A Level H1 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key — Version 2 of 5

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Section A

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Question 1 [2 marks]

$\bar{x} = \frac{12 + 15 + 10 + 14 + 11 + 13}{6} = \frac{75}{6} = 12.5$

Answer: 12.5 hours

Marking: M1 for correct substitution into the formula. A1 for correct answer.

Teaching note: The sample mean $\bar{x}$ is the best point estimate of the population mean $\mu$. We sum all observations and divide by the number of observations $n$. This is the standard unbiased estimator for the population mean.

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Question 2 [2 marks]

First compute each deviation from the mean $\bar{x} = 12.5$:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	$-0.5$	0.25
15	2.5	6.25
10	$-2.5$	6.25
14	1.5	2.25
11	$-1.5$	2.25
13	0.5	0.25

$\sum(x_i - \bar{x})^2 = 0.25 + 6.25 + 6.25 + 2.25 + 2.25 + 0.25 = 17.5$

$s^2 = \frac{17.5}{6-1} = \frac{17.5}{5} = 3.50$

Answer: 3.50 hours²

Marking: M1 for correct deviations/squared deviations. A1 for correct answer with $n-1$ denominator.

Common mistake: Using $n = 6$ in the denominator gives $\frac{17.5}{6} = 2.92$, which is the biased sample variance. The unbiased estimator uses $n-1 = 5$ because one degree of freedom is "used up" when estimating the sample mean. This correction (Bessel's correction) ensures that on average, $s^2$ equals the true population variance $\sigma^2$.

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Question 3 [2 marks]

$X \sim \mathrm{B}(12, 0.35)$

$\mathrm{P}(X = 5) = \binom{12}{5}(0.35)^5(0.65)^7$

$= 792 \times 0.005252 \times 0.049022$

$= 792 \times 0.0002575 = 0.2039$

Answer: 0.204 (3 s.f.)

Marking: M1 for correct binomial probability formula with $\binom{12}{5}$, $p = 0.35$, $q = 0.65$. A1 for correct answer to 3 s.f.

Teaching note: For $X \sim \mathrm{B}(n, p)$, the probability of exactly $r$ successes is $\mathrm{P}(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$. The graphing calculator's binomialPdf function can be used directly.

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Question 4 [2 marks]

Let $X$ be the number of sixes in 8 rolls. Then $X \sim \mathrm{B}\left(8, \frac{1}{6}\right)$.

$\mathrm{P}(X = 3) = \binom{8}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^5$

$= 56 \times \frac{1}{216} \times \frac{3125}{7776}$

$= 56 \times 0.0046296 \times 0.40188$

$= 0.1042$

Answer: 0.104 (3 s.f.)

Marking: M1 for identifying $n = 8$, $p = \frac{1}{6}$ and using the binomial formula. A1 for correct answer.

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Question 5 [3 marks]

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

(a) [2 marks]

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

$= \Phi(1.0) - \Phi(-0.8) = \Phi(1.0) - [1 - \Phi(0.8)]$

$= 0.8413 - (1 - 0.7881) = 0.8413 - 0.2119 = 0.6294$

Answer: 0.629 (3 s.f.)

Marking: M1 for standardising and using normal tables/calculator. A1 for correct answer.

(b) [1 mark]

$\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50-42}{5}\right) = \mathrm{P}(Z > 1.6) = 1 - \Phi(1.6) = 1 - 0.9452 = 0.0548$

Expected number $= 200 \times 0.0548 = 10.96$

Answer: 11 plants (approximately)

Marking: A1 for correct calculation and rounding.

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Question 6 [3 marks]

(a) [2 marks]

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

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

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

Setting equal to 1: $k \cdot \frac{32}{3} = 1$, so $k = \frac{3}{32}$. $\quad \blacksquare$

(b) [1 mark]

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

$= \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4 = \frac{3}{32}\left[\frac{256}{3} - 64\right] = \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \times \frac{64}{3} = 2$

Answer: 2

Marking: M1 for setting up the expectation integral. A1 for correct answer.

Teaching note: For a continuous random variable with PDF $f(x)$, the expected value is $\mathrm{E}(X) = \int x\,f(x)\,dx$ over the domain. The symmetry of $f(x) = \frac{3}{32}x(4-x)$ about $x = 2$ confirms $\mathrm{E}(X) = 2$.

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Question 7 [3 marks]

(a) [1 mark]

The class with the highest frequency is $20 \le t < 30$ (frequency 16).

Answer: $20 \le t < 30$

(b) [2 marks]

Using midpoints: 5, 15, 25, 40, 65.

$\bar{x} = \frac{6(5) + 14(15) + 16(25) + 10(40) + 4(65)}{50} = \frac{30 + 210 + 400 + 400 + 260}{50} = \frac{1300}{50} = 26.0$

Answer: 26.0 minutes

Marking: M1 for using midpoints correctly. A1 for correct answer.

Teaching note: For grouped data, we use the midpoint of each class as the representative value. The last class $50 \le t < 80$ has midpoint $\frac{50+80}{2} = 65$.

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Question 8 [2 marks]

Check independence: $A$ and $B$ are independent if and only if $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B)$.

First find $\mathrm{P}(A \cap B)$ using the addition rule:

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

$0.76 = 0.6 + 0.4 - \mathrm{P}(A \cap B)$

$\mathrm{P}(A \cap B) = 1.0 - 0.76 = 0.24$

Now check:

$\mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24$

Since $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.24$, the events are independent.

Answer: Yes, A and B are independent because P(A ∩ B) = P(A) × P(B) = 0.24.

Marking: M1 for finding P(A ∩ B) using the addition formula. A1 for correct conclusion with justification.

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Question 9 [3 marks]

Total balls = 5 + 4 + 3 = 12. Number of ways to draw 3 balls from 12: $\binom{12}{3} = 220$.

All red: $\binom{5}{3} = 10$
All blue: $\binom{4}{3} = 4$
All green: $\binom{3}{3} = 1$

$\mathrm{P}(\text{all same colour}) = \frac{10 + 4 + 1}{220} = \frac{15}{220} = \frac{3}{44} = 0.0682$

Answer: $\frac{3}{44}$ or 0.0682 (3 s.f.)

Marking: M1 for using combinations for each colour. M1 for summing and dividing by total. A1 for correct answer.

Teaching note: Since the balls are drawn without replacement, we use combinations (not permutations) because the order of selection does not matter. The events "all red", "all blue", and "all green" are mutually exclusive, so we add their probabilities.

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Question 10 [3 marks]

From standard normal tables:

• $\mathrm{P}(Y < 30) = 0.1587$ means $\mathrm{P}\left(Z < \frac{30-\mu}{\sigma}\right) = 0.1587$, so $\frac{30-\mu}{\sigma} = -1.0$ (since $\Phi(-1) = 0.1587$)

• $\mathrm{P}(Y > 50) = 0.0228$ means $\mathrm{P}\left(Z > \frac{50-\mu}{\sigma}\right) = 0.0228$, so $\frac{50-\mu}{\sigma} = 2.0$ (since $\Phi(2) = 0.9772$)

Solving the simultaneous equations:

$30 - \mu = -\sigma \quad \Rightarrow \quad \mu = 30 + \sigma \quad \text{...(i)}$

$50 - \mu = 2\sigma \quad \text{...(ii)}$

Substitute (i) into (ii):

$50 - (30 + \sigma) = 2\sigma$

$20 - \sigma = 2\sigma$

$20 = 3\sigma \quad \Rightarrow \quad \sigma = \frac{20}{3} = 6.67$

$\mu = 30 + \frac{20}{3} = \frac{110}{3} = 36.67$

Answer: $\mu = 36\frac{2}{3}$ (or 36.7), $\sigma = 6\frac{2}{3}$ (or 6.67)

Marking: M1 for setting up both equations from the given probabilities. M1 for solving the simultaneous equations. A1 for both correct values.

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Section B

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Question 11 [8 marks]

(a) [4 marks]

$\bar{x} = \frac{120 + 180 + 95 + 210 + 150 + 165 + 130 + 195}{8} = \frac{1245}{8} = 155.625$

For the unbiased variance, compute $\sum(x_i - \bar{x})^2$:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
120	$-35.625$	1269.14
180	24.375	594.14
95	$-60.625$	3675.39
210	54.375	2956.64
150	$-5.625$	31.64
165	9.375	87.89
130	$-25.625$	656.64
195	39.375	1550.39

$\sum(x_i - \bar{x})^2 = 10821.875$

$s^2 = \frac{10821.875}{8-1} = \frac{10821.875}{7} = 1545.98$

Answer: Mean = $155.6 (or $155.63), Variance = 1550 (3 s.f.)

Marking: M1 for correct mean. M1 for correct squared deviations (at least 3 correct). M1 for using $n-1 = 7$ in the denominator. A1 for correct variance to 3 s.f.

(b) [2 marks]

The sample mean is \bar{x} = \155.63, which is greater than \140. This provides some support for the researcher's claim that the population mean exceeds $140. However, with only a sample of 8, there is considerable sampling variability, and a formal hypothesis test would be needed to draw a statistically significant conclusion.

Marking: B1 for noting the sample mean exceeds $140. B1 for acknowledging the limitation of the small sample size.

(c) [2 marks]

Using $n$ in the denominator gives the average of the squared deviations from the sample mean. However, the sample mean $\bar{x}$ is itself estimated from the same data, so the deviations $(x_i - \bar{x})$ tend to be slightly smaller than the true deviations $(x_i - \mu)$ from the population mean. Dividing by $n-1$ instead of $n$ corrects this downward bias, making $s^2$ an unbiased estimator of $\sigma^2$. This correction is known as Bessel's correction.

Marking: B1 for explaining that the sample mean causes deviations to be smaller. B1 for stating that $n-1$ corrects the bias (Bessel's correction).

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Question 12 [9 marks]

Let $X$ be the number of calls per minute. $X \sim \mathrm{Po}(4.2)$.

(a) [2 marks]

$\mathrm{P}(X = 5) = \frac{e^{-4.2}(4.2)^5}{5!} = \frac{e^{-4.2} \times 1306.91}{120} = \frac{1306.91}{120} \times 0.0150 = 0.1634$

Answer: 0.163 (3 s.f.)

Marking: M1 for correct Poisson formula with $\lambda = 4.2$. A1 for correct answer.

(b) [3 marks]

$\mathrm{P}(X \ge 3) = 1 - \mathrm{P}(X \le 2) = 1 - [\mathrm{P}(X=0) + \mathrm{P}(X=1) + \mathrm{P}(X=2)]$

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

$\mathrm{P}(X = 1) = 4.2 \times e^{-4.2} = 4.2 \times 0.0150 = 0.0630$

$\mathrm{P}(X = 2) = \frac{4.2^2}{2} \times e^{-4.2} = 8.82 \times 0.0150 = 0.1323$

$\mathrm{P}(X \ge 3) = 1 - (0.0150 + 0.0630 + 0.1323) = 1 - 0.2103 = 0.7897$

Answer: 0.790 (3 s.f.)

Marking: M1 for using the complement method. M1 for calculating individual probabilities correctly. A1 for correct final answer.

(c) [4 marks]

Over 10 minutes, $\lambda = 4.2 \times 10 = 42$. Let $Y \sim \mathrm{Po}(42)$.

Since $\lambda = 42$ is large, use the normal approximation: $Y \approx \mathrm{N}(42, 42)$.

Assumption: The Poisson rate is constant over time, and calls in different minutes are independent.

With continuity correction:

$\mathrm{P}(Y < 35) = \mathrm{P}(Y \le 34) \approx \mathrm{P}\left(Z < \frac{34.5 - 42}{\sqrt{42}}\right) = \mathrm{P}\left(Z < \frac{-7.5}{6.481}\right) = \mathrm{P}(Z < -1.157)$

$= 1 - \Phi(1.157) = 1 - 0.8764 = 0.1236$

Answer: 0.124 (3 s.f.)

Marking: M1 for $\lambda = 42$ and stating the normal approximation. M1 for stating assumptions (constant rate, independence). M1 for continuity correction. A1 for correct answer.

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Question 13 [9 marks]

Let $X \sim \mathrm{N}(800, 100^2)$.

(a) [3 marks]

$\mathrm{P}(700 < X < 950) = \mathrm{P}\left(\frac{700-800}{100} < Z < \frac{950-800}{100}\right) = \mathrm{P}(-1.0 < Z < 1.5)$

$= \Phi(1.5) - \Phi(-1.0) = \Phi(1.5) - [1 - \Phi(1.0)]$

$= 0.9332 - (1 - 0.8413) = 0.9332 - 0.1587 = 0.7745$

Answer: 0.775 (3 s.f.)

Marking: M1 for standardising both bounds. M1 for correct use of $\Phi$ values. A1 for correct answer.

(b) [3 marks]

We need $\mathrm{P}(X < k) = 0.05$.

From tables, $\Phi(-1.645) = 0.05$, so:

$\frac{k - 800}{100} = -1.645$

$k = 800 - 164.5 = 635.5$

Answer: 636 hours (or 635 hours)

Marking: M1 for setting up $\mathrm{P}(X < k) = 0.05$. M1 for using the correct $z$-value ($-1.645$). A1 for correct answer.

(c) [3 marks]

By the Central Limit Theorem, the sample mean $\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right) = \mathrm{N}\left(800, \frac{100^2}{16}\right) = \mathrm{N}(800, 625)$, so $\sigma_{\bar{X}} = 25$.

$\mathrm{P}(\bar{X} > 820) = \mathrm{P}\left(Z > \frac{820 - 800}{25}\right) = \mathrm{P}(Z > 0.8) = 1 - \Phi(0.8) = 1 - 0.7881 = 0.2119$

Answer: 0.212 (3 s.f.)

Marking: M1 for correct distribution of sample mean with $\sigma/\sqrt{n} = 25$. M1 for standardising. A1 for correct answer.

Teaching note: The distribution of the sample mean has the same mean $\mu$ but a reduced standard deviation $\sigma/\sqrt{n}$ (the standard error). This reflects the fact that sample means vary less than individual observations.

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Question 14 [9 marks]

(a) [2 marks]

$\mathrm{P}(\text{Female and MRT}) = \frac{15}{120} = \frac{1}{8}$

Answer: $\frac{1}{8}$ or 0.125

Marking: M1 for identifying the correct cell (Female, MRT = 15). A1 for correct probability.

(b) [2 marks]

$\mathrm{P}(\text{Male} \mid \text{Walk}) = \frac{\mathrm{P}(\text{Male and Walk})}{\mathrm{P}(\text{Walk})} = \frac{15/120}{35/120} = \frac{15}{35} = \frac{3}{7}$

Answer: $\frac{3}{7}$ or 0.429 (3 s.f.)

Marking: M1 for using conditional probability formula. A1 for correct answer.

(c) [3 marks]

Check if $\mathrm{P}(\text{Male} \cap \text{Bus}) = \mathrm{P}(\text{Male}) \times \mathrm{P}(\text{Bus})$.

$\mathrm{P}(\text{Male} \cap \text{Bus}) = \frac{20}{120} = \frac{1}{6} = 0.1667$

$\mathrm{P}(\text{Male}) \times \mathrm{P}(\text{Bus}) = \frac{70}{120} \times \frac{32}{120} = \frac{70 \times 32}{14400} = \frac{2240}{14400} = 0.1556$

Since $0.1667 \ne 0.1556$, the events are not independent.

Answer: The events "male" and "prefers bus" are NOT independent because P(Male ∩ Bus) ≠ P(Male) × P(Bus).

Marking: M1 for calculating both sides. M1 for comparing. A1 for correct conclusion.

(d) [2 marks]

$\mathrm{P}(\text{both MRT}) = \frac{40}{120} \times \frac{39}{119} = \frac{1}{3} \times \frac{39}{119} = \frac{39}{357} = \frac{13}{119} = 0.1092$

Answer: $\frac{13}{119}$ or 0.109 (3 s.f.)

Marking: M1 for multiplying probabilities without replacement. A1 for correct answer.

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Mark Summary

Question	Marks
1	2
2	2
3	2
4	2
5	3
6	3
7	3
8	2
9	3
10	3
Section A Total	25
11	8
12	9
13	9
14	9
Section B Total	35
Grand Total	60