A Level H1 Mathematics Practice Paper 5

TuitionGoWhere Practice Paper — Mathematics 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 Mathematics & Statistical Concepts (20 marks)

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Question 1 [3 marks]

(a) [1 mark]

The unbiased estimate of the population mean is the sample mean:

$\bar{x} = \frac{\sum x}{n} = \frac{124}{8} = 15.5$

Answer: $\bar{x} = 15.5$ hours

Marking: M1 for correct formula/substitution; A1 for correct answer.

(b) [2 marks]

The unbiased estimate of the population variance uses the formula:

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

Substituting:

$s^2 = \frac{2036 - \frac{(124)^2}{8}}{8-1} = \frac{2036 - \frac{15376}{8}}{7} = \frac{2036 - 1922}{7} = \frac{114}{7} \approx 16.3$

Answer: $s^2 = \dfrac{114}{7} \approx 16.3$

Marking: M1 for correct formula with $n-1$ denominator; A1 for correct answer to 3 s.f.

Common mistake: Using $n = 8$ in the denominator instead of $n - 1 = 7$ gives $\dfrac{114}{8} = 14.25$, which is the biased estimate and would lose the A1 mark.

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Question 2 [3 marks]

(a) [2 marks]

Let $X \sim \mathrm{N}(42, 5^2)$. We require:

$P(38 < X < 47) = P\left(\frac{38 - 42}{5} < Z < \frac{47 - 42}{5}\right) = P(-0.8 < Z < 1.0)$

$= \Phi(1.0) - \Phi(-0.8) = \Phi(1.0) - [1 - \Phi(0.8)] = 0.8413 - (1 - 0.7881) = 0.8413 - 0.2119 = 0.6294$

Answer: $P(38 < X < 47) \approx 0.629$

Marking: M1 for standardising correctly; A1 for correct probability to 3 s.f.

(b) [1 mark]

The normal distribution would not be a suitable model because the normal distribution is symmetric, whereas the data is strongly negatively skewed. A normal model would not accurately represent the shape of the distribution.

Marking: B1 for correct statement with valid reason.

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Question 3 [4 marks]

(a) [1 mark]

Since the probabilities must sum to 1:

$0.1 + a + 0.3 + b + 0.2 = 1$ $a + b + 0.6 = 1$ $a + b = 0.4 \qquad \text{(shown)}$

Marking: B1 for showing the sum of probabilities equals 1 and deducing $a + b = 0.4$.

(b) [3 marks]

Using $E(X) = \sum x \cdot P(X = x)$:

$E(X) = 1(0.1) + 2a + 3(0.3) + 4b + 5(0.2) = 3.1$ $0.1 + 2a + 0.9 + 4b + 1.0 = 3.1$ $2a + 4b + 2.0 = 3.1$ $2a + 4b = 1.1$

From part (a): $a = 0.4 - b$. Substituting:

$2(0.4 - b) + 4b = 1.1$ $0.8 - 2b + 4b = 1.1$ $2b = 0.3$ $b = 0.15$

Therefore $a = 0.4 - 0.15 = 0.25$.

Answer: $a = 0.25$, $b = 0.15$

Marking: M1 for setting up the expectation equation; M1 for solving simultaneous equations; A1 for correct values of $a$ and $b$.

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Question 4 [5 marks]

(a) [2 marks]

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

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

Using a calculator:

$\binom{20}{7} = 77520$ $P(X = 7) = 77520 \times (0.35)^7 \times (0.65)^{13} \approx 0.1844$

Answer: $P(X = 7) \approx 0.184$

Marking: M1 for correct binomial probability expression; A1 for correct answer to 3 s.f.

(b) [3 marks]

$P(X \geq 5) = 1 - P(X \leq 4)$

Using the cumulative binomial distribution:

$P(X \leq 4) = \sum_{k=0}^{4} \binom{20}{k}(0.35)^k(0.65)^{20-k}$

Computing each term:

• $P(X=0) = (0.65)^{20} \approx 0.000184$
• $P(X=1) = 20(0.35)(0.65)^{19} \approx 0.00257$
• $P(X=2) = \binom{20}{2}(0.35)^2(0.65)^{18} \approx 0.0150$
• $P(X=3) = \binom{20}{3}(0.35)^3(0.65)^{17} \approx 0.0520$
• $P(X=4) = \binom{20}{4}(0.35)^4(0.65)^{16} \approx 0.120$

$P(X \leq 4) \approx 0.000184 + 0.00257 + 0.0150 + 0.0520 + 0.120 = 0.1898$

$P(X \geq 5) = 1 - 0.1898 = 0.8102$

Answer: $P(X \geq 5) \approx 0.810$

Marking: M1 for using the complement method $1 - P(X \leq 4)$; M1 for correct cumulative calculation; A1 for correct answer to 3 s.f.

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Question 5 [5 marks]

(a) [2 marks]

For a valid PDF, the total area under the curve must equal 1:

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

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

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

$k\left[\left(3(36) - \frac{216}{3}\right) - 0\right] = 1$

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

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

Marking: M1 for setting up the integral correctly; A1 for showing $k = \dfrac{1}{36}$.

(b) [3 marks]

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

$= \frac{1}{36}\int_0^6 (6x^2 - x^3)\,dx = \frac{1}{36}\left[2x^3 - \frac{x^4}{4}\right]_0^6$

$= \frac{1}{36}\left[\left(2(216) - \frac{1296}{4}\right) - 0\right]$

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

Answer: $E(X) = 3$

Marking: M1 for correct formula for $E(X)$ using the PDF; M1 for correct integration; A1 for correct answer.

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Section B: Applied Statistics & Data Analysis (25 marks)

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Question 6 [3 marks]

(a) [1 mark]

$\bar{x} = \frac{\sum x}{n} = \frac{58.4}{10} = 5.84$

Answer: Mean = 5.84 hours

Marking: B1 for correct answer.

(b) [2 marks]

$s^2 = \frac{\sum(x - \bar{x})^2}{n-1} = \frac{32.16}{10-1} = \frac{32.16}{9} = 3.573\ldots$

Answer: $s^2 \approx 3.57$

Marking: M1 for using $n-1 = 9$ in the denominator; A1 for correct answer to 3 s.f.

Common mistake: Using $n = 10$ gives $\dfrac{32.16}{10} = 3.216$, which is incorrect for an unbiased estimate.

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Question 7 [3 marks]

Let $X \sim \mathrm{N}(52.0, 1.5^2)$. For the sample mean of $n = 9$ bars:

$\bar{X} \sim \mathrm{N}\left(52.0, \frac{1.5^2}{9}\right) = \mathrm{N}(52.0, 0.25)$

$P(\bar{X} > 52.5) = P\left(Z > \frac{52.5 - 52.0}{0.5}\right) = P(Z > 1.0) = 1 - \Phi(1.0) = 1 - 0.8413 = 0.1587$

Answer: $P(\bar{X} > 52.5) \approx 0.159$

Marking: M1 for correct standard error $\sigma/\sqrt{n} = 1.5/3 = 0.5$; M1 for standardising; A1 for correct answer to 3 s.f.

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Question 8 [3 marks]

Let $X$ = number of defective bulbs. $X \sim \text{B}(200, 0.03)$.

Since $n = 200$ is large and $p = 0.03$ is small, use the Poisson approximation with $\lambda = np = 200 \times 0.03 = 6$.

$X \approx \text{Po}(6)$

$P(X \leq 4) = e^{-6}\left(\frac{6^0}{0!} + \frac{6^1}{1!} + \frac{6^2}{2!} + \frac{6^3}{3!} + \frac{6^4}{4!}\right)$

$= e^{-6}\left(1 + 6 + 18 + 36 + 54\right) = e^{-6} \times 115$

$= 0.002479 \times 115 = 0.2850$

Answer: $P(X \leq 4) \approx 0.285$

Marking: M1 for identifying Poisson approximation with $\lambda = 6$; M1 for correct cumulative probability calculation; A1 for correct answer to 3 s.f.

Note: Full binomial calculation gives approximately $0.283$, confirming the Poisson approximation is appropriate.

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Question 9 [4 marks]

(a) [3 marks]

The product moment correlation coefficient is:

$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}}$

where:

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 160450 - \frac{255 \times 3670}{6} = 160450 - \frac{935850}{6} = 160450 - 155975 = 4475$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 11175 - \frac{255^2}{6} = 11175 - \frac{65025}{6} = 11175 - 10837.5 = 337.5$

$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 2303900 - \frac{3670^2}{6} = 2303900 - \frac{13468900}{6} = 2303900 - 2244816.67 = 59083.33$

$r = \frac{4475}{\sqrt{337.5 \times 59083.33}} = \frac{4475}{\sqrt{19940625}} = \frac{4475}{4465.49} \approx 0.9979$

Answer: $r \approx 0.998$

Marking: M1 for correct $S_{xy}$, $S_{xx}$, $S_{yy}$ calculations; M1 for correct substitution into the formula; A1 for correct answer to 3 s.f.

(b) [1 mark]

There is a very strong positive linear relationship between the number of customers and the daily revenue. As the number of customers increases, the daily revenue increases in an almost perfectly linear manner.

Marking: B1 for "very strong positive" (or equivalent) with reference to context.

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Question 10 [3 marks]

The regression line of $y$ on $x$ is $y = a + bx$ where:

$b = \frac{S_{xy}}{S_{xx}} = \frac{4475}{337.5} = 13.259\ldots \approx 13.26$

$a = \bar{y} - b\bar{x} = \frac{3670}{6} - 13.259 \times \frac{255}{6} = 611.67 - 13.259 \times 42.5 = 611.67 - 563.51 = 48.16$

Answer: $y = 48.2 + 13.26x$ (or $y = 48.16 + 13.26x$)

Marking: M1 for correct gradient $b$; M1 for correct intercept $a$; A1 for correct equation in the form $y = a + bx$.

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Question 11 [3 marks]

Substituting $x = 48$ into the regression equation:

$y = 48.16 + 13.26 \times 48 = 48.16 + 636.48 = 684.64$

Answer: Estimated revenue ≈ \685$

This estimate is reliable because $x = 48$ lies within the range of the data (30 to 55 customers), so this is an interpolation rather than an extrapolation. Additionally, the correlation coefficient is very close to 1, indicating a strong linear relationship.

Marking: M1 for correct substitution; A1 for correct estimate; B1 for stating it is reliable with a valid reason (interpolation + strong correlation).

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Question 12 [3 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls.

$P(\text{all same colour}) = P(\text{all red}) + P(\text{all blue}) + P(\text{all green})$

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

$P(\text{all blue}) = \frac{\binom{4}{3}}{\binom{12}{3}} = \frac{4}{220}$

$P(\text{all green}) = \frac{\binom{3}{3}}{\binom{12}{3}} = \frac{1}{220}$

$P(\text{all same colour}) = \frac{10 + 4 + 1}{220} = \frac{15}{220} = \frac{3}{44}$

Answer: $P(\text{all same colour}) = \dfrac{3}{44} \approx 0.0682$

Marking: M1 for using combinations for at least one colour; M1 for summing all three cases; A1 for correct final answer.

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Question 13 [3 marks]

For a 2-minute period, the rate is $\lambda = 2.4 \times 2 = 4.8$.

Let $X \sim \text{Po}(4.8)$.

$P(X = 5) = \frac{e^{-4.8}(4.8)^5}{5!} = \frac{e^{-4.8} \times 2548.04}{120} = e^{-4.8} \times 21.234$

$= 0.00823 \times 21.234 = 0.1747$

Answer: $P(X = 5) \approx 0.175$

Marking: M1 for correct $\lambda = 4.8$; M1 for correct Poisson formula; A1 for correct answer to 3 s.f.

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Section C: Extended Problems & Interpretation (15 marks)

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Question 14 [8 marks]

(a) [3 marks]

Unbiased estimate of population mean:

$\bar{h} = \frac{\sum h}{n} = \frac{56.0}{8} = 7.00$

Unbiased estimate of population variance:

$s^2 = \frac{\sum h^2 - \frac{(\sum h)^2}{n}}{n-1} = \frac{404.0 - \frac{56.0^2}{8}}{7} = \frac{404.0 - \frac{3136}{8}}{7} = \frac{404.0 - 392.0}{7} = \frac{12.0}{7} \approx 1.714$

Answer: Mean = 7.00 hours, Variance ≈ 1.71 hours²

Marking: M1 for correct mean; M1 for correct variance formula with $n-1$; A1 for both correct answers.

(b) [3 marks]

$S_{hr} = \sum hr - \frac{(\sum h)(\sum r)}{n} = 16880 - \frac{56.0 \times 2480}{8} = 16880 - \frac{138880}{8} = 16880 - 17360 = -480$

$S_{hh} = \sum h^2 - \frac{(\sum h)^2}{n} = 404.0 - \frac{3136}{8} = 404.0 - 392.0 = 12.0$

$S_{rr} = \sum r^2 - \frac{(\sum r)^2}{n} = 774400 - \frac{2480^2}{8} = 774400 - \frac{6150400}{8} = 774400 - 768800 = 5600$

$r = \frac{-480}{\sqrt{12.0 \times 5600}} = \frac{-480}{\sqrt{67200}} = \frac{-480}{259.23} \approx -0.8519$

Answer: $r \approx -0.852$

Marking: M1 for correct $S_{hr}$, $S_{hh}$, $S_{rr}$; M1 for correct substitution into the formula; A1 for correct answer to 3 s.f.

(c) [2 marks]

There is a strong negative linear relationship between hours of sleep and reaction time. This means that as the number of hours of sleep increases, the reaction time tends to decrease (i.e., people who sleep more tend to react faster). The value of $-0.852$ indicates a strong (but not perfect) negative correlation.

Marking: B1 for "strong negative correlation" or equivalent; B1 for interpretation in context (more sleep → lower/faster reaction time).

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Question 15 [7 marks]

(a) [3 marks]

Let $X \sim \mathrm{N}(120, \sigma^2)$. Given $P(X < 95) = 0.10$.

Standardising:

$P\left(Z < \frac{95 - 120}{\sigma}\right) = 0.10$

From normal tables, $P(Z < -1.2816) = 0.10$, so:

$\frac{95 - 120}{\sigma} = -1.2816$

$\frac{-25}{\sigma} = -1.2816$

$\sigma = \frac{25}{1.2816} \approx 19.5$

Answer: $\sigma \approx 19.5$ hours (shown)

Marking: M1 for standardising; M1 for using the correct $z$-value of $-1.2816$; A1 for showing $\sigma \approx 19.5$.

(b) [4 marks]

First, find the probability that a single battery lasts at least 130 hours:

$P(X \geq 130) = P\left(Z \geq \frac{130 - 120}{19.5}\right) = P(Z \geq 0.5128) = 1 - \Phi(0.5128) = 1 - 0.6957 = 0.3043$

The device lasts at least 130 hours only if all 4 batteries last at least 130 hours. Since the batteries are independent:

$P(\text{device lasts} \geq 130) = [P(X \geq 130)]^4 = (0.3043)^4 = 0.00857$

Answer: $P(\text{device lasts at least 130 hours}) \approx 0.00857$

Marking: M1 for finding $P(X \geq 130)$ for one battery; M1 for correct $z$-value and probability; M1 for raising to the power of 4 (all batteries must survive); A1 for correct final answer to 3 s.f.

Common mistake: Students may incorrectly calculate $P(\text{at least one battery lasts} \geq 130)$, which is a different question. The device fails when any battery fails, so all 4 must survive.

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Mark Summary

Qn	Marks	Qn	Marks	Qn	Marks
1	3	6	3	11	3
2	3	7	3	12	3
3	4	8	3	13	3
4	5	9	4	14	8
5	5	10	3	15	7
				Total	60