A Level Mathematics 9709 Paper 63, Oct/Nov 2024

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9709/63 · Oct/Nov 2024

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The heights of a certain species of deer are known to have standard deviation 0.35 m. A zoologist takes a random sample of 150 of these deer and finds that the mean height of the deer in the sample is 1.42 m.

Calculate a 96% confidence interval for the population mean height.

[3]

Bubay says that 96% of deer of this species are likely to have heights that are within this confidence interval. Explain briefly whether Bubay is correct.

[1]

The masses, in kilograms, of small and large bags of wheat have the independent distributions $N (16.0, 0.4)$ and $N (51.0, 0.9)$ respectively. Find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag.

[5]

The times, $T$ minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by $Σ t = 230$ and $Σ t^{2} = 930$ .

Calculate unbiased estimates of the population mean and variance of $T$ .

[3]

You should now assume that your estimates from part (a) are the true values of the population mean and variance of $T$ . The times taken by another random sample of 75 students were noted, and the sample mean, $\overline{T}$ , was found. Find the value of $a$ such that $P (\overline{T} > a) = 0.234$ .

[3]

A random variable $X$ has probability density function $f$ defined by $f (x) = {\begin{cases} \frac{a}{x^{2}} - \frac{18}{x^{3}} & 2 ⩽ x ⩽ 3, \\ 0 & otherwise, \end{cases}$ where $a$ is a constant.

Show that $a = \frac{27}{2}$ .

[3]

Show that $E (X) = \frac{27}{2} \ln \frac{3}{2} - 3$ .

[3]

The lengths, in centimetres, of worms of a certain kind are normally distributed with mean $μ$ and standard deviation 2.3. An article in a magazine states that the value of $μ$ is 12.7. A scientist wishes to test whether this value is correct. He measures the lengths, $x cm$ , of a random sample of 50 worms of this kind and finds that $Σ x = 597.1$ . He plans to carry out a test, at the 1% significance level, of whether the true value of $μ$ is different from 12.7.

State, with a reason, whether he should use a one-tailed or a two-tailed test.

[1]

Carry out the test.

[5]

The numbers of customers arriving at service desks $A$ and $B$ during a 10-minute period have the independent distributions $P o (1.8)$ and $P o (2.1)$ respectively.

Find the probability that during a randomly chosen 15-minute period more than 2 customers will arrive at desk $A$ .

[2]

Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4.

[3]

An inspector waits at desk $B$ . She wants to wait long enough to be 90% certain of seeing at least one customer arrive at the desk. Find the minimum time for which she should wait, giving your answer correct to the nearest minute.

[4]

The number of accidents per year on a certain road has the distribution $P o (λ)$ . In the past the value of $λ$ was 3.3. Recently, a new speed limit was imposed and the council wishes to test whether the value of $λ$ has decreased. The council notes the total number, $X$ , of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the 5% significance level.

Calculate the probability of a Type I error.

[4]

Given that $X = 2$ , carry out the test.

[3]

The council decides to carry out another similar test at the 5% significance level using the same hypotheses and two different randomly chosen years. Given that the true value of $λ$ is 0.6, calculate the probability of a Type II error.

[3]

Using $λ = 0.6$ and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years.

[4]

Mark scheme

9709/63 · Oct/Nov 2024

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1a · 3 marks

B1
z = 2.054 or 2.055
M1
1.42 ± z 0.35 / √150
A1
1.36 to 1.48 [m] (3 sf)

1b · 1 mark

B1
No. CI is about mean, not individual values.

2 · 5 marks

B1
E(S₁ + S₂ + S₃ – L) = 16 x 3 – 51 [= −3 ]
M1
Var(S₁ + S₂ + S₃ – L) = 3 × 0.4 + 0.9 [= 2.1]
M1
(0 − (−3)) / √'2.1' [= 2.070]
M1
1 − Φ(‘2.070’)
A1
= 0.0192 (3 sf)

3a · 3 marks

B1
t̄ = 230 / 75 [= 3.0666… or 3.07 (3 sf)] [ 0r 46/15 ]
M1
s² = 75/74 (930/75 − (230/75)²) or 1/74(930 – 230²/75)
A1
= 3.0360… or 3.04 (3 sf) or = 337/111

3b · 3 marks

B1
[ Φ⁻¹(1 − 0.234) ] = 0.726
M1
± (a − '3.0667') / √('3.04'/75) = ± ‘0.726’
A1
a = 3.21 (3 sf)

4a · 3 marks

M1
∫₂³ (a/x² − 18/x³) dx = 1
A1
[−a/x + 9/x²]₂³ = 1
A1
[−a/3 + 1 + a/2 − 9/4] = 1 ⇒ a = 27/2 (AG)

4b · 3 marks

M1
∫₂³ (27/(2x) − 18/x²) dx
A1
[27/2 ln x + 18/x]₂³ or [27/2 ln 2x + 18/x]₂³
A1
= 27/2 ln 3 + 6 − 27/2 ln 2 − 9 = 27/2 ln(3/2) − 3 AG

5a · 1 mark

B1
Two-tailed because looking for difference

5b · 5 marks

B1
H₀: μ = 12.7 H₁: μ ≠ 12.7
M1
(597.1/50 − 12.7) / (2.3 / √50)
A1
= −2.330 Accept – 2.336 or – 2.337
M1
‘−2.330’ > −2.576 or ‘2.330’ < 2.576 or ‘0.00989’ > 0.005 or 0.0097 > 0.005
A1
[Not reject H₀] There is insufficient evidence to suggest that µ is not 12.7

6a · 2 marks

M1
[λ = 2.7] 1 − e⁻²·⁷(1 + 2.7 + 2.7²/2) or 1 − e⁻²·⁷(1 + 2.7 + 3.645) or 1- ( 0.06721 + 0.1815 + 0.2450 )
A1
= 0.506 (3 sf)

6b · 3 marks

B1
λ = 1.95
M1
e⁻¹·⁹⁵(1 + 1.95 + 1.95²/2 + 1.95³/3!) or e⁻¹·⁹⁵(1 + 1.95 + 1.90125 + 1.2358) or 0.1423 + 0.2774+ 0.2705 + 0.1758
A1
= 0.866

6c · 4 marks

M1
1 – e⁻²·¹ˣ ⩾ 0.90 or 1 – e⁻ᵞ ⩾ 0.90
M1
[e⁻²·¹ˣ ≤ 0.1] or e⁻ᵞ ≤ 0.1 −2.1x ≤ ln0.1 or −λ ≤ ln0.1 [ λ ≥ 2.3026, 2.3026/2.1 ]
A1
1.096 or 10.96 accept 1.097 or 10.97
A1
She must wait for at least 11 minutes

7a · 4 marks

B1
λ = 6.6
M1
P(X ≤ 2) = e⁻⁶·⁶(1 + 6.6 + 6.6²/2) [= 0.0400] [ < 0.05 ] or e⁻⁶·⁶(1 + 6.6 + 21.78 ) or 0.001360 + 0.008978 + 0.02963
B1
P(X ≤ 3) = e⁻⁶·⁶(1 + 6.6 + 6.6²/2 + 6.6³/3!) or 0.0400 + e⁻⁶·⁶ × 6.6³/3! = 0.105 [ > 0.05 ]
A1
P(Type I error) = 0.0400 (3 sf)

7b · 3 marks

B1
H₀: λ = 6.6, H₁: λ < 6.6
M1
[P(X ≤ 2) = 0.0400] 0.04 < 0.05
A1
[Reject H₀] There is evidence to suggest that mean number of accidents has decreased

7c · 3 marks

M1
P(X > 2) attempted, with any λ
M1
P(X > 2) = 1 − e⁻¹·²(1 + 1.2 + 1.2²/2) or = 1 − e⁻¹·²(1 + 1.2 +0.72) or = 1 – ( 0.3012 + 0.3614 + 0.2169 )
A1
0.121 (3 sf) or 0.120

7d · 4 marks

B1
N(18, 18) seen or implied
M1
(10.5 − 18) / √18 [= −1.768]
M1
P(X > ‘−1.768’) = Φ(‘1.768’)
A1
= 0.961 or 0.962 (3 sf)

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