A Level H1 Mathematics Practice Paper 1

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