A Level H1 Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics H1 Level: A-Level Paper: Practice Paper — 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 are 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 marks for this paper is 60.
• The number of marks is shown in brackets [ ] at the end of each question or part-question.

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Section A: Pure Statistics (30 marks)

Answer all questions in this section.

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Question 1

A random sample of 8 students recorded the number of hours they spent on revision in a week:

$12,\ 15,\ 10,\ 18,\ 14,\ 11,\ 16,\ 13$

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

[4]

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Question 2

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

(a) Find $\mathrm{P}(X = 7)$.

[2]

(b) Find $\mathrm{P}(X \geq 6)$.

[2]

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Question 3

A factory produces light bulbs, and 5% are defective. A random sample of 20 bulbs is selected.

(a) State two conditions under which a binomial model is appropriate for the number of defective bulbs.

[2]

(b) Using a binomial distribution, find the probability that exactly 2 bulbs are defective.

[2]

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Question 4

The heights of adult women in a city are normally distributed with mean $162\text{ cm}$ and standard deviation $5.4\text{ cm}$.

(a) Find the probability that a randomly selected woman has a height between $158\text{ cm}$ and $168\text{ cm}$.

[3]

(b) A random sample of 10 women is selected. Find the probability that at least 8 of them have heights between $158\text{ cm}$ and $168\text{ cm}$.

[3]

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Question 5

A researcher collects data on the daily screen time (in hours) of 10 teenagers:

$4.2,\ 5.8,\ 3.5,\ 6.1,\ 7.3,\ 4.9,\ 5.2,\ 6.5,\ 3.8,\ 5.6$

(a) Calculate the median and interquartile range of the data.

[3]

(b) Determine whether there are any outliers using the $1.5 \times \mathrm{IQR}$ rule. Show your working clearly.

[3]

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Question 6

The following table shows the cumulative frequency distribution of the masses (in kg) of 80 packages:

Mass (kg)	Cumulative Frequency
$0 < m \leq 2$	8
$0 < m \leq 4$	22
$0 < m \leq 6$	45
$0 < m \leq 8$	65
$0 < m \leq 10$	80

Generated graph for Q6.

(a) Draw a cumulative frequency curve to represent the data.

[2]

(b) Use your graph to estimate the median mass.

[1]

(c) Use your graph to estimate the 90th percentile.

[1]

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Section B: Probability & Distributions (30 marks)

Answer all questions in this section.

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Question 7

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

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

(a) Find the value of $a$.

[1]

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

[4]

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Question 8

The number of emails received by an employee per hour follows a Poisson distribution with mean 4.2.

(a) Find the probability that the employee receives exactly 5 emails in a given hour.

[2]

(b) Find the probability that the employee receives at least 3 emails in a given hour.

[3]

(c) Find the probability that the employee receives fewer than 2 emails in each of two consecutive hours.

[2]

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Question 9

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

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

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

[2]

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

[2]

(c) Find $\mathrm{P}(X > 4)$.

[3]

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Question 10

In a large population, the time taken to complete a certain task is normally distributed with mean 45 minutes and standard deviation 8 minutes.

(a) Find the probability that a randomly selected person takes more than 50 minutes.

[2]

(b) Find the value of $t$ such that $\mathrm{P}(X < t) = 0.75$.

[3]

(c) A random sample of 25 people is selected. Using the Central Limit Theorem, find the probability that the sample mean time is less than 43 minutes.

[3]

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Question 11

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.

[2]

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

[3]

(c) Given that at least one ball is red, find the probability that exactly two balls are red.

[3]

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Question 12

A market researcher surveys 200 adults to investigate whether there is an association between age group and preference for online shopping. The results are summarised in the table below:

	Prefer Online	Prefer In-Store	Total
Under 40	62	28	90
40 and over	48	62	110
Total	110	90	200

(a) Calculate the expected frequency for the cell corresponding to "Under 40" and "Prefer Online" under the assumption of no association.

[2]

(b) Perform a chi-squared test at the 5% significance level to determine whether there is evidence of association between age group and shopping preference. State your hypotheses clearly.

[6]

(c) State your conclusion in context.

[1]

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End of Paper

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TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Subject: Mathematics H1 Paper: Practice Paper — Statistics & Probability Total Marks: 60

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Section A: Pure Statistics (30 marks)

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Question 1 [4 marks]

Data: 12, 15, 10, 18, 14, 11, 16, 13; $n = 8$

Unbiased estimate of the population mean:

$\bar{x} = \frac{\sum x_i}{n} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Unbiased estimate of the population variance:

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

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	−1.625	2.640625
15	1.375	1.890625
10	−3.625	13.140625
18	4.375	19.140625
14	0.375	0.140625
11	−2.625	6.890625
16	2.375	5.640625
13	−0.625	0.390625

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{7} = 7.125$

Answers:

• Unbiased estimate of mean = 13.6 (or 13.625 hours)
• Unbiased estimate of variance = 7.13 (or 7.125 hours²)

Marking:

• [1] Correct calculation of $\bar{x} = 13.625$
• [1] Correct setup of $s^2$ formula with $n - 1 = 7$ in denominator
• [1] Correct sum of squared deviations (or correct use of $\sum x_i^2 - n\bar{x}^2$ method)
• [1] Correct final answer $s^2 = 7.125$

Common mistakes:

• Using $n = 8$ instead of $n - 1 = 7$ in the variance denominator (this gives the biased estimate, not the unbiased estimate).
• Rounding too early; keep full precision in intermediate steps.

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Question 2 [4 marks]

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

(a) $\mathrm{P}(X = 7) = \binom{20}{7}(0.35)^7(0.65)^{13}$

$= 77520 \times (0.35)^7 \times (0.65)^{13}$

$= 77520 \times 0.0006434... \times 0.009041...$

$= 0.184 \quad \text{(3 s.f.)}$

(b) $\mathrm{P}(X \geq 6) = 1 - \mathrm{P}(X \leq 5)$

Using calculator/binomial tables:

$\mathrm{P}(X \leq 5) = \sum_{k=0}^{5} \binom{20}{k}(0.35)^k(0.65)^{20-k} = 0.2454...$

$\mathrm{P}(X \geq 6) = 1 - 0.2454 = 0.755 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct binomial probability formula setup; [1] Correct answer 0.184
• (b) [1] Correct use of complement $1 - \mathrm{P}(X \leq 5)$; [1] Correct answer 0.755

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Question 3 [4 marks]

(a) Two conditions for a binomial model:

1. Each trial (each bulb) has only two outcomes: defective or not defective.
2. The probability of a bulb being defective is constant (5%) for each bulb, and the bulbs are independent of each other.

(b) Let $X \sim \mathrm{B}(20, 0.05)$ be the number of defective bulbs.

$\mathrm{P}(X = 2) = \binom{20}{2}(0.05)^2(0.95)^{18}$

$= 190 \times 0.0025 \times 0.3972...$

$= 0.189 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Two correct conditions stated (any valid pair: fixed trials, two outcomes, constant probability, independence)
• (b) [1] Correct binomial setup; [1] Correct answer 0.189

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Question 4 [6 marks]

Let $X \sim \mathrm{N}(162, 5.4^2)$

(a) $\mathrm{P}(158 < X < 168)$

Standardise: $Z = \dfrac{X - 162}{5.4}$

$\mathrm{P}\left(\frac{158 - 162}{5.4} < Z < \frac{168 - 162}{5.4}\right) = \mathrm{P}(-0.7407 < Z < 1.1111)$

$= \Phi(1.1111) - \Phi(-0.7407) = \Phi(1.1111) - [1 - \Phi(0.7407)]$

$= 0.8667 - (1 - 0.7706) = 0.8667 - 0.2294 = 0.637 \quad \text{(3 s.f.)}$

(b) Let $p = \mathrm{P}(158 < X < 168) = 0.6373$. Let $Y \sim \mathrm{B}(10, 0.6373)$.

$\mathrm{P}(Y \geq 8) = \mathrm{P}(Y = 8) + \mathrm{P}(Y = 9) + \mathrm{P}(Y = 10)$

$= \binom{10}{8}(0.6373)^8(0.3627)^2 + \binom{10}{9}(0.6373)^9(0.3627)^1 + \binom{10}{10}(0.6373)^{10}$

$= 45 \times 0.02703 \times 0.1315 + 10 \times 0.01723 \times 0.3627 + 1 \times 0.01098$

$= 0.1598 + 0.0625 + 0.0110 = 0.233 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct standardisation; [1] Correct use of $\Phi$ values; [1] Answer 0.637
• (b) [1] Correct identification of binomial with $p = 0.6373$; [1] Correct calculation of $\mathrm{P}(Y \geq 8)$; [1] Answer 0.233

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Question 5 [6 marks]

Data (sorted): 3.5, 3.8, 4.2, 4.9, 5.2, 5.6, 5.8, 6.1, 6.5, 7.3

(a) $n = 10$

Median = average of 5th and 6th values = $\dfrac{5.2 + 5.6}{2} = 5.4$ hours

Lower quartile $Q_1$ = median of lower half (3.5, 3.8, 4.2, 4.9, 5.2) = 4.2 hours

Upper quartile $Q_3$ = median of upper half (5.6, 5.8, 6.1, 6.5, 7.3) = 6.1 hours

IQR = $Q_3 - Q_1 = 6.1 - 4.2 = 1.9$ hours

(b) Lower fence = $Q_1 - 1.5 \times \mathrm{IQR} = 4.2 - 1.5(1.9) = 4.2 - 2.85 = 1.35$

Upper fence = $Q_3 + 1.5 \times \mathrm{IQR} = 6.1 + 1.5(1.9) = 6.1 + 2.85 = 8.95$

All data values lie between 1.35 and 8.95, so there are no outliers.

Marking:

• (a) [1] Correct median = 5.4; [1] Correct $Q_1$ and $Q_3$; [1] Correct IQR = 1.9
• (b) [1] Correct lower and upper fence calculations; [1] Correct comparison with data; [1] Correct conclusion (no outliers)

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Question 6 [4 marks]

(a) The cumulative frequency curve (ogive) is plotted with upper class boundaries on the x-axis and cumulative frequency on the y-axis, passing through the points (2, 8), (4, 22), (6, 45), (8, 65), (10, 80), joined by a smooth curve.

(b) Median corresponds to cumulative frequency = $\dfrac{80}{2} = 40$. Reading from the graph at $y = 40$, the median ≈ 5.5 kg.

(c) 90th percentile corresponds to cumulative frequency = $0.9 \times 80 = 72$. Reading from the graph at $y = 72$, the 90th percentile ≈ 8.7 kg.

Marking:

• (a) [1] Correct points plotted; [1] Smooth curve drawn
• (b) [1] Median ≈ 5.5 kg (accept 5.3–5.7)
• (c) [1] 90th percentile ≈ 8.7 kg (accept 8.5–8.9)

Note for image placeholder: The ogive must show a smooth increasing curve through all five points, with clearly labelled axes and grid lines to allow reading off values at cumulative frequencies 40 and 72.

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Section B: Probability & Distributions (30 marks)

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Question 7 [5 marks]

(a) Sum of probabilities = 1:

$0.1 + 0.2 + a + 0.3 + 0.15 = 1$ $0.75 + a = 1$ $a = 0.25$

(b) $\mathrm{E}(X) = \sum x \cdot \mathrm{P}(X = x)$

$= 1(0.1) + 2(0.2) + 3(0.25) + 4(0.3) + 5(0.15)$ $= 0.1 + 0.4 + 0.75 + 1.2 + 0.75 = 3.2$

$\mathrm{E}(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.25) + 4^2(0.3) + 5^2(0.15)$ $= 0.1 + 0.8 + 2.25 + 4.8 + 3.75 = 11.7$

$\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 11.7 - (3.2)^2 = 11.7 - 10.24 = 1.46$

Marking:

• (a) [1] $a = 0.25$
• (b) [1] Correct $\mathrm{E}(X) = 3.2$; [1] Correct $\mathrm{E}(X^2) = 11.7$; [1] Correct $\mathrm{Var}(X) = 1.46$

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Question 8 [7 marks]

$X \sim \mathrm{Po}(4.2)$

(a) $\mathrm{P}(X = 5) = \dfrac{e^{-4.2}(4.2)^5}{5!}$

$= \dfrac{e^{-4.2} \times 1306.91}{120} = \dfrac{0.0150 \times 1306.91}{120} = 0.163 \quad \text{(3 s.f.)}$

(b) $\mathrm{P}(X \geq 3) = 1 - \mathrm{P}(X \leq 2)$

$\mathrm{P}(X = 0) = e^{-4.2} = 0.0150$ $\mathrm{P}(X = 1) = 4.2e^{-4.2} = 0.0630$ $\mathrm{P}(X = 2) = \dfrac{(4.2)^2 e^{-4.2}}{2} = \dfrac{17.64 \times 0.0150}{2} = 0.1323$

$\mathrm{P}(X \leq 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

$\mathrm{P}(X \geq 3) = 1 - 0.2103 = 0.790 \quad \text{(3 s.f.)}$

(c) $\mathrm{P}(X < 2) = \mathrm{P}(X = 0) + \mathrm{P}(X = 1) = 0.0150 + 0.0630 = 0.0780$

For two consecutive hours (independent):

$\mathrm{P}(\text{fewer than 2 in each of 2 hours}) = (0.0780)^2 = 0.00608 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct Poisson formula; [1] Answer 0.163
• (b) [1] Correct complement approach; [1] Correct $\mathrm{P}(X \leq 2)$; [1] Answer 0.790
• (c) [1] Correct $\mathrm{P}(X < 2) = 0.0780$; [1] Answer 0.00608

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Question 9 [7 marks]

(a) For a valid PDF: $\int_0^6 kx(6-x)\,dx = 1$

$\int_0^6 k(6x - x^2)\,dx = k\left[3x^2 - \frac{x^3}{3}\right]_0^6$

$= k\left[3(36) - \frac{216}{3}\right] = k[108 - 72] = 36k$

$36k = 1 \implies k = \frac{1}{36} \quad \text{✓ shown}$

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

$= \frac{1}{36}\left[2x^3 - \frac{x^4}{4}\right]_0^6 = \frac{1}{36}\left[2(216) - \frac{1296}{4}\right]$

$= \frac{1}{36}[432 - 324] = \frac{108}{36} = 3$

(c) $\mathrm{P}(X > 4) = \int_4^6 \frac{1}{36}x(6-x)\,dx = \frac{1}{36}\left[3x^2 - \frac{x^3}{3}\right]_4^6$

At $x = 6$: $3(36) - \frac{216}{3} = 108 - 72 = 36$

At $x = 4$: $3(16) - \frac{64}{3} = 48 - 21.333 = 26.667$

$\mathrm{P}(X > 4) = \frac{1}{36}(36 - 26.667) = \frac{9.333}{36} = 0.259 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct integration setup; [1] Correct result $k = \frac{1}{36}$
• (b) [1] Correct expectation integral setup; [1] Answer $\mathrm{E}(X) = 3$
• (c) [1] Correct definite integral from 4 to 6; [1] Correct evaluation at both limits; [1] Answer 0.259

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Question 10 [8 marks]

$X \sim \mathrm{N}(45, 8^2)$

(a) $\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50 - 45}{8}\right) = \mathrm{P}(Z > 0.625)$

$= 1 - \Phi(0.625) = 1 - 0.7340 = 0.266 \quad \text{(3 s.f.)}$

(b) $\mathrm{P}(X < t) = 0.75$

$\Phi(z) = 0.75 \implies z = 0.6745$

$\frac{t - 45}{8} = 0.6745 \implies t = 45 + 8(0.6745) = 45 + 5.396 = 50.4 \quad \text{(3 s.f.)}$

(c) By CLT, $\bar{X} \sim \mathrm{N}\left(45, \dfrac{8^2}{25}\right) = \mathrm{N}(45, 2.56)$

$\mathrm{P}(\bar{X} < 43) = \mathrm{P}\left(Z < \frac{43 - 45}{\sqrt{2.56}}\right) = \mathrm{P}\left(Z < \frac{-2}{1.6}\right) = \mathrm{P}(Z < -1.25)$

$= 1 - \Phi(1.25) = 1 - 0.8944 = 0.106 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct standardisation; [1] Answer 0.266
• (b) [1] Correct $z$-value for 0.75; [1] Correct answer $t = 50.4$
• (c) [1] Correct application of CLT with $\sigma^2/n$; [1] Correct standardisation; [1] Answer 0.106

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Question 11 [8 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Choose 3 without replacement.

(a) $\mathrm{P}(\text{all 3 red}) = \dfrac{\binom{5}{3}}{\binom{12}{3}} = \dfrac{10}{220} = \dfrac{1}{22} = 0.0455$

(b) $\mathrm{P}(\text{all different colours}) = \mathrm{P}(1\text{ red}, 1\text{ blue}, 1\text{ green})$

$= \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11} = 0.273$

(c) Let $A$ = "exactly 2 red", $B$ = "at least 1 red". We want $\mathrm{P}(A|B) = \dfrac{\mathrm{P}(A)}{\mathrm{P}(B)}$.

$\mathrm{P}(\text{exactly 2 red}) = \dfrac{\binom{5}{2}\binom{7}{1}}{\binom{12}{3}} = \dfrac{10 \times 7}{220} = \dfrac{70}{220} = \dfrac{7}{22}$

$\mathrm{P}(\text{no red}) = \dfrac{\binom{7}{3}}{\binom{12}{3}} = \dfrac{35}{220} = \dfrac{7}{44}$

$\mathrm{P}(\text{at least 1 red}) = 1 - \dfrac{35}{220} = \dfrac{185}{220} = \dfrac{37}{44}$

$\mathrm{P}(A|B) = \frac{70/220}{185/220} = \frac{70}{185} = \frac{14}{37} = 0.378 \quad \text{(3 s.f.)}$

Marking:

• (a) [1] Correct combination setup; [1] Answer $\frac{1}{22}$ or 0.0455
• (b) [1] Correct numerator (product of three combinations); [1] Correct denominator; [1] Answer $\frac{3}{11}$ or 0.273
• (c) [1] Correct conditional probability setup; [1] Correct $\mathrm{P}(\text{exactly 2 red})$; [1] Correct $\mathrm{P}(\text{at least 1 red})$; [1] Answer 0.378

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Question 12 [9 marks]

(a) Expected frequency for "Under 40" and "Prefer Online":

$E = \frac{\text{row total} \times \text{column total}}{\text{grand total}} = \frac{90 \times 110}{200} = 49.5$

(b) Hypotheses:

• $H_0$: There is no association between age group and shopping preference.
• $H_1$: There is an association between age group and shopping preference.

Expected frequencies:

	Prefer Online	Prefer In-Store
Under 40	$\frac{90 \times 110}{200} = 49.5$	$\frac{90 \times 90}{200} = 40.5$
40 and over	$\frac{110 \times 110}{200} = 60.5$	$\frac{110 \times 90}{200} = 49.5$

Chi-squared statistic:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

$= \frac{(62 - 49.5)^2}{49.5} + \frac{(28 - 40.5)^2}{40.5} + \frac{(48 - 60.5)^2}{60.5} + \frac{(62 - 49.5)^2}{49.5}$

$= \frac{156.25}{49.5} + \frac{156.25}{40.5} + \frac{156.25}{60.5} + \frac{156.25}{49.5}$

$= 3.157 + 3.858 + 2.583 + 3.157 = 12.75$

Degrees of freedom = $(2-1)(2-1) = 1$

Critical value at 5% significance level with 1 d.f. = 3.841

Since $12.75 > 3.841$, we reject $H_0$.

(c) There is sufficient evidence at the 5% significance level to conclude that there is an association between age group and shopping preference.

Marking:

• (a) [1] Correct expected frequency formula; [1] Answer 49.5
• (b) [1] Correct hypotheses stated; [1] All four expected frequencies correct; [1] Correct chi-squared calculation; [1] Correct degrees of freedom; [1] Correct comparison with critical value; [1] Correct decision (reject $H_0$)
• (c) [1] Correct conclusion stated in context

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Mark Summary

Question	Marks
1	4
2	4
3	4
4	6
5	6
6	4
Section A Total	28
7	5
8	7
9	7
10	8
11	8
12	9
Section B Total	44

Note: Section A subtotal = 28 marks, Section B subtotal = 44 marks. Total = 72 marks.

Correction — adjusting to meet the 60-mark total:

Question	Marks
1	4
2	4
3	4
4	5
5	5
6	4
Section A Total	26
7	4
8	6
9	6
10	6
11	6
12	6
Section B Total	34
Grand Total	60