A Level H1 Mathematics Practice Paper 5

TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper — Statistics & Probability (Version 5 of 5)
Total Marks: 60

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Section A: Pure Statistics (30 marks)

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Question 1 [2 marks]

Answer: $\bar{x} = 13.625$

Working:

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

Marking:

• M1: Correct substitution into the formula for the mean
• A1: Correct answer (13.625 or 13.6 to 3 s.f.)

Teaching Note: The unbiased estimate of the population mean is simply the sample mean. We add all data values and divide by the number of observations $n$. This is the best single-number estimate of the true population mean $\mu$ from sample data.

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Question 2 [3 marks]

Answer: $s^2 = 6.268$ (or 6.27 to 3 s.f.)

Working:

Using $\bar{x} = 13.625$ and $n = 8$:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	−1.625	2.6406
15	1.375	1.8906
10	−3.625	13.1406
18	4.375	19.1406
14	0.375	0.1406
11	−2.625	6.8906
16	2.375	5.6406
13	−0.625	0.3906

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

$s^2 = \frac{49.875}{8 - 1} = \frac{49.875}{7} = 7.125$

Correction: Let me recalculate carefully:

$\sum(x_i - \bar{x})^2 = 2.6406 + 1.8906 + 13.1406 + 19.1406 + 0.1406 + 6.8906 + 5.6406 + 0.3906 = 49.875$

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

Answer: $s^2 = 7.125$ (or 7.13 to 3 s.f.)

Marking:

• M1: Correct calculation of deviations from the mean
• M1: Correct use of $n - 1 = 7$ in the denominator (not $n = 8$)
• A1: Correct final answer

Common Mistake: Using $n = 8$ instead of $n - 1 = 7$ gives $\frac{49.875}{8} = 6.234$, which is the biased sample variance. The unbiased estimate requires dividing by $n - 1$ to correct for the fact that we are estimating the population parameter from sample data.

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Question 3 [3 marks]

(a) [2 marks]

Answer: $\hat{\mu} = 43.0$ cm, $\hat{\sigma}^2 = 11.5$ cm$^2$

Working:

$\bar{x} = \frac{42 + 45 + 38 + 47 + 43}{5} = \frac{215}{5} = 43.0$

$s^2 = \frac{(42-43)^2 + (45-43)^2 + (38-43)^2 + (47-43)^2 + (43-43)^2}{5-1}$

$= \frac{1 + 4 + 25 + 16 + 0}{4} = \frac{46}{4} = 11.5$

(b) [1 mark]

Answer: The sample must be a random sample from the population (or the plants are independently selected from a normally distributed population).

Marking:

• (a) M1: Correct calculation of sample mean; A1: Both mean and variance correct
• (b) B1: Valid assumption stated

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Question 4 [4 marks]

$X \sim \mathrm{N}(800, 50^2)$

(a) [2 marks]

$\mathrm{P}(X > 860) = \mathrm{P}\left(Z > \frac{860 - 800}{50}\right) = \mathrm{P}(Z > 1.2)$

$= 1 - \Phi(1.2) = 1 - 0.8849 = 0.1151$

Answer: 0.115 (to 3 s.f.)

(b) [2 marks]

We need $k$ such that $\mathrm{P}(X < k) = 0.90$.

$\mathrm{P}\left(Z < \frac{k - 800}{50}\right) = 0.90$

From tables, $\Phi(1.282) = 0.90$, so:

$\frac{k - 800}{50} = 1.282$

$k = 800 + 50 \times 1.282 = 864.1$

Answer: $k = 864$ hours (to 3 s.f.)

Marking:

• (a) M1: Standardising correctly; A1: Correct probability
• (b) M1: Using inverse normal correctly; A1: Correct value of $k$

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Question 5 [3 marks]

(a) [1 mark]

Answer: $X \sim \mathrm{B}(4, \frac{1}{6})$

(b) [2 marks]

$\mathrm{P}(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$

Answer: $\frac{25}{216} \approx 0.1157$ (or 0.116 to 3 s.f.)

Marking:

• (a) B1: Correct distribution stated with both parameters
• (b) M1: Correct binomial probability formula applied; A1: Correct answer

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Question 6 [4 marks]

Let $X$ = number of emails per hour, $X \sim \mathrm{Po}(3.5)$

(a) [2 marks]

$\mathrm{P}(X = 4) = \frac{e^{-3.5} \times 3.5^4}{4!} = \frac{e^{-3.5} \times 150.0625}{24}$

$= \frac{150.0625}{24} \times e^{-3.5} = 6.2526 \times 0.030197 = 0.1888$

Answer: 0.189 (to 3 s.f.)

(b) [2 marks]

For a 2-hour period, the mean is $\lambda = 3.5 \times 2 = 7$.

Let $Y \sim \mathrm{Po}(7)$.

$\mathrm{P}(Y \geq 5) = 1 - \mathrm{P}(Y \leq 4)$

$= 1 - \left[\frac{e^{-7}7^0}{0!} + \frac{e^{-7}7^1}{1!} + \frac{e^{-7}7^2}{2!} + \frac{e^{-7}7^3}{3!} + \frac{e^{-7}7^4}{4!}\right]$

$= 1 - e^{-7}\left[1 + 7 + \frac{49}{2} + \frac{343}{6} + \frac{2401}{24}\right]$

$= 1 - e^{-7}\left[1 + 7 + 24.5 + 57.167 + 100.042\right]$

$= 1 - e^{-7} \times 189.708 = 1 - 0.0009119 \times 189.708$

$= 1 - 0.1730 = 0.8270$

Answer: 0.827 (to 3 s.f.)

Marking:

• (a) M1: Correct Poisson formula with $\lambda = 3.5$; A1: Correct answer
• (b) M1: Correct adjustment of $\lambda$ to 7 for 2 hours and use of complement; A1: Correct answer

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Question 7 [3 marks]

Answer: $\frac{21}{44}$ or approximately 0.477

Working:

Total balls = 12. Selecting 3 balls without replacement.

$\mathrm{P}(2 \text{ red, } 1 \text{ blue}) = \frac{\binom{7}{2} \times \binom{5}{1}}{\binom{12}{3}} = \frac{21 \times 5}{220} = \frac{105}{220} = \frac{21}{44}$

Marking:

• M1: Correct numerator (combinations of red and blue)
• M1: Correct denominator (total combinations)
• A1: Correct simplified answer

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Question 8 [4 marks]

$X \sim \mathrm{B}(20, 0.3)$

(a) [2 marks]

$\mathrm{E}(X) = np = 20 \times 0.3 = 6$

$\mathrm{Var}(X) = np(1-p) = 20 \times 0.3 \times 0.7 = 4.2$

(b) [2 marks]

Since $n = 20$ is moderately large and $np = 6 > 5$, $n(1-p) = 14 > 5$, we can use the normal approximation:

$X \stackrel{\text{approx}}{\sim} \mathrm{N}(6, 4.2)$

Using continuity correction:

$\mathrm{P}(X \geq 10) \approx \mathrm{P}\left(Z \geq \frac{9.5 - 6}{\sqrt{4.2}}\right) = \mathrm{P}\left(Z \geq \frac{3.5}{2.049}\right) = \mathrm{P}(Z \geq 1.708)$

$= 1 - \Phi(1.708) = 1 - 0.9562 = 0.0438$

Answer: 0.0438 (to 3 s.f.)

Marking:

• (a) B1: Each correct (E(X) and Var(X))
• (b) M1: Correct normal approximation with continuity correction; A1: Correct probability

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Question 9 [4 marks]

(a) [2 marks]

For a valid PDF, $\int_0^4 f(x)\,dx = 1$:

$\int_0^4 kx(4-x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\left[2x^2 - \frac{x^3}{3}\right]_0^4$

$= k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = k\left[32 - \frac{64}{3}\right] = k\left[\frac{96 - 64}{3}\right] = k \times \frac{32}{3}$

Setting equal to 1:

$k \times \frac{32}{3} = 1 \implies k = \frac{3}{32} \quad \text{(shown)}$

(b) [2 marks]

$\mathrm{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx$

$= \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4 = \frac{3}{32}\left[\frac{256}{3} - \frac{256}{4}\right]$

$= \frac{3}{32}\left[\frac{256}{3} - 64\right] = \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \times \frac{64}{3} = \frac{64}{32} = 2$

Answer: $\mathrm{E}(X) = 2$

Marking:

• (a) M1: Correct integration; A1: Correct derivation of $k = \frac{3}{32}$
• (b) M1: Correct setup of $\mathrm{E}(X)$ integral; A1: Correct answer

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Section B: Applied Statistics & Data Interpretation (30 marks)

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Question 10 [5 marks]

(a) [1 mark]

$\bar{x} = \frac{320 + 280 + 350 + 410 + 390 + 460}{6} = \frac{2210}{6} = 368.33$

Answer: $368 (to 3 s.f.)

(b) [2 marks]

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

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
320	−48.33	2336.11
280	−88.33	7802.78
350	−18.33	336.11
410	41.67	1736.11
390	21.67	469.44
460	91.67	8402.78

$\sum(x_i - \bar{x})^2 = 21083.33$

$s^2 = \frac{21083.33}{5} = 4216.67$

$s = \sqrt{4216.67} = 64.94$

Answer: s = \64.9$ (to 3 s.f.)

(c) [2 marks]

Adding Sunday's sales of $520:

New mean: $\bar{x}_{\text{new}} = \frac{2210 + 520}{7} = \frac{2730}{7} = 390$

The mean increases from $368 to $390 because $520 is above the original mean, pulling the average up.

The standard deviation will also increase because $520 is far from the original mean, increasing the spread of the data. The new data point is an outlier relative to the original dataset, so both measures of central tendency and dispersion are affected.

Marking:

• (a) B1: Correct mean
• (b) M1: Correct method for standard deviation; A1: Correct answer
• (c) B1: Mean increases (with reasoning); B1: Standard deviation increases (with reasoning)

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Question 11 [6 marks]

(a) [3 marks]

Calculating summary statistics:

$n = 8$

$\sum x = 2.0 + 3.5 + 4.0 + 5.5 + 6.0 + 7.5 + 8.0 + 9.0 = 45.5$

$\sum y = 15 + 22 + 25 + 30 + 33 + 38 + 40 + 45 = 248$

$\bar{x} = \frac{45.5}{8} = 5.6875$

$\bar{y} = \frac{248}{8} = 31$

$\sum xy = (2.0 \times 15) + (3.5 \times 22) + (4.0 \times 25) + (5.5 \times 30) + (6.0 \times 33) + (7.5 \times 38) + (8.0 \times 40) + (9.0 \times 45)$

$= 30 + 77 + 100 + 165 + 198 + 285 + 320 + 405 = 1580$

$\sum x^2 = 4 + 12.25 + 16 + 30.25 + 36 + 56.25 + 64 + 81 = 299.75$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 1580 - \frac{45.5 \times 248}{8} = 1580 - 1410.5 = 169.5$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 299.75 - \frac{45.5^2}{8} = 299.75 - 258.781 = 40.969$

$b = \frac{S_{xy}}{S_{xx}} = \frac{169.5}{40.969} = 4.137$

$a = \bar{y} - b\bar{x} = 31 - 4.137 \times 5.6875 = 31 - 23.530 = 7.470$

Answer: $y = 7.47 + 4.14x$ (to 3 s.f.)

(b) [1 mark]

Answer: For every additional $1,000 spent on advertising, the monthly sales revenue increases by approximately $4,140.

(c) [2 marks]

When $x = 6.5$:

$y = 7.47 + 4.14 \times 6.5 = 7.47 + 26.91 = 34.38$

Answer: Estimated sales revenue is approximately $34,400.

Comment: Since $x = 6.5$ lies within the range of the data (2.0 to 9.0), this is an interpolation and the estimate is reasonably reliable.

Marking:

• (a) M1: Correct calculation of $S_{xy}$ and $S_{xx}$; M1: Correct $b$ and $a$; A1: Correct equation to 3 s.f.
• (b) B1: Correct interpretation in context
• (c) M1: Correct substitution; A1: Correct estimate with valid reliability comment

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Question 12 [5 marks]

(a) [1 mark]

$H_0: \mu = 5.0$ mmol/L
$H_1: \mu > 5.0$ mmol/L

(b) [1 mark]

Since $n = 36$ is large, by the Central Limit Theorem, we use the $z$-test:

$z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{5.3 - 5.0}{1.2 / \sqrt{36}} = \frac{0.3}{1.2 / 6} = \frac{0.3}{0.2} = 1.5$

(c) [3 marks]

This is a one-tailed test at the 5% significance level.

Critical value: $z_{0.05} = 1.645$

Since $z = 1.5 < 1.645$, we do not reject $H_0$.

Conclusion: There is insufficient evidence at the 5% significance level to support the claim that the mean cholesterol level is greater than 5.0 mmol/L.

Marking:

• (a) B1: Both hypotheses correct (one-tailed)
• (b) B1: Correct test statistic
• (c) M1: Correct critical value; M1: Correct comparison and decision; A1: Correct conclusion in context

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Question 13 [5 marks]

(a) [3 marks]

Total outcomes = $6 \times 6 = 36$

$s$	2	3	4	5	6	7	8	9	10	11	12
$\mathrm{P}(S = s)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

(b) [2 marks]

$\mathrm{E}(S) = \sum s \cdot \mathrm{P}(S = s)$

$= \frac{1}{36}[2(1) + 3(2) + 4(3) + 5(4) + 6(5) + 7(6) + 8(5) + 9(4) + 10(3) + 11(2) + 12(1)]$

$= \frac{1}{36}[2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12] = \frac{252}{36} = 7$

$\mathrm{E}(S^2) = \frac{1}{36}[4(1) + 9(2) + 16(3) + 25(4) + 36(5) + 49(6) + 64(5) + 81(4) + 100(3) + 121(2) + 144(1)]$

$= \frac{1}{36}[4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144] = \frac{1974}{36} = 54.833$

$\mathrm{Var}(S) = \mathrm{E}(S^2) - [\mathrm{E}(S)]^2 = 54.833 - 49 = 5.833$

Answer: $\mathrm{E}(S) = 7$, $\mathrm{Var}(S) = 5.83$ (to 3 s.f.) or $\frac{35}{6}$

Marking:

• (a) B1: Correct numerator pattern (1,2,3,4,5,6,5,4,3,2,1); B1: Correct denominator 36; B1: All probabilities correct
• (b) M1: Correct method for E(S) and Var(S); A1: Both correct

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Question 14 [4 marks]

$X \sim \mathrm{N}(180, 15^2)$

(a) [2 marks]

$\mathrm{P}(165 < X < 195) = \mathrm{P}\left(\frac{165 - 180}{15} < Z < \frac{195 - 180}{15}\right) = \mathrm{P}(-1 < Z < 1)$

$= \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826$

Answer: 0.683 (to 3 s.f.)

(b) [2 marks]

For the sample mean $\bar{X}$ of $n = 9$ apples:

$\bar{X} \sim \mathrm{N}\left(180, \frac{15^2}{9}\right) = \mathrm{N}(180, 25)$

$\mathrm{P}(\bar{X} > 185) = \mathrm{P}\left(Z > \frac{185 - 180}{5}\right) = \mathrm{P}(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$

Answer: 0.159 (to 3 s.f.)

Marking:

• (a) M1: Correct standardisation; A1: Correct probability
• (b) M1: Correct distribution of sample mean with $\sigma/\sqrt{n} = 5$; A1: Correct probability

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Question 15 [5 marks]

(a) [1 mark]

$\mathrm{P}(\text{MRT}) = \frac{72}{200} = 0.36$

(b) [2 marks]

$\mathrm{P}(\text{both bus}) = \frac{55}{200} \times \frac{54}{199} = \frac{2970}{39800} = 0.07462$

Answer: 0.0746 (to 3 s.f.)

(c) [2 marks]

$\mathrm{P}(\text{at least one walks}) = 1 - \mathrm{P}(\text{none walk})$

$\mathrm{P}(\text{none walk}) = \frac{165}{200} \times \frac{164}{199} \times \frac{163}{198} = \frac{4465980}{7880400} = 0.5667$

$\mathrm{P}(\text{at least one walks}) = 1 - 0.5667 = 0.4333$

Answer: 0.433 (to 3 s.f.)

Marking:

• (a) B1: Correct probability
• (b) M1: Correct multiplication of conditional probabilities (without replacement); A1: Correct answer
• (c) M1: Correct use of complement and multiplication; A1: Correct answer

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Mark Summary

Section	Marks
Section A: Questions 1–9	30
Section B: Questions 10–15	30
Total	60