A Level H1 Mathematics Practice Paper 3

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics H1
Level: A-Level
Paper: Practice Paper — Version 3 of 5
Topic Focus: Statistics & Probability
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ___________________________
Class: ___________________________
Date: ___________________________

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Instructions

• Write your answers in the spaces provided.
• Show all working clearly. Marks may be awarded for correct method even if the final answer is wrong.
• Give answers correct to 3 significant figures unless otherwise stated.
• A graphing calculator may be used where appropriate.
• The total mark for this paper is 60.
• The number of marks for each question or part-question is shown in brackets [ ].

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Section A: Pure Mathematics Foundations [20 marks]

Answer all questions in this section.

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Question 1 [2 marks]

The random variable $X \sim \mathrm{B}(20, 0.35)$. Find $\mathrm{P}(X = 8)$.

$\text{Answer: } \underline{\hspace{8cm}}$

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Question 2 [3 marks]

A discrete random variable $Y$ has the following probability distribution:

$y$	1	2	3	4	5
$\mathrm{P}(Y=y)$	0.1	0.2	$a$	0.3	0.15

(a) Find the value of $a$.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Find $\mathrm{E}(Y)$.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 3 [3 marks]

The heights of a certain species of plant are normally distributed with mean 42 cm and standard deviation 5 cm.

(a) Find the probability that a randomly selected plant has a height between 38 cm and 47 cm.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) In a random sample of 200 plants, how many would you expect to have a height greater than 50 cm?

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 4 [4 marks]

A random sample of 10 students recorded the following daily screen times (in hours):

$4.2,\ 5.8,\ 3.1,\ 6.5,\ 4.9,\ 7.2,\ 3.8,\ 5.1,\ 6.0,\ 4.5$

(a) Calculate the unbiased estimate of the population mean.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Calculate the unbiased estimate of the population variance.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 5 [4 marks]

The masses of a certain brand of chocolate bar are normally distributed with mean 52 g and standard deviation 1.5 g.

(a) Find the probability that a randomly chosen chocolate bar has a mass less than 49.5 g.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) A random sample of 8 chocolate bars is selected. Find the probability that exactly 3 of them have a mass greater than 53 g.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 6 [4 marks]

A continuous random variable $X$ has probability density function given by

$f(x) = \begin{cases} kx(6 - x) & 0 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $k = \dfrac{1}{36}$.

$\text{Working: } \underline{\hspace{8cm}}$

(b) Find $\mathrm{E}(X)$.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Section B: Statistics & Probability — Applied [40 marks]

Answer all questions in this section.

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Question 7 [5 marks]

A factory produces light bulbs. The lifetime of a light bulb (in hours) follows a normal distribution with mean 1200 hours and standard deviation 150 hours.

(a) Find the probability that a randomly selected light bulb has a lifetime between 1000 and 1350 hours.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) The factory offers a warranty for bulbs that fail before 900 hours. In a batch of 500 bulbs, how many would you expect to be replaced under warranty?

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 8 [6 marks]

A researcher is investigating whether a new teaching method improves students' test scores. A random sample of 12 students taught using the new method achieved the following scores:

$78,\ 85,\ 92,\ 71,\ 88,\ 76,\ 95,\ 82,\ 90,\ 74,\ 87,\ 80$

The national mean score using the traditional method is 79. Assume the population of scores is normally distributed.

(a) Calculate the unbiased estimates of the population mean and variance for the new method.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Test, at the 5% significance level, whether the new method produces a higher mean score than the traditional method. State your hypotheses clearly.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 9 [6 marks]

A call centre receives calls at an average rate of 4.2 calls per minute. The number of calls received in a given time period follows a Poisson distribution.

(a) Find the probability that exactly 6 calls are received in a 2-minute interval.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Find the probability that at least 3 calls are received in a 30-second interval.

$\text{Answer (b): } \underline{\hspace{6cm}}$

(c) Over a 5-day working week (8 hours per day), estimate the number of 1-minute intervals in which no calls are received.

$\text{Answer (c): } \underline{\hspace{6cm}}$

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Question 10 [5 marks]

The following table shows the daily commute times (in minutes) for a sample of 80 employees at a company.

Commute time (min)	Frequency
$0 \le t < 10$	8
$10 \le t < 20$	15
$20 \le t < 30$	22
$30 \le t < 50$	25
$50 \le t < 80$	10

(a) Estimate the mean commute time.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Estimate the standard deviation of the commute times.

$\text{Answer (b): } \underline{\hspace{6cm}}$

(c) On a separate diagram, a histogram is drawn to represent this data. State the class interval that has the highest frequency density.

$\text{Answer (c): } \underline{\hspace{6cm}}$

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Question 11 [6 marks]

A bag contains 5 red balls, 4 blue balls, and 3 green balls. Three balls are drawn at random without replacement.

(a) Find the probability that all three balls are red.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Find the probability that exactly two balls are red and one is blue.

$\text{Answer (b): } \underline{\hspace{6cm}}$

(c) Find the probability that all three balls are of different colours.

$\text{Answer (c): } \underline{\hspace{6cm}}$

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Question 12 [5 marks]

The weights of adult male cats at a veterinary clinic are normally distributed with mean 4.8 kg and standard deviation 0.6 kg.

(a) Find the probability that a randomly selected adult male cat weighs between 4.0 kg and 5.5 kg.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) The heaviest 10% of cats are classified as overweight. Find the minimum weight for a cat to be classified as overweight.

$\text{Answer (b): } \underline{\hspace{6cm}}$

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Question 13 [7 marks]

A market researcher collected data on the weekly advertising spend (in thousands of dollars) and the corresponding weekly sales revenue (in thousands of dollars) for 8 small businesses.

Advertising spend, $x$	1.2	2.0	2.8	3.5	4.0	4.8	5.5	6.2
Sales revenue, $y$	15	22	28	33	38	42	48	55

(a) Calculate the equation of the least squares regression line of $y$ on $x$, giving your answer in the form $y = a + bx$.

$\text{Answer (a): } \underline{\hspace{6cm}}$

(b) Interpret the value of $b$ in context.

$\text{Answer (b): } \underline{\hspace{6cm}}$

(c) Estimate the sales revenue when the advertising spend is $3.0$ thousand dollars. Comment on the reliability of this estimate.

$\text{Answer (c): } \underline{\hspace{6cm}}$

(d) Calculate the product moment correlation coefficient between $x$ and $y$.

$\text{Answer (d): } \underline{\hspace{6cm}}$

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End of Paper

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Section Mark Summary

Section	Marks
Section A: Questions 1–6	20
Section B: Questions 7–13	40
Total	60

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TuitionGoWhere Practice Paper — Answer Key

Subject: Mathematics H1 (A-Level)
Paper: Practice Paper — Version 3 of 5
Topic Focus: Statistics & Probability
Total Marks: 60

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Section A: Pure Mathematics Foundations [20 marks]

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Question 1 [2 marks]

Answer: $\mathrm{P}(X = 8) = 0.161$ (to 3 s.f.)

Working:

$X \sim \mathrm{B}(20, 0.35)$

$\mathrm{P}(X = 8) = \binom{20}{8}(0.35)^8(0.65)^{12}$

$= 125970 \times (0.35)^8 \times (0.65)^{12}$

$= 0.16128...$

$\approx 0.161 \text{ (3 s.f.)}$

Marking notes:

• M1: Correct binomial probability formula with $n=20$, $p=0.35$, $r=8$
• A1: Correct answer 0.161 (3 s.f.)

Common mistake: Using $(0.35)^{12}(0.65)^8$ instead of $(0.35)^8(0.65)^{12}$ — the powers must correspond to the number of successes and failures respectively.

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Question 2 [3 marks]

(a) [1 mark]

Answer: $a = 0.25$

Working:

All probabilities must sum to 1:

$0.1 + 0.2 + a + 0.3 + 0.15 = 1$

$0.75 + a = 1$

$a = 0.25$

(b) [2 marks]

Answer: $\mathrm{E}(Y) = 3.15$

Working:

$\mathrm{E}(Y) = \sum y \cdot \mathrm{P}(Y=y)$

$= 1(0.1) + 2(0.2) + 3(0.25) + 4(0.3) + 5(0.15)$

$= 0.1 + 0.4 + 0.75 + 1.2 + 0.75$

$= 3.20$

Correction: Let me recalculate:

$= 0.1 + 0.4 + 0.75 + 1.2 + 0.75 = 3.20$

Answer: $\mathrm{E}(Y) = 3.20$

Marking notes:

• (a) A1: Correct value $a = 0.25$
• (b) M1: Correct formula for expected value applied; A1: Correct answer 3.20

Common mistake: Forgetting to multiply each value by its probability before summing.

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Question 3 [3 marks]

(a) [2 marks]

Answer: $\mathrm{P}(38 < X < 47) = 0.641$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(42, 5^2)$

Standardise:

$z_1 = \frac{38 - 42}{5} = -0.80, \quad z_2 = \frac{47 - 42}{5} = 1.00$

$\mathrm{P}(38 < X < 47) = \mathrm{P}(-0.80 < Z < 1.00)$

$= \Phi(1.00) - \Phi(-0.80)$

$= 0.8413 - 0.2119$

$= 0.6294$

$\approx 0.629 \text{ (3 s.f.)}$

Answer: $0.629$ (to 3 s.f.)

(b) [1 mark]

Answer: Approximately 11 plants.

Working:

$\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50-42}{5}\right) = \mathrm{P}(Z > 1.60) = 1 - \Phi(1.60) = 1 - 0.9452 = 0.0548$

Expected number: $200 \times 0.0548 = 10.96 \approx 11$ plants.

Marking notes:

• (a) M1: Correct standardisation and use of normal tables; A1: Correct answer 0.629
• (b) A1: Correct answer 11 (accept 10.96 or 11)

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Question 4 [4 marks]

(a) [1 mark]

Answer: $\bar{x} = 5.11$ hours (to 3 s.f.)

Working:

$\bar{x} = \frac{4.2 + 5.8 + 3.1 + 6.5 + 4.9 + 7.2 + 3.8 + 5.1 + 6.0 + 4.5}{10}$

$= \frac{51.1}{10} = 5.11$

(b) [3 marks]

Answer: $s^2 = 1.57$ (to 3 s.f.)

Working:

$s^2 = \frac{1}{n-1}\sum(x_i - \bar{x})^2 = \frac{1}{9}\sum(x_i - 5.11)^2$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
4.2	−0.91	0.8281
5.8	0.69	0.4761
3.1	−2.01	4.0401
6.5	1.39	1.9321
4.9	−0.21	0.0441
7.2	2.09	4.3681
3.8	−1.31	1.7161
5.1	−0.01	0.0001
6.0	0.89	0.7921
4.5	−0.61	0.3721

$\sum(x_i - \bar{x})^2 = 14.569$

$s^2 = \frac{14.569}{9} = 1.6188... \approx 1.62 \text{ (3 s.f.)}$

Answer: $s^2 = 1.62$ (to 3 s.f.)

Marking notes:

• (a) A1: Correct mean 5.11
• (b) M1: Correct formula for unbiased variance with $n-1$ denominator; M1: Correct computation of squared deviations; A1: Correct answer 1.62

Common mistake: Using $n = 10$ instead of $n - 1 = 9$ in the denominator. The unbiased estimate requires $n-1$.

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Question 5 [4 marks]

(a) [2 marks]

Answer: $\mathrm{P}(X < 49.5) = 0.0478$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(52, 1.5^2)$

$z = \frac{49.5 - 52}{1.5} = \frac{-2.5}{1.5} = -1.667$

$\mathrm{P}(X < 49.5) = \Phi(-1.667) = 1 - \Phi(1.667) = 1 - 0.9522 = 0.0478$

(b) [2 marks]

Answer: $0.0579$ (to 3 s.f.)

Working:

First find $\mathrm{P}(\text{one bar} > 53)$:

$z = \frac{53 - 52}{1.5} = 0.667$

$\mathrm{P}(X > 53) = 1 - \Phi(0.667) = 1 - 0.7476 = 0.2524$

Let $Y \sim \mathrm{B}(8, 0.2524)$. Then:

$\mathrm{P}(Y = 3) = \binom{8}{3}(0.2524)^3(0.7476)^5$

$= 56 \times 0.01608 \times 0.2337$

$= 0.2099$

$\approx 0.210 \text{ (3 s.f.)}$

Marking notes:

• (a) M1: Correct standardisation; A1: Correct answer 0.0478
• (b) M1: Correct probability for one bar > 53g, then correct binomial setup; A1: Correct answer 0.210

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Question 6 [4 marks]

(a) [2 marks]

Working:

For a valid PDF, $\int_{-\infty}^{\infty} f(x)\,dx = 1$:

$\int_0^6 kx(6-x)\,dx = 1$

$k\int_0^6 (6x - x^2)\,dx = 1$

$k\left[3x^2 - \frac{x^3}{3}\right]_0^6 = 1$

$k\left[\left(108 - 72\right) - 0\right] = 1$

$k \times 36 = 1$

$k = \frac{1}{36} \quad \text{(shown)}$

(b) [2 marks]

Answer: $\mathrm{E}(X) = 3$

Working:

$\mathrm{E}(X) = \int_0^6 x \cdot \frac{1}{36}x(6-x)\,dx = \frac{1}{36}\int_0^6 (6x^2 - x^3)\,dx$

$= \frac{1}{36}\left[2x^3 - \frac{x^4}{4}\right]_0^6$

$= \frac{1}{36}\left[\left(432 - 324\right) - 0\right]$

$= \frac{1}{36} \times 108 = 3$

Marking notes:

• (a) M1: Correct integration setup; A1: Correct derivation showing $k = \frac{1}{36}$
• (b) M1: Correct expectation integral; A1: Correct answer 3

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Section B: Statistics & Probability — Applied [40 marks]

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Question 7 [5 marks]

(a) [3 marks]

Answer: $\mathrm{P}(1000 < X < 1350) = 0.749$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(1200, 150^2)$

$z_1 = \frac{1000 - 1200}{150} = -1.333, \quad z_2 = \frac{1350 - 1200}{150} = 1.000$

$\mathrm{P}(1000 < X < 1350) = \Phi(1.000) - \Phi(-1.333)$

$= 0.8413 - (1 - 0.9088) = 0.8413 - 0.0912 = 0.7501$

$\approx 0.750 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: Approximately 11 bulbs.

Working:

$\mathrm{P}(X < 900) = \mathrm{P}\left(Z < \frac{900-1200}{150}\right) = \mathrm{P}(Z < -2.00) = 1 - \Phi(2.00) = 1 - 0.9772 = 0.0228$

Expected number: $500 \times 0.0228 = 11.4 \approx 11$ bulbs.

Marking notes:

• (a) M1: Correct standardisation; M1: Correct use of normal tables; A1: Correct answer 0.750
• (b) M1: Correct probability calculation; A1: Correct answer 11 (accept 11.4)

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Question 8 [6 marks]

(a) [2 marks]

Answer: $\bar{x} = 83.17$, $s^2 = 54.52$ (to 3 s.f.)

Working:

$\bar{x} = \frac{78+85+92+71+88+76+95+82+90+74+87+80}{12} = \frac{998}{12} = 83.166... \approx 83.2$

$s^2 = \frac{1}{11}\sum(x_i - 83.167)^2$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
78	−5.167	26.698
85	1.833	3.361
92	8.833	78.028
71	−12.167	148.036
88	4.833	23.361
76	−7.167	51.361
95	11.833	140.028
82	−1.167	1.361
90	6.833	46.694
74	−9.167	84.036
87	3.833	14.694
80	−3.167	10.028

$\sum(x_i - \bar{x})^2 = 627.686$

$s^2 = \frac{627.686}{11} = 57.062... \approx 57.1 \text{ (3 s.f.)}$

So $\bar{x} = 83.2$, $s^2 = 57.1$ (to 3 s.f.)

(b) [4 marks]

Answer: There is sufficient evidence at the 5% level to conclude that the new method produces a higher mean score.

Working:

Step 1: Hypotheses

$H_0: \mu = 79$ (the mean score equals the national mean) $H_1: \mu > 79$ (the new method gives a higher mean — one-tailed test)

Step 2: Test statistic

Using $t$-test (population variance unknown, small sample):

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.167 - 79}{\sqrt{57.062}/\sqrt{12}} = \frac{4.167}{2.176} = 1.915$

Step 3: Critical value

Degrees of freedom $= n - 1 = 11$. At 5% significance (one-tailed), $t_{0.05, 11} = 1.796$.

Step 4: Conclusion

Since $t = 1.915 > 1.796$, we reject $H_0$.

There is sufficient evidence at the 5% significance level to conclude that the new teaching method produces a higher mean score than the traditional method.

Marking notes:

• (a) A1: Correct mean 83.2; A1: Correct variance 57.1
• (b) B1: Correct hypotheses (one-tailed); B1: Correct test statistic calculation; B1: Correct critical value or comparison; B1: Correct conclusion in context

Common mistake: Using a two-tailed test when the question asks whether the new method is higher (one-tailed). Also, using $z$-test instead of $t$-test when population variance is unknown.

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Question 9 [6 marks]

(a) [2 marks]

Answer: $0.114$ (to 3 s.f.)

Working:

For a 2-minute interval, $\lambda = 4.2 \times 2 = 8.4$.

$X \sim \mathrm{Po}(8.4)$

$\mathrm{P}(X = 6) = \frac{e^{-8.4}(8.4)^6}{6!} = \frac{e^{-8.4} \times 351298.032}{720}$

$= e^{-8.4} \times 487.914 = 0.0002246 \times 487.914$

$= 0.1096 \approx 0.110 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: $0.350$ (to 3 s.f.)

Working:

For a 30-second interval, $\lambda = 4.2 \times 0.5 = 2.1$.

$Y \sim \mathrm{Po}(2.1)$

$\mathrm{P}(Y \ge 3) = 1 - \mathrm{P}(Y \le 2)$

$= 1 - \left[\frac{e^{-2.1}(2.1)^0}{0!} + \frac{e^{-2.1}(2.1)^1}{1!} + \frac{e^{-2.1}(2.1)^2}{2!}\right]$

$= 1 - e^{-2.1}\left[1 + 2.1 + \frac{4.41}{2}\right]$

$= 1 - e^{-2.1} \times 5.305$

$= 1 - 0.1225 \times 5.305$

$= 1 - 0.6498 = 0.3502$

$\approx 0.350 \text{ (3 s.f.)}$

(c) [2 marks]

Answer: Approximately 1 interval.

Working:

Total 1-minute intervals in a week: $5 \times 8 \times 60 = 2400$ intervals.

For a 1-minute interval, $\lambda = 4.2$.

$\mathrm{P}(X = 0) = e^{-4.2} = 0.0150$

Expected number: $2400 \times 0.0150 = 36$ intervals.

Answer: Approximately 36 intervals.

Marking notes:

• (a) M1: Correct $\lambda = 8.4$ for 2 minutes; A1: Correct answer 0.110
• (b) M1: Correct $\lambda = 2.1$ and complement method; A1: Correct answer 0.350
• (c) M1: Correct $\mathrm{P}(X=0) = e^{-4.2}$; A1: Correct answer 36

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Question 10 [5 marks]

(a) [2 marks]

Answer: $31.6$ minutes (to 3 s.f.)

Working:

Midpoints: 5, 15, 25, 40, 65

$\bar{x} = \frac{8(5) + 15(15) + 22(25) + 25(40) + 10(65)}{80}$

$= \frac{40 + 225 + 550 + 1000 + 650}{80} = \frac{2465}{80} = 30.8125$

$\approx 30.8 \text{ minutes (to 3 s.f.)}$

(b) [2 marks]

Answer: $s = 16.9$ minutes (to 3 s.f.)

Working:

$s^2 = \frac{8(5^2) + 15(15^2) + 22(25^2) + 25(40^2) + 10(65^2)}{80} - (30.8125)^2$

$= \frac{200 + 3375 + 13750 + 40000 + 42250}{80} - 949.410$

$= \frac{99575}{80} - 949.410 = 1244.688 - 949.410 = 295.278$

$s = \sqrt{295.278} = 17.184... \approx 17.2 \text{ (3 s.f.)}$

(c) [1 mark]

Answer: The class $0 \le t < 10$ has the highest frequency density.

Working:

Frequency density = frequency ÷ class width:

• $0 \le t < 10$: $8/10 = 0.80$
• $10 \le t < 20$: $15/10 = 1.50$
• $20 \le t < 30$: $22/10 = 2.20$
• $30 \le t < 50$: $25/20 = 1.25$
• $50 \le t < 80$: $10/30 = 0.33$

The highest frequency density is $2.20$ in the class $20 \le t < 30$.

Answer: $20 \le t < 30$

Marking notes:

• (a) M1: Correct midpoints used; A1: Correct answer 30.8
• (b) M1: Correct variance formula for grouped data; A1: Correct answer 17.2
• (c) A1: Correct class $20 \le t < 30$

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Question 11 [5 marks]

(a) [2 marks]

Answer: $\dfrac{1}{22}$

Working:

Total balls = 12. Drawing 3 without replacement.

$\mathrm{P}(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22}$

(b) [2 marks]

Answer: $\dfrac{2}{11}$

Working:

$\mathrm{P}(\text{2 red, 1 blue}) = \frac{\binom{5}{2}\binom{4}{1}}{\binom{12}{3}} = \frac{10 \times 4}{220} = \frac{40}{220} = \frac{2}{11}$

(c) [1 mark]

Answer: $\dfrac{3}{11}$

Working:

$\mathrm{P}(\text{1 red, 1 blue, 1 green}) = \frac{\binom{5}{1}\binom{4}{1}\binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11}$

Marking notes:

• (a) M1: Correct combination approach; A1: Correct answer $\frac{1}{22}$
• (b) M1: Correct numerator and denominator; A1: Correct answer $\frac{2}{11}$
• (c) M1: Correct product of combinations; A1: Correct answer $\frac{3}{11}$

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Question 12 [5 marks]

(a) [3 marks]

Answer: $0.749$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(4.8, 0.6^2)$

$z_1 = \frac{4.0 - 4.8}{0.6} = -1.333, \quad z_2 = \frac{5.5 - 4.8}{0.6} = 1.167$

$\mathrm{P}(4.0 < X < 5.5) = \Phi(1.167) - \Phi(-1.333)$

$= 0.8784 - (1 - 0.9088) = 0.8784 - 0.0912 = 0.7872$

$\approx 0.787 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: $5.57$ kg (to 3 s.f.)

Working:

We need $w$ such that $\mathrm{P}(X > w) = 0.10$, i.e., $\mathrm{P}(X \le w) = 0.90$.

From tables, $\Phi(z) = 0.90$ gives $z \approx 1.282$.

$w = 4.8 + 1.282 \times 0.6 = 4.8 + 0.7692 = 5.5692$

$\approx 5.57 \text{ kg (to 3 s.f.)}$

Marking notes:

• (a) M1: Correct standardisation; M1: Correct use of normal tables; A1: Correct answer 0.787
• (b) M1: Correct $z$-value for 90th percentile; A1: Correct answer 5.57

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Question 13 [7 marks]

(a) [3 marks]

Answer: $y = 7.00 + 7.86x$ (to 3 s.f.)

Working:

$n = 8$

$\sum x = 1.2 + 2.0 + 2.8 + 3.5 + 4.0 + 4.8 + 5.5 + 6.2 = 30.0$

$\sum y = 15 + 22 + 28 + 33 + 38 + 42 + 48 + 55 = 281$

$\bar{x} = 30.0/8 = 3.75$, $\bar{y} = 281/8 = 35.125$

$\sum x^2 = 1.44 + 4.00 + 7.84 + 12.25 + 16.00 + 23.04 + 30.25 + 38.44 = 133.26$

$\sum xy = 18.0 + 44.0 + 78.4 + 115.5 + 152.0 + 201.6 + 264.0 + 341.0 = 1214.5$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 133.26 - \frac{900}{8} = 133.26 - 112.5 = 20.76$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 1214.5 - \frac{30.0 \times 281}{8} = 1214.5 - 1053.75 = 160.75$

$b = \frac{S_{xy}}{S_{xx}} = \frac{160.75}{20.76} = 7.743... \approx 7.74 \text{ (3 s.f.)}$

$a = \bar{y} - b\bar{x} = 35.125 - 7.743 \times 3.75 = 35.125 - 29.036 = 6.089 \approx 6.09$

Answer: $y = 6.09 + 7.74x$ (to 3 s.f.)

(b) [1 mark]

Answer: For every additional thousand dollars spent on advertising, the weekly sales revenue increases by approximately $7,740.

(c) [2 marks]

Answer: Estimated revenue = $29.3 thousand. This is an interpolation (since $x = 3.0$ lies within the data range), so the estimate is reliable.

Working:

$y = 6.09 + 7.74(3.0) = 6.09 + 23.22 = 29.31 \approx 29.3 \text{ (thousand dollars)}$

Since $x = 3.0$ lies within the range of the data ($1.2 \le x \le 6.2$), this is interpolation and the estimate is reasonably reliable.

(d) [1 mark]

Answer: $r = 0.997$ (to 3 s.f.)

Working:

$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n}$

$\sum y^2 = 225 + 484 + 784 + 1089 + 1444 + 1764 + 2304 + 3025 = 11119$

$S_{yy} = 11119 - \frac{281^2}{8} = 11119 - 9870.125 = 1248.875$

$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{160.75}{\sqrt{20.76 \times 1248.875}} = \frac{160.75}{\sqrt{25926.6}} = \frac{160.75}{161.018} = 0.9983...$

$\approx 0.998 \text{ (3 s.f.)}$

Marking notes:

• (a) M1: Correct calculation of $S_{xx}$ and $S_{xy}$; M1: Correct $b$ and $a$ values; A1: Correct equation $y = 6.09 + 7.74x$
• (b) B1: Correct interpretation in context (mentioning units: thousand dollars)
• (c) M1: Correct substitution; A1: Correct estimate with valid reliability comment
• (d) A1: Correct answer 0.998

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Section Mark Summary

Section	Marks
Section A: Questions 1–6	20
Section B: Questions 7–13	40
Total	60