A Level Mathematics 9709 Paper 53, Oct/Nov 2024

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9709/53 · Oct/Nov 2024

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30% of the residents of Wimfield own an electric car.

Three residents are chosen at random. Find the probability that either all three own an electric car or none of them owns an electric car.

[2]

A random sample of 125 of the residents of Wimfield is selected. Use a suitable approximation to find the probability that more than 45 of these residents own an electric car.

[5]

A red fair six-sided dice has faces labelled 1, 1, 1, 2, 2, 2. A blue fair six-sided dice has faces labelled 1, 1, 2, 2, 3, 3. Both dice are thrown. The random variable $X$ is the product of the scores on the two dice.

Draw up the probability distribution table for $X$ .

[3]

Find $E (X)$ .

[1]

In Molimba, the heights, in cm, of adult males are normally distributed with mean $176 cm$ and standard deviation $4.8 cm$ .

Find the probability that a randomly chosen adult male in Molimba has a height greater than $170 cm$ .

[3]

60% of adult males in Molimba have a height between $170 cm$ and $k cm$ , where $k$ is greater than 170. Find the value of $k$ , giving your answer correct to 1 decimal place.

[4]

On a certain day, the heights of 150 sunflower plants grown by children at a local school are measured, correct to the nearest cm. These heights are summarised in the following table.

Height (cm)	$10 – 19$	$20 – 29$	$30 – 39$	$40 – 44$	$45 – 49$	$50 – 54$	$55 – 59$
Frequency	$10$	$18$	$32$	$42$	$28$	$14$	$6$

Draw a cumulative frequency graph to illustrate the data.

[4]

Use your graph to estimate the 30th percentile of the heights of the sunflower plants.

[2]

Calculate estimates for the mean and the standard deviation of the heights of the 150 sunflower plants.

[5]

A factory produces chocolates. 30% of the chocolates are wrapped in gold foil, 25% are wrapped in red foil and the remainder are unwrapped.

Indigo chooses 8 chocolates at random from the production line. Find the probability that she obtains no more than 2 chocolates that are wrapped in red foil.

[3]

Jake chooses chocolates one at a time at random from the production line. Find the probability that the first time he obtains a chocolate that is wrapped in red foil is before the 7th choice.

[2]

Keifa chooses chocolates one at a time at random from the production line. Find the probability that the second chocolate chosen is the first one wrapped in gold foil given that the fifth chocolate chosen is the first unwrapped chocolate.

[5]

Find the number of different arrangements of the 9 letters in the word HAPPINESS.

[1]

Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the first and last letters are not the same as each other.

[3]

Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the two Ps are together and there are exactly two letters between the two Ss.

[4]

The 9 letters in the word HAPPINESS are divided at random into a group of 5 and a group of 4. Find the probability that both Ps are in one group and both Ss are in the other group.

[3]

Mark scheme

9709/53 · Oct/Nov 2024

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1a · 2 marks

M1
0.3³ + 0.7³ or 1 – (3 × 0.3² × 0.7 + 3 × 0.3 × 0.7²) = [1 – 0.63]
A1
0.37[0]

1b · 5 marks

B1
[Mean = 125 × 0.3 =] 37.5 [Variance = 125 × 0.3 × 0.7 =] 26.25
M1
Substituting their mean and their positive standard deviation into the ±standardising formula (any number for 45.5), not their σ², not their √σ
M1
Use continuity corrections 44.5 or 45.5 in their standardisation formula
M1
[1 – Φ(their 1.5614)] = 1 – their 0.9407
A1
0.0593

2a · 3 marks

B1
Table with correct x values and at least one correct probability linked with the correct x-value.
B1
4 correct probabilities linked with the correct x-values, need not be in table, accept unsimplified.
B1
5 correct probabilities linked with correct x-values, may not be in table. Decimals correct to at least 3 SF.

2b · 1 mark

B1
[E(X) = 1/36 (6 + 24 + 18 + 24 + 36) =] 3

3a · 3 marks

M1
[P(X > 170) = P(Z > (170 – 176) / 4.8)]
M1
[Φ(1.25)] = 0.894
A1
0.894

3b · 4 marks

M1
P(h < 170) = 1 – 0.8944 = 0.1056
B1
(k – 176) / 4.8 [= Φ⁻¹(0.1056 + 0.6)] = 0.541
M1
Use of ±standardisation formula with k, 176, 4.8 equated to a z-value
A1
k = 178.6

4a · 4 marks

B1
Cf values 28, 60, 102, 130, 144, 150 seen, Condone omission of 10. May be implied by accurate plotting (scale no less than 1cm = 10). May be by data table.
B1
Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 150) and height (h) and centimetres (cm) (from 9.5 to 59.5) with at least 3 values identified on each. Axes can be the other way round.
M1
At least 4 points plotted at upper boundary ± 0.5, (e.g. allow (19, 19.5 or 20, 10) etc.) on correctly scaled axes. (9.5,0), (19.5,10), (29.5, 28), (39.5, 60), (44.5,102), (49.5, 130), (54.5, 144), (59.5,150).
A1
All points plotted correctly, curve drawn (within tolerance), joined to (9.5, 0) and not going beyond above 150 vertically. A0 if straight line segments used.

4b · 2 marks

M1
[150 × 0.3 = 45]
A1
Line drawn from 45 on cf axis to meet graph at h = 36

4c · 5 marks

B1
Midpoints 14.5, 24.5, 34.5, 42, 47, 52, 57
M1
Mean = (10×14.5 + 18×24.5 + 32×34.5 + 42×42 + 28×47 + 14×52 + 6×57) / 150 [= (145 + 441 + 1104 + 1764 + 1316 + 728 + 342) / 150]
A1
= 5840/150, 584/15, 38 14/15, 38.9
M1
sd² = (10×14.5² + 18×24.5² + 32×34.5² + 42×42² + 28×47² + 14×52² + 6×57²) / 150 – (their 5840/150)² [= 244285/150 – (their 5840/150)²]
A1
[= 112.76] standard deviation [√112.76] = 10.6

5a · 3 marks

M1
One term ⁸Cₓ (p)ˣ (1 – p)⁸⁻ˣ, 0 < p < 1, 0 < x < 8
A1
Correct expression, accept unsimplified, no terms omitted leading to final answer
B1
= 0.679

5b · 2 marks

M1
1 – 0.75⁶ (or summing 6 or 7 terms, or 1 – 0.75⁷ – 0.25×0.75⁶)
A1
= 0.822, 3367/4096

5c · 5 marks

M1
0.25 × 0.3 × 0.55 × 0.55 × 0.45 [= 0.01021] on its own or as a numerator
A1
5 correct probabilities multiplied (or all probabilities correct and attempt to sum the 4 scenarios)
B1
[P(5th is first unwrapped) =] (0.55)⁴ (0.45) [= 0.041178]
M1
[P(2nd is first gold | 5th is first unwrapped) =] their P(2nd gold ∩ 5th is first unwrapped) / their P(5th is first unwrapped)
A1
= 0.248, 30/121

6a · 1 mark

B1
[9! / (2!2!) =] 90720

6b · 3 marks

M1
a – 7!/2! × b, a = their 6(a) or correct, b = 1,2
M1
a – 7!/c! × 2, a = their 6(a) or correct, c = 1,2
A1
85680

6c · 4 marks

M1
6! + d, d an integer ≥ 1, may be implied
M1
e + 5! × f, e, f integers ≥ 1, may be implied
M1
e + g × (5 × 4 or ⁵P₂), e an integer ≥ 1, g = 4, 5, 6
A1
[Total ]= 3120

6d · 3 marks

M1
a × ⁵C₂, a × ⁵C₃, or ⁵C₂ + ⁵C₃ seen as a numerator of one or two fractions where a is 1 or 2, no extra terms (or a × 5 × 4 × 4 × 3 seen as a numerator of a fraction where a = 1 or 2)
M1
⁹C₅ or ⁹C₄ seen (no addition, multiplication) as a denominator of one or two fractions (or 9 × 8 × 7 × 6 seen as a denominator of a fraction)
A1
Probability = 20/126, 10/63, 0.159

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