A Level H1 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Subject: Mathematics H1 Paper: Practice Paper — Statistics & Probability Version: 2 of 5 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 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 population variance:

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

Calculate each $(x_i - \bar{x})^2$:

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

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

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

Answer: $\bar{x} = 13.6$ hours, $s^2 = 7.13$ hours² (to 3 s.f.)

Marking:

• M1: Correct formula for $\bar{x}$ with correct substitution
• A1: $\bar{x} = 13.625$ or 13.6
• M1: Correct formula for $s^2$ using $n - 1 = 7$ in denominator
• A1: $s^2 = 7.125$ or 7.13

Common mistake: Using $n = 8$ instead of $n - 1 = 7$ gives $\frac{49.875}{8} = 6.23$, which is the biased estimator. This loses the A1 mark.

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Question 2 [4 marks]

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

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

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

$= 77520 \times 0.00064339... \times 0.005479...$

$= 0.164 \text{ (to 3 s.f.)}$

Marking: M1 for correct binomial probability formula with $n=20, p=0.35, r=7$; A1 for answer 0.164.

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

Using calculator/binomial tables:

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

Computing each term:

• $\mathrm{P}(X=0) = (0.65)^{20} = 0.000182$
• $\mathrm{P}(X=1) = 20(0.35)(0.65)^{19} = 0.002098$
• $\mathrm{P}(X=2) = 190(0.35)^2(0.65)^{18} = 0.01157$
• $\mathrm{P}(X=3) = 1140(0.35)^3(0.65)^{17} = 0.03834$
• $\mathrm{P}(X=4) = 4845(0.35)^4(0.65)^{16} = 0.08918$

$\mathrm{P}(X \leq 4) = 0.1414$

$\mathrm{P}(X \geq 5) = 1 - 0.1414 = 0.859 \text{ (to 3 s.f.)}$

Marking: M1 for using complement $1 - \mathrm{P}(X \leq 4)$; A1 for answer 0.859.

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Question 3 [4 marks]

(a) For a valid PDF, $\int_{-\infty}^{\infty} f(x)\,dx = 1$:

$\int_0^4 kx(4-x)\,dx = 1$

$k\int_0^4 (4x - x^2)\,dx = 1$

$k\left[2x^2 - \frac{x^3}{3}\right]_0^4 = 1$

$k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = 1$

$k\left[32 - \frac{64}{3}\right] = 1$

$k\left[\frac{96 - 64}{3}\right] = 1$

$k \times \frac{32}{3} = 1$

$k = \frac{3}{32} \quad \text{✓ shown}$

Marking: M1 for setting up the integral equal to 1; M1 for correct integration; A1 for showing $k = \frac{3}{32}$.

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

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

$= \frac{3}{32}\left[\frac{4(64)}{3} - \frac{256}{4}\right]$

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

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

$= \frac{3}{32} \times \frac{64}{3} = \frac{64}{32} = 2$

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

Marking: M1 for correct expectation integral setup; M1 for correct integration; A1 for answer 2.

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Question 4 [5 marks]

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

(a) $\mathrm{P}(38 < X < 47)$

Standardise: $Z = \dfrac{X - 42}{5}$

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

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

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

Answer: 0.629 (to 3 s.f.)

Marking: M1 for standardising; M1 for using correct probability expression; A1 for answer 0.629.

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

Expected number in 200 plants: $200 \times 0.0548 = 10.96$

Answer: 11 plants (to nearest whole number)

Marking: M1 for finding $\mathrm{P}(X > 50)$; M1 for multiplying by 200; A1 for answer 11.

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Question 5 [6 marks]

Step 1: State hypotheses

$H_0: \mu = 6.5$ (mean daily screen time is 6.5 hours) $H_1: \mu \neq 6.5$ (mean daily screen time differs from 6.5 hours)

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

Step 2: Test statistic

Since $n = 50$ is large, by CLT we use the $z$-test:

$z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{7.2 - 6.5}{2.1 / \sqrt{50}} = \frac{0.7}{0.29698} = 2.357$

Step 3: Critical value / p-value

For a two-tailed test at 5%, critical values are $z = \pm 1.96$.

Since $2.357 > 1.96$, we reject $H_0$.

Alternatively, p-value $= 2 \times \mathrm{P}(Z > 2.357) = 2 \times (1 - 0.9908) = 2 \times 0.0092 = 0.0184$

Since $0.0184 < 0.05$, we reject $H_0$.

Step 4: Conclusion

There is sufficient evidence at the 5% significance level to conclude that the mean daily screen time of teenagers differs from 6.5 hours.

Marking:

• M1: Correct hypotheses stated (two-tailed)
• M1: Correct test statistic formula and substitution
• A1: $z = 2.36$ (to 3 s.f.)
• M1: Comparison with critical value or p-value comparison with 0.05
• A1: Correct decision (reject $H_0$)
• B1: Conclusion in context

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Question 6 [7 marks]

(a) Calculate the mean:

Midpoints: 9.5, 29.5, 49.5, 69.5, 90

Class	Midpoint $m$	Frequency $f$	$fm$
0–19	9.5	4	38
20–39	29.5	10	295
40–59	49.5	18	891
60–79	69.5	16	1112
80–100	90	12	1080

$\bar{x} = \frac{\sum fm}{\sum f} = \frac{38 + 295 + 891 + 1112 + 1080}{60} = \frac{3416}{60} = 56.93$

Answer: Mean = 56.9 (to 3 s.f.)

Marking: M1 for using midpoints; M1 for correct calculation; A1 for answer 56.9.

(b) The modal class is 40–59 (highest frequency of 18).

Marking: B1 for correct modal class.

(c) Histogram:

Frequency density = frequency ÷ class width. All class widths are 20.

Class	Frequency density
0–19	4/20 = 0.20
20–39	10/20 = 0.50
40–59	18/20 = 0.90
60–79	16/20 = 0.80
80–100	12/20 = 0.60

Image pending generation: chart for Q6(c).

Marking: M1 for calculating frequency densities; M1 for drawing bars with correct heights; A1 for fully correct histogram with labels.

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Section B: Probability & Distributions (30 marks)

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Question 7 [8 marks]

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

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

Marking: M1 for using combinations; A1 for $\frac{1}{22}$ or 0.0455.

(b) $\mathrm{P}(\text{2 red, 1 blue}) = \frac{\binom{5}{2} \times \binom{4}{1}}{\binom{12}{3}} = \frac{10 \times 4}{220} = \frac{40}{220} = \frac{2}{11} = 0.182$

Marking: M1 for correct numerator (selecting 2 red from 5 AND 1 blue from 4); A1 for $\frac{2}{11}$ or 0.182.

(c) $\mathrm{P}(\text{all different colours}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11} = 0.273$

Marking: M1 for selecting 1 of each colour; A1 for $\frac{3}{11}$ or 0.273.

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Question 8 [8 marks]

$X \sim \mathrm{Po}(4.2)$ where $X$ = number of emails per hour.

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

$(4.2)^5 = 1306.91232$

$5! = 120$

$\mathrm{P}(X = 5) = \frac{e^{-4.2} \times 1306.91232}{120} = \frac{0.014996 \times 1306.91232}{120} = \frac{19.594}{120} = 0.1633$

Answer: 0.163 (to 3 s.f.)

Marking: M1 for correct Poisson formula; A1 for answer 0.163.

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

$\mathrm{P}(X = 0) = e^{-4.2} = 0.014996$

$\mathrm{P}(X = 1) = 4.2 \times e^{-4.2} = 0.06298$

$\mathrm{P}(X = 2) = \frac{(4.2)^2}{2} \times e^{-4.2} = \frac{17.64}{2} \times 0.014996 = 8.82 \times 0.014996 = 0.13227$

$\mathrm{P}(X \leq 2) = 0.014996 + 0.06298 + 0.13227 = 0.21025$

$\mathrm{P}(X \geq 3) = 1 - 0.21025 = 0.790$

Answer: 0.790 (to 3 s.f.)

Marking: M1 for using complement; M1 for computing individual probabilities; A1 for answer 0.790.

(c) For 30 minutes, mean = $4.2 \times 0.5 = 2.1$. Let $Y \sim \mathrm{Po}(2.1)$.

$\mathrm{P}(Y < 2) = \mathrm{P}(Y = 0) + \mathrm{P}(Y = 1)$

$\mathrm{P}(Y = 0) = e^{-2.1} = 0.12246$

$\mathrm{P}(Y = 1) = 2.1 \times e^{-2.1} = 0.25716$

$\mathrm{P}(Y < 2) = 0.12246 + 0.25716 = 0.380$

Answer: 0.380 (to 3 s.f.)

Marking: M1 for halving the mean; M1 for computing $\mathrm{P}(Y=0) + \mathrm{P}(Y=1)$; A1 for answer 0.380.

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Question 9 [9 marks]

$X \sim \mathrm{N}(150, \sigma^2)$

(a) $\mathrm{P}(X > 165) = 0.08$

Standardising: $\mathrm{P}\left(Z > \frac{165 - 150}{\sigma}\right) = 0.08$

$\mathrm{P}\left(Z < \frac{15}{\sigma}\right) = 0.92$

From tables, $\Phi(1.405) \approx 0.92$, so:

$\frac{15}{\sigma} = 1.405$

$\sigma = \frac{15}{1.405} = 10.68$

Answer: $\sigma = 10.7$ g (to 3 s.f.)

Marking: M1 for standardising; M1 for using $\Phi^{-1}(0.92) \approx 1.405$; A1 for $\sigma = 10.7$.

(b) $\mathrm{P}(X < 130) = \mathrm{P}\left(Z < \frac{130 - 150}{10.68}\right) = \mathrm{P}(Z < -1.873) = 1 - \Phi(1.873) = 1 - 0.9695 = 0.0305$

Answer: 0.0305 (to 3 s.f.)

Marking: M1 for standardising with found $\sigma$; A1 for answer 0.0305.

(c) Let $W$ = number of "small" apples in sample of 10. $W \sim \mathrm{B}(10, 0.0305)$.

$\mathrm{P}(W \geq 2) = 1 - \mathrm{P}(W = 0) - \mathrm{P}(W = 1)$

$\mathrm{P}(W = 0) = (0.9695)^{10} = 0.7374$

$\mathrm{P}(W = 1) = 10 \times 0.0305 \times (0.9695)^9 = 10 \times 0.0305 \times 0.7606 = 0.2320$

$\mathrm{P}(W \geq 2) = 1 - 0.7374 - 0.2320 = 0.0306$

Answer: 0.0306 (to 3 s.f.)

Marking: M1 for identifying binomial with $n=10, p=0.0305$; M1 for using complement; A1 for answer 0.0306.

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Question 10 [6 marks]

Let $X$ = number of sixes in 4 rolls. $X \sim \mathrm{B}(4, \frac{1}{6})$.

(a) $\mathrm{P}(X = 2) = \binom{4}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216} = 0.1157$

Answer: 0.116 (to 3 s.f.)

Marking: M1 for correct binomial formula; A1 for answer 0.116.

(b) $\mathrm{P}(X \geq 1) = 1 - \mathrm{P}(X = 0) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} = 0.5177$

Answer: 0.518 (to 3 s.f.)

Marking: M1 for using complement; A1 for answer 0.518.

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Question 11 [7 marks]

$T \sim \mathrm{Exp}(\lambda)$ with mean $\mathrm{E}(T) = \frac{1}{\lambda} = 8000$, so $\lambda = \frac{1}{8000} = 0.000125$.

(a) $f(t) = \lambda e^{-\lambda t} = 0.000125\, e^{-0.000125t}$ for $t \geq 0$

Marking: B1 for correct PDF with correct $\lambda$.

(b) $\mathrm{P}(T > 10000) = e^{-0.000125 \times 10000} = e^{-1.25} = 0.2865$

Answer: 0.287 (to 3 s.f.)

Marking: M1 for using survival function of exponential; A1 for answer 0.287.

(c) Let $Y$ = number of bulbs (out of 5) lasting more than 10,000 hours.

$p = 0.2865$, $Y \sim \mathrm{B}(5, 0.2865)$

$\mathrm{P}(Y = 3) = \binom{5}{3}(0.2865)^3(1 - 0.2865)^2 = 10 \times 0.02352 \times 0.5091 = 0.1197$

Answer: 0.120 (to 3 s.f.)

Marking: M1 for identifying binomial; M1 for correct substitution; A1 for answer 0.120.

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Question 12 [6 marks]

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

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

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

Answer: $\mathrm{P}(A \cap B) = 0.24$

Marking: M1 for correct addition rule; A1 for answer 0.24.

(b) If independent: $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24$

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

Marking: M1 for computing $\mathrm{P}(A) \times \mathrm{P}(B)$; A1 for correct conclusion with justification.

(c) $\mathrm{P}(A' \mid B) = \frac{\mathrm{P}(A' \cap B)}{\mathrm{P}(B)}$

Since $A$ and $B$ are independent, $\mathrm{P}(A' \mid B) = \mathrm{P}(A') = 1 - 0.6 = 0.4$

Alternatively: $\mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.4 - 0.24 = 0.16$

$\mathrm{P}(A' \mid B) = \frac{0.16}{0.4} = 0.4$

Answer: $\mathrm{P}(A' \mid B) = 0.4$

Marking: M1 for correct conditional probability formula or independence argument; A1 for answer 0.4.

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Question 13 [5 marks]

Let $X$ = number of calls in 1 hour. $X \sim \mathrm{Po}(2.5 \times 60) = \mathrm{Po}(150)$.

Since $\lambda = 150$ is large, use normal approximation:

$X \approx \mathrm{N}(150, 150)$

With continuity correction:

$\mathrm{P}(X < 130) = \mathrm{P}(X \leq 129) \approx \mathrm{P}\left(Z < \frac{129.5 - 150}{\sqrt{150}}\right)$

$= \mathrm{P}\left(Z < \frac{-20.5}{12.247}\right) = \mathrm{P}(Z < -1.674)$

$= 1 - \Phi(1.674) = 1 - 0.9530 = 0.0470$

Answer: 0.0470 (to 3 s.f.)

Marking:

• M1: Correct Poisson mean $\lambda = 150$
• M1: Normal approximation $X \approx \mathrm{N}(150, 150)$
• M1: Continuity correction (using 129.5)
• A1: Correct $z$-value
• A1: Final answer 0.0470

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Question 14 [8 marks]

Class	Frequency	Cumulative frequency
0–9	6	6
10–19	14	20
20–29	22	42
30–39	20	62
40–49	12	74
50–59	6	80

$n = 80$

(a) Median position = $\frac{n}{2} = \frac{80}{2} = 40$th value.

The 40th value lies in the class 20–29 (cumulative frequency reaches 42).

Using linear interpolation:

$\text{Median} = 20 + \frac{40 - 20}{22} \times 10 = 20 + \frac{20}{22} \times 10 = 20 + 9.09 = 29.09$

Answer: Median ≈ 29.1 minutes (to 3 s.f.)

Marking: M1 for identifying median class; M1 for interpolation formula; A1 for answer 29.1.

(b) Lower quartile $Q_1$: position = $\frac{n}{4} = \frac{80}{4} = 20$th value.

The 20th value lies at the upper boundary of class 10–19 (cumulative frequency = 20).

$Q_1 = 19.5 \text{ (or by interpolation: } 10 + \frac{20 - 6}{14} \times 10 = 10 + 10 = 20\text{)}$

Using interpolation: $Q_1 = 10 + \frac{20 - 6}{14} \times 10 = 10 + 10 = 20.0$

Upper quartile $Q_3$: position = $\frac{3n}{4} = \frac{3 \times 80}{4} = 60$th value.

The 60th value lies in class 30–39 (cumulative frequency reaches 62).

$Q_3 = 30 + \frac{60 - 42}{20} \times 10 = 30 + \frac{18}{20} \times 10 = 30 + 9 = 39.0$

$\mathrm{IQR} = Q_3 - Q_1 = 39.0 - 20.0 = 19.0$

Answer: IQR = 19.0 minutes

Marking: M1 for identifying $Q_1$ and $Q_3$ classes; M1 for interpolation; A1 for $Q_1$; A1 for $Q_3$; A1 for IQR = 19.0.

(c) The distribution is approximately symmetric (or very slightly right-skewed). The median (29.1) is roughly in the middle of the range, and the frequencies rise to a central peak at 20–29 then decrease in a similar pattern. $Q_3 - \text{median} = 39.0 - 29.1 = 9.9$ and $\text{median} - Q_1 = 29.1 - 20.0 = 9.1$, which are approximately equal, suggesting approximate symmetry.

Marking: B1 for stating shape; B1 for justification using quartiles or frequency pattern.

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Question 15 [7 marks]

$p = 0.05$ (probability of defective). Let $X$ = number of tests until first defective. $X \sim \mathrm{Geometric}(p = 0.05)$.

(a) $\mathrm{P}(X = 5) = (1 - p)^{4} \times p = (0.95)^4 \times 0.05 = 0.8145 \times 0.05 = 0.0407$

Answer: 0.0407 (to 3 s.f.)

Marking: M1 for geometric distribution formula; A1 for answer 0.0407.

(b) $\mathrm{E}(X) = \frac{1}{p} = \frac{1}{0.05} = 20$

Answer: Expected number of tests = 20

Marking: B1 for correct formula and answer.

(c) Let $Y$ = number of defectives in 100 components. $Y \sim \mathrm{B}(100, 0.05)$.

Using Poisson approximation with $\lambda = np = 100 \times 0.05 = 5$:

$\mathrm{P}(Y = 3) \approx \frac{e^{-5}(5^3)}{3!} = \frac{0.006738 \times 125}{6} = \frac{0.84225}{6} = 0.1404$

Answer: 0.140 (to 3 s.f.)

Marking: M1 for identifying Poisson approximation with $\lambda = 5$; M1 for correct Poisson formula; A1 for answer 0.140.

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Question 16 [6 marks]

$\mathrm{P}(X = x, Y = y) = \frac{x + y}{30}$, for $x = 1, 2, 3$ and $y = 1, 2, 3$.

(a) $\mathrm{P}(X = 2, Y = 3) = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$

Answer: $\frac{1}{6}$ or 0.167

Marking: B1 for correct substitution.

(b) $\mathrm{P}(X = 2) = \sum_{y=1}^{3} \mathrm{P}(X = 2, Y = y)$

$= \frac{2+1}{30} + \frac{2+2}{30} + \frac{2+3}{30} = \frac{3}{30} + \frac{4}{30} + \frac{5}{30} = \frac{12}{30} = \frac{2}{5}$

Answer: $\frac{2}{5}$ or 0.4

Marking: M1 for summing over all $y$ values; A1 for answer $\frac{2}{5}$.

(c) First find the full marginal distribution of $X$:

$\mathrm{P}(X = 1) = \frac{1+1}{30} + \frac{1+2}{30} + \frac{1+3}{30} = \frac{2+3+4}{30} = \frac{9}{30} = \frac{3}{10}$

$\mathrm{P}(X = 2) = \frac{12}{30} = \frac{2}{5}$ (from part b)

$\mathrm{P}(X = 3) = \frac{3+1}{30} + \frac{3+2}{30} + \frac{3+3}{30} = \frac{4+5+6}{30} = \frac{15}{30} = \frac{1}{2}$

Check: $\frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{12}{10}$... Let me recheck.

$\mathrm{P}(X=1) = \frac{9}{30}$, $\mathrm{P}(X=2) = \frac{12}{30}$, $\mathrm{P}(X=3) = \frac{15}{30}$

Sum: $\frac{9+12+15}{30} = \frac{36}{30} = \frac{6}{5} \neq 1$

Wait — let me verify the total probability over all 9 cells:

$\sum_{x=1}^{3}\sum_{y=1}^{3} \frac{x+y}{30} = \frac{1}{30}\sum_{x=1}^{3}\sum_{y=1}^{3}(x+y)$

For each $x$: $\sum_{y=1}^3(x+y) = 3x + (1+2+3) = 3x + 6$

Total: $\sum_{x=1}^3(3x+6) = (3+6)+(6+6)+(9+6) = 9+12+15 = 36$

So total probability = $\frac{36}{30} = \frac{6}{5} > 1$. This is not a valid joint probability distribution as stated.

Correction for the question: The distribution should be $\mathrm{P}(X = x, Y = y) = \frac{x + y}{36}$ for the probabilities to sum to 1.

With the corrected denominator of 36:

(a) $\mathrm{P}(X = 2, Y = 3) = \frac{2 + 3}{36} = \frac{5}{36}$

(b) $\mathrm{P}(X = 2) = \frac{3}{36} + \frac{4}{36} + \frac{5}{36} = \frac{12}{36} = \frac{1}{3}$

(c) $\mathrm{P}(X = 1) = \frac{2+3+4}{36} = \frac{9}{36} = \frac{1}{4}$

$\mathrm{P}(X = 2) = \frac{12}{36} = \frac{1}{3}$

$\mathrm{P}(X = 3) = \frac{4+5+6}{36} = \frac{15}{36} = \frac{5}{12}$

Check: $\frac{9+12+15}{36} = \frac{36}{36} = 1$ ✓

$\mathrm{E}(X) = 1 \times \frac{9}{36} + 2 \times \frac{12}{36} + 3 \times \frac{15}{36} = \frac{9 + 24 + 45}{36} = \frac{78}{36} = \frac{13}{6} = 2.167$

Answer: $\mathrm{E}(X) = \frac{13}{6}$ or 2.17 (to 3 s.f.)

Marking:

• M1: Correcting the denominator to 36 (or noting the distribution must sum to 1)
• M1: Finding marginal probabilities by summing over $y$
• A1: Correct marginal probabilities
• M1: Using $\mathrm{E}(X) = \sum x \cdot \mathrm{P}(X=x)$
• A1: $\mathrm{E}(X) = \frac{13}{6}$

Note to student: Always verify that a joint probability distribution sums to 1 over all possible values. If it doesn't, there may be an error in the question or the normalising constant.

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Question 17 [5 marks]

$X \sim \mathrm{N}(\mu, \sigma^2)$

$\mathrm{P}(X < 25) = 0.1587$

From standard normal tables, $\Phi(-1.00) = 0.1587$, so:

$\frac{25 - \mu}{\sigma} = -1.00 \quad \Rightarrow \quad 25 - \mu = -\sigma \quad \Rightarrow \quad \mu - \sigma = 25 \quad \text{...(i)}$

$\mathrm{P}(X > 45) = 0.0228$

$\mathrm{P}(X < 45) = 1 - 0.0228 = 0.9772$

From tables, $\Phi(2.00) = 0.9772$, so:

$\frac{45 - \mu}{\sigma} = 2.00 \quad \Rightarrow \quad 45 - \mu = 2\sigma \quad \text{...(ii)}$

From (i): $\mu = 25 + \sigma$

Substitute into (ii): $45 - (25 + \sigma) = 2\sigma$

$20 - \sigma = 2\sigma$

$20 = 3\sigma$

$\sigma = \frac{20}{3} = 6.667$

$\mu = 25 + \frac{20}{3} = \frac{75 + 20}{3} = \frac{95}{3} = 31.67$

Answer: $\mu = 31.7$, $\sigma = 6.67$ (to 3 s.f.)

Marking:

• M1: Converting to $z$-scores using standard normal table values
• A1: Correct $z$-values (−1.00 and 2.00)
• M1: Setting up simultaneous equations
• M1: Solving the equations
• A1: $\mu = 31.7$, $\sigma = 6.67$

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Question 18 [7 marks]

Sample space for sum of two dice: 36 outcomes.

Sum	Outcomes	Count
2	(1,1)	1
3	(1,2),(2,1)	2
4	(1,3),(2,2),(3,1)	3
5	(1,4),(2,3),(3,2),(4,1)	4
6	(1,5),(2,4),(3,3),(4,2),(5,1)	5
7	(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)	6
8	(2,6),(3,5),(4,4),(5,3),(6,2)	5
9	(3,6),(4,5),(5,4),(6,3)	4
10	(4,6),(5,5),(6,4)	3
11	(5,6),(6,5)	2
12	(6,6)	1

(a) $\mathrm{P}(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$

Marking: B1 for correct probability.

(b) $\mathrm{P}(\text{sum} > 9) = \mathrm{P}(\text{sum} = 10, 11, \text{or } 12) = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$

Marking: B1 for correct probability.

(c) Let $W$ = winnings.

Outcome	Winnings	Probability
Sum = 7	$10	$\frac{6}{36}$
Sum > 9	$5	$\frac{6}{36}$
Otherwise	−$3	$\frac{24}{36}$

$\mathrm{P}(\text{otherwise}) = 1 - \frac{6}{36} - \frac{6}{36} = \frac{24}{36} = \frac{2}{3}$

$\mathrm{E}(W) = 10 \times \frac{6}{36} + 5 \times \frac{6}{36} + (-3) \times \frac{24}{36}$

$= \frac{60}{36} + \frac{30}{36} - \frac{72}{36} = \frac{18}{36} = 0.50$

Answer: Expected winnings = $0.50 per game

Marking: M1 for identifying all three outcomes and probabilities; M1 for correct expectation formula; A1 for answer $0.50.

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Question 19 [7 marks]

$n = 10$, $\sum x = 156$, $\sum x^2 = 2478$

(a) Unbiased estimate of mean:

$\bar{x} = \frac{\sum x}{n} = \frac{156}{10} = 15.6$

Unbiased estimate of variance:

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(2478 - \frac{156^2}{10}\right)$

$= \frac{1}{9}\left(2478 - \frac{24336}{10}\right) = \frac{1}{9}(2478 - 2433.6) = \frac{1}{9}(44.4) = 4.933$

Answer: $\bar{x} = 15.6$, $s^2 = 4.93$ (to 3 s.f.)

Marking: M1 for correct mean; M1 for correct variance formula (using $n-1$); A1 for $\bar{x} = 15.6$; A1 for $s^2 = 4.93$.

(b) 95% confidence interval for $\mu$:

Since $\sigma$ is unknown and $n = 10$ is small, use $t$-distribution with $n - 1 = 9$ degrees of freedom.

$t_{0.025, 9} = 2.262$

$\text{CI} = \bar{x} \pm t_{0.025,9} \times \frac{s}{\sqrt{n}} = 15.6 \pm 2.262 \times \frac{\sqrt{4.933}}{\sqrt{10}}$

$= 15.6 \pm 2.262 \times \frac{2.221}{3.162} = 15.6 \pm 2.262 \times 0.7024$

$= 15.6 \pm 1.589$

$= (14.01, 17.19)$

Answer: 95% CI = $(14.0, 17.2)$ (to 3 s.f.)

Marking: M1 for using $t$-distribution with 9 d.f.; M1 for correct critical value 2.262; M1 for correct standard error; A1 for correct interval.

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Question 20 [6 marks]

$F(x) = \begin{cases} 0 & x < 0 \\ \dfrac{x^3}{64} & 0 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$

(a) $f(x) = F'(x)$

For $0 \leq x \leq 4$: $f(x) = \frac{d}{dx}\left(\frac{x^3}{64}\right) = \frac{3x^2}{64}$

$f(x) = \begin{cases} \dfrac{3x^2}{64} & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$

Marking: M1 for differentiating $F(x)$; A1 for correct PDF.

(b) $\mathrm{P}(1 < X < 3) = F(3) - F(1) = \frac{27}{64} - \frac{1}{64} = \frac{26}{64} = \frac{13}{32} = 0.40625$

Answer: 0.406 (to 3 s.f.)

Marking: M1 for using $F(3) - F(1)$; A1 for answer 0.406.

(c) Median $m$ satisfies $F(m) = 0.5$:

$\frac{m^3}{64} = 0.5$

$m^3 = 32$

$m = \sqrt[3]{32} = 2\sqrt[3]{4} = 3.1748$

Answer: Median = 3.17 (to 3 s.f.)

Marking: M1 for setting $F(m) = 0.5$; M1 for solving $m^3 = 32$; A1 for answer 3.17.

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Mark Summary

Section	Marks
Section A (Questions 1–6)	30
Section B (Questions 7–20)	30
Total	60