Allow one end error.
Any λ.
or e−4 (8 + 10.67 + 10.67)
Expression must be seen.
or 0.1465 + 0.19537 + 0.19537
= 0.537 (3sf) A1 SC B1 B1 for unsupported correct answer.
SC B2 for use of Binomial leading to 0.537.
Note: use of normal could score B1 only for mean = 4.
3
1(b) n = 4000 > 50 and either np = 4 < 5 or p = 0.001 < 0.1 B1 Explicit values seen.
1
The widths, w cm, of a random sample of 150 leaves of a certain kind were measured. The sample mean of w was found to be 3.12 cm.
Using this sample, an approximate 95% confidence interval for the population mean of the widths in centimetres was found to be [3.01, 3.23].
(a) Calculate an estimate of the population standard deviation. [3]
(b) Explain whether it was necessary to use the Central Limit theorem in your answer to part (a). [1]
Mark scheme
2(a) σ
3.12 + z × 150 = 3.23 M1 OE – correct expression.
Any z, but must be a z value.
z = 1.96 B1
σ = 0.687 (3sf) [cm] A1
3
2(b) Yes, B1 Or ‘underlying distribution’ instead of population.
because population [of widths] not given to be normally distributed Allow ‘yes, because population distribution not known’.
Need both statements.
1
The masses in kilograms of large and small bags of cement have the independent distributions
N(50, 2.4) and N(26, 1.8) respectively.
Find the probability that the total mass of 5 randomly chosen large bags of cement is greater than the total mass of 10 randomly chosen small bags of cement. [5]
Mark scheme
3 Diff ~ N(5×50 − 10×26, 5×2.4 + 10×1.8) [= N(−10, 30) ] B1 For N and mean = ± (5×50 − 10×26) SOI
B1 For var = 5×2.4 + 10×1.8 SOI
0 − (' −10')
[= 1.826] M1 Standardising with their values.
'30'
1 − Φ(‘1.826’) M1 For area consistent with their values.
= 0.0339 or 0.034[0] (3sf) A1
5
In this question you should not use an approximating distribution.
At an election in Menham last year, 24% of voters supported the Today Party. A student wishes to test whether support for the Today Party has decreased since last year. He chooses a random sample of
25 voters in Menham and finds that exactly 2 of them say that they support the Today Party.
Test at the 5% significance level whether support for the Today Party has decreased. [5]
Mark scheme
4 H0: p = 0.24 B1
H1: p < 0.24
P(X < 2) = 0.7625 + 25×0.7624×0.24 + 25C2×0.7623×0.242 M1 Expression must be seen.
or 0.0010479 + 0.0082732 + 0.0313513 No end errors.
= 0.0407 A1 SC B1 for unsupported 0.0407.
0.0407 < 0.05 M1 For valid comparison.
[Evidence to reject H0.] There is sufficient evidence to suggest that A1FT FT their probability.
the support for the Today Party has decreased. In context, not definite, no contradictions.
SC: if H1: p ≠ 0.24 and compare with 0.025;
max B0 M1 A1 M1 A0.
5
A random variable X has probability density function f given by
ax - x 3 0 G x G 2, f (x) = )
0 otherwise,
where a is a constant.
(a) Show that a = 2 . [3]
(b) Find the median of X. [4]
(c) Find the exact value of E(X ). [3]
Mark scheme
5(a) 2 M1 Attempted integration of f(x) and equated to 1.
0
(ax − x3 )dx = 1
2 A1 Correct integration and substitute correct limits.
a x22 − x44 = 1
0
a − 44 = 1
a=2 A1 AG
Convincingly obtained and no errors seen.
3
( 2 x − x ) dx =
5(b) m
3 M1 Attempt integrate f(x) with limits 0 to m (or m to 2) and equate to
= 12
0 1
2
.
4
m 2 − m4 = 12 A1 For correct quartic in any form.
m4 − 4m2 + 2 = 0 m2 = 4± 216−8 [= 2 ± 2] M1 For solving their three term quartic to find m2.
A1
m= 2 − 2 or 0.765 (3sf)
4
5(c) 2 M1 Attempt to integrate xf(x). Ignore limits.
0
(2 x 2 − x 4 )dx
2 A1 Correct integration and correct limits.
23x3 − x55
0
[= 4 3 2 − 4 5 2 ] = 158 2 A1 OE
For single exact term.
3
The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table.
Number of green sweets 0 1 2 3 23
Number of packets 32 50 97 21 0
(a) Calculate an unbiased estimate for the population mean of the number of green sweets in a packet of Frutos, and show that an unbiased estimate of the population variance is 0.783 correct to
3 significant figures. [3]
The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65 .
Anji believes that the true value of the mean, n , is less than 1.65 . She uses the results from the 200 randomly chosen packets to test the manufacturers’ claim.
(b) State suitable null and alternative hypotheses for the test. [1]
(c) Show that the result of Anji’s test is significant at the 5% level but not at the 1% level. [4]
(d) It is given that Anji made a Type I error.
Explain how this shows that the significance level that Anji used in her test was not 1%. [1]
Mark scheme
6(a) μ̂ = 307
200
or 1.535 B1 Accept 3sf if nothing better seen (1.53 or 1.54).
Σx2f = 627, Est(σ 2 ) = 199 ( 200 − '1.535'2 ) or 199
200 '627 ' 1
( 2
'627'− '307'
200 ) M1 Use of a correct formula with their values.
= 0.783 A1 AG
Correctly obtained with no errors seen.
3
6(b) H0: µ = 1.65 B1 Accept ‘population mean’ but not just ‘mean’.
H1: µ < 1.65
1
6(c) '1.535' −1.65 M1 Standardising with their mean.
0.783÷ 200
= −1.838 or −1.84 A1*
Φ(0.05) and Φ(0.01) attempted M1 Or P(z < −`1.838`) attempted.
SC: Condone Φ(0.025) = 2.807 and Φ(0.005) = 3.291 following
two-tailed test in (b).
−1.645 > −1.838 > −2.326 DA1 AG
= 0.033 and 0.05 > 0.033 > 0.01
SC: use of 1.54 or 1.53 for the mean leading to -1.645 > –1.758 > –
[Hence significant at 5% but not 1% level] 2.326
or –1.645 > -1.918 > –2.326
or 0.95 < 0.9606 or 0.9724 < 0.99
scores M1 M1 A1.
Accept use of critical value method
1.535 < 1.547
or accept 1.65 > 1.638.
4
6(d) At the 1% level H0 is not rejected B1 OE
Or a Type I error can only occur if H0 is rejected.
1
The independent random variables X and Y have the distributions Po(1.9) and Po(2.2) respectively.
(a) Find P (X + Y 1 4) . [3]
(b) Find the probability that X = 2 given that X + Y 1 4 . [4]
(c) A sample of 60 randomly chosen pairs of values of X and Y is taken, and the value of X + Y is calculated for each pair. The sample mean of these 60 values is found.
Find the probability that the sample mean of X + Y is less than 4.0 . [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) λ = 1.9 + 2.2 [= 4.1] B1
2
e−4.1(1 + 4.1 + 4.12! + 4.1 )
3
M1 Allow any λ.
3!
−4.1 Allow one end error.
or e (1 + 4.1 + 8.405 + 11.487)
Must see expression.
or 0.01657 + 0.06795 + 0.13929 + 0.19037
= 0.414 (3sf) A1 SC: unsupported answer 0.414 scores B1 B1.
3
7(b) P(X + Y < 4 and X = 2) = P(2, 0 or 2, 1) M1 Stated or implied.
2
= e−1.9× 1.92 (e−2.2 + e−2.2×2.2) [= M1
0.0957]
P(X = 2 | X + Y < 4) = '0.0957
'0.414 '
' M1 Attempt P ( X +Y < 4 and X = 2) .
P( X +Y < 4)
Prob for denominator can be found in (a).
0.231 (3sf) A1
4
7(c) E(X + Y) = 4.1 Var(X + Y) = 4.1 B1 SOI
or Po(246)
Normal and var = 4.1
60
M1
Or normal and var = 246
4.0− 4.1 M1 No mixed methods.
4.1÷60 4.0 − 120
1 − 4.1
Or continuity correction: (4.1÷ 60)
− 246
or totals method 240246 or 239.5− 246
.
246
or use of continuity correction Condone incorrect continuity correction for M1.
= −0.383 (3sf) A1 = −0.414
Φ(‘−0.383’) = 1 − Φ(‘0.383’) M1 Φ(‘−0.414’) = 1 − Φ(‘0.414’)
= 0.351 (3sf) A1 = 0.340 or 0.339
6
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Mathematics 9709/63 for May/June 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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