A Level H1 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key — Statistics & Probability (Version 3 of 5)

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Question 1 [3]

Unbiased estimate of the population mean:

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

Unbiased estimate of the population variance:

First calculate $\sum(x_i - \bar{x})^2$:

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

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

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

Answer: Unbiased estimate of mean $= 13.625$, unbiased estimate of variance $= 7.125$

[1] for correct mean, [1] for correct sum of squared deviations or correct method, [1] for correct variance using $n-1$ denominator.

Common mistake: Using $n = 8$ in the denominator instead of $n-1 = 7$. This gives the biased sample variance, not the unbiased estimate of the population variance.

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Question 2 [2]

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

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

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

$= 77520 \times 0.00064339297 \times 0.0056880009$

$\approx 0.283$

Answer: $\mathrm{P}(X = 7) \approx 0.283$

[1] for correct binomial formula with correct substitution, [1] for correct numerical answer to 3 s.f.

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Question 3 [3]

Probability of rolling a number greater than 4 on a single roll: $\mathrm{P}(\text{> 4}) = \mathrm{P}(5 \text{ or } 6) = \frac{2}{6} = \frac{1}{3}$

Let $Y$ be the number of rolls (out of 4) showing a number greater than 4. Then $Y \sim \mathrm{B}\!\left(4, \frac{1}{3}\right)$.

$\mathrm{P}(Y = 2) = \binom{4}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2 = 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{24}{81} = \frac{8}{27}$

$\approx 0.296$

Answer: $\dfrac{8}{27}$ or $0.296$

[1] for identifying $p = \frac{1}{3}$, [1] for correct binomial expression, [1] for correct final answer.

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Question 4(a) [2]

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

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

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

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

$\approx 0.629$

Answer: $0.629$

[1] for correct standardisation, [1] for correct answer to 3 s.f.

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Question 4(b) [3]

From part (a), $p = 0.6294$ is the probability that a single plant has height between 38 cm and 47 cm.

Let $Y$ be the number of plants (out of 5) with heights in this range. $Y \sim \mathrm{B}(5, 0.6294)$.

$\mathrm{P}(Y \ge 4) = \mathrm{P}(Y = 4) + \mathrm{P}(Y = 5)$

$\mathrm{P}(Y = 4) = \binom{5}{4}(0.6294)^4(0.3706)^1 = 5 \times 0.1569 \times 0.3706 \approx 0.2907$

$\mathrm{P}(Y = 5) = (0.6294)^5 \approx 0.0987$

$\mathrm{P}(Y \ge 4) \approx 0.2907 + 0.0987 = 0.389$

Answer: $0.389$

[1] for identifying the binomial distribution with correct $p$ from (a), [1] for correct calculation of $\mathrm{P}(Y=4)$ and $\mathrm{P}(Y=5)$, [1] for correct final answer.

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Question 5(a) [1]

Since probabilities sum to 1:

$0.2 + 0.3 + a + 0.1 = 1$ $a = 0.4$

Answer: $a = 0.4$

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Question 5(b) [3]

$\mathrm{E}(Y) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4$

$\mathrm{E}(Y^2) = 1^2(0.2) + 2^2(0.3) + 3^2(0.4) + 4^2(0.1) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6$

$\mathrm{Var}(Y) = \mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2 = 6.6 - (2.4)^2 = 6.6 - 5.76 = 0.84$

Answer: $\mathrm{E}(Y) = 2.4$, $\mathrm{Var}(Y) = 0.84$

[1] for correct $\mathrm{E}(Y)$, [1] for correct $\mathrm{E}(Y^2)$, [1] for correct $\mathrm{Var}(Y)$.

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Question 6(a) [2]

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

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

$= k\left[\left(3(36) - \frac{216}{3}\right) - 0\right] = k\left[108 - 72\right] = 36k$

Setting $36k = 1$ gives $k = \dfrac{1}{36}$. $\quad \blacksquare$

[1] for correct integration, [1] for showing $k = \frac{1}{36}$.

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Question 6(b) [2]

$\mathrm{E}(X) = \int_0^6 x \cdot \frac{1}{36}x(6-x)\,dx = \frac{1}{36}\int_0^6 (6x^2 - x^3)\,dx$

$= \frac{1}{36}\left[2x^3 - \frac{x^4}{4}\right]_0^6 = \frac{1}{36}\left[\left(2(216) - \frac{1296}{4}\right) - 0\right]$

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

Answer: $\mathrm{E}(X) = 3$

[1] for correct integral setup, [1] for correct evaluation.

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Question 7(a) [2]

Let $X \sim \mathrm{B}(15, 0.3)$.

$\mathrm{P}(X = 5) = \binom{15}{5}(0.3)^5(0.7)^{10} = 3003 \times 0.00243 \times 0.02825 \approx 0.206$

Answer: $0.206$

[1] for correct binomial expression, [1] for correct answer.

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Question 7(b) [3]

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

$\mathrm{P}(X=0) = (0.7)^{15} \approx 0.004748$

$\mathrm{P}(X=1) = 15(0.3)(0.7)^{14} \approx 15 \times 0.3 \times 0.006782 \approx 0.030520$

$\mathrm{P}(X=2) = \binom{15}{2}(0.3)^2(0.7)^{13} = 105 \times 0.09 \times 0.009689 \approx 0.091560$

$\mathrm{P}(X \le 2) \approx 0.004748 + 0.030520 + 0.091560 = 0.126828$

$\mathrm{P}(X \ge 3) \approx 1 - 0.126828 = 0.873$

Answer: $0.873$

[1] for using the complement method, [1] for correct individual probabilities, [1] for correct final answer.

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Question 8(a) [2]

$\mathrm{P}(X < 65) = 0.08$ gives $\dfrac{65 - \mu}{\sigma} = z_{0.08}$ where $\Phi(z_{0.08}) = 0.08$.

From tables, $z_{0.08} \approx -1.405$ (since $\Phi(-1.405) \approx 0.08$).

$\frac{65 - \mu}{\sigma} = -1.405 \quad \Rightarrow \quad 65 - \mu = -1.405\sigma \quad \cdots (1)$

$\mathrm{P}(X > 82) = 0.15$ gives $\mathrm{P}(X < 82) = 0.85$, so $\dfrac{82 - \mu}{\sigma} = z_{0.85} \approx 1.036$.

$\frac{82 - \mu}{\sigma} = 1.036 \quad \Rightarrow \quad 82 - \mu = 1.036\sigma \quad \cdots (2)$

Answer: Equations are $65 - \mu = -1.405\sigma$ and $82 - \mu = 1.036\sigma$

[1] for each correct equation.

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Question 8(b) [3]

Subtracting equation (1) from equation (2):

$(82 - \mu) - (65 - \mu) = 1.036\sigma - (-1.405\sigma)$ $17 = 2.441\sigma$ $\sigma = \frac{17}{2.441} \approx 6.96$

From equation (1): $\mu = 65 + 1.405\sigma = 65 + 1.405(6.96) = 65 + 9.78 \approx 74.8$

Answer: $\mu \approx 74.8$, $\sigma \approx 6.96$

[1] for correct $\sigma$, [1] for correct $\mu$, [1] for both to 3 s.f.

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Question 9(a) [2]

Unbiased estimate of variance of $x$:

$s_x^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(412.36 - \frac{(62.4)^2}{10}\right)$

$= \frac{1}{9}\left(412.36 - \frac{3893.76}{10}\right) = \frac{1}{9}(412.36 - 389.376) = \frac{1}{9}(22.984) = 2.5538$

$\approx 2.55$

Answer: $2.55$

[1] for correct formula, [1] for correct answer.

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Question 9(b) [3]

$S_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 338.7 - \frac{62.4 \times 58}{10} = 338.7 - 361.92 = -23.22$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 412.36 - 389.376 = 22.984$

$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 362 - \frac{58^2}{10} = 362 - 336.4 = 25.6$

$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{-23.22}{\sqrt{22.984 \times 25.6}} = \frac{-23.22}{\sqrt{588.390}} = \frac{-23.22}{24.257}$

$\approx -0.957$

Answer: $r \approx -0.957$

[1] for correct $S_{xy}$, [1] for correct $S_{xx}$ and $S_{yy}$, [1] for correct $r$.

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Question 9(c) [1]

There is a strong negative linear correlation between daily screen time and sleep quality score. This means that as daily screen time increases, sleep quality tends to decrease.

[1] for correct interpretation mentioning strong negative correlation in context.

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Question 10(a) [2]

Let $X \sim \mathrm{Po}(4.2)$.

$\mathrm{P}(X = 6) = \frac{e^{-4.2}(4.2)^6}{6!} = \frac{e^{-4.2} \times 5489.031744}{720}$

$e^{-4.2} \approx 0.014996$

$\mathrm{P}(X = 6) = \frac{0.014996 \times 5489.032}{720} = \frac{82.305}{720} \approx 0.114$

Answer: $0.114$

[1] for correct Poisson formula, [1] for correct answer.

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Question 10(b) [3]

For a 30-second interval, the mean is $\lambda = 4.2 \times 0.5 = 2.1$.

Let $Y \sim \mathrm{Po}(2.1)$.

$\mathrm{P}(Y \ge 2) = 1 - \mathrm{P}(Y=0) - \mathrm{P}(Y=1)$

$\mathrm{P}(Y=0) = e^{-2.1} \approx 0.1225$

$\mathrm{P}(Y=1) = 2.1 \times e^{-2.1} \approx 2.1 \times 0.1225 = 0.2572$

$\mathrm{P}(Y \ge 2) = 1 - 0.1225 - 0.2572 = 0.620$

Answer: $0.620$

[1] for correct $\lambda = 2.1$, [1] for correct complement calculation, [1] for correct answer.

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Question 10(c) [1]

Answer: Calls occur independently (or at random), at a constant average rate, and the probability of more than one call in a very small time interval is negligible.

[1] for any valid assumption (independence or constant rate).

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Question 11(a) [3]

$X \sim \mathrm{N}(1200, 150^2)$

$\mathrm{P}(1000 < X < 1350) = \mathrm{P}\!\left(\frac{1000-1200}{150} < Z < \frac{1350-1200}{150}\right) = \mathrm{P}(-1.333 < Z < 1.0)$

$= \Phi(1.0) - \Phi(-1.333) = \Phi(1.0) - (1 - \Phi(1.333))$

$= 0.8413 - (1 - 0.9088) = 0.8413 - 0.0912 = 0.750$

Answer: $0.750$

[1] for correct standardisation, [1] for correct use of $\Phi$ values, [1] for correct answer.

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Question 11(b) [3]

We need $\mathrm{P}(X < t) = 0.03$.

From normal tables, $\Phi(z) = 0.03$ gives $z \approx -1.881$.

$\frac{t - 1200}{150} = -1.881$ $t = 1200 - 1.881 \times 150 = 1200 - 282.15 = 917.85$

$\approx 918 \text{ hours}$

Answer: $t \approx 918$ hours

[1] for correct $z$-value, [1] for correct equation, [1] for correct answer.

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Question 12(a) [2]

Total balls = 12. Number of ways to choose 3 from 12: $\binom{12}{3} = 220$.

Number of ways to choose 3 red from 5: $\binom{5}{3} = 10$.

$\mathrm{P}(\text{all red}) = \frac{10}{220} = \frac{1}{22}$

Answer: $\dfrac{1}{22}$

[1] for correct numerator and denominator, [1] for correct simplified answer.

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Question 12(b) [3]

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

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

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

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

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

Answer: $\dfrac{3}{44}$

[1] for recognising three cases, [1] for correct calculation of each case, [1] for correct final answer.

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Question 12(c) [3]

Exactly 2 red and 1 non-red:

Number of ways: $\binom{5}{2} \times \binom{7}{1} = 10 \times 7 = 70$

$\mathrm{P}(\text{exactly 2 red}) = \frac{70}{220} = \frac{7}{22}$

Answer: $\dfrac{7}{22}$

[1] for choosing 2 red from 5, [1] for choosing 1 non-red from 7, [1] for correct final answer.

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Question 13(a) [2]

$\mathrm{P}(X > 3) = 1 - \mathrm{P}(X \le 3) = 1 - F(3) = 1 - \frac{3^3}{125} = 1 - \frac{27}{125} = \frac{98}{125} = 0.784$

Answer: $0.784$

[1] for using $1 - F(3)$, [1] for correct answer.

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Question 13(b) [2]

$f(x) = F'(x) = \frac{d}{dx}\left(\frac{x^3}{125}\right) = \frac{3x^2}{125}, \quad 0 \le x \le 5$

Answer: $f(x) = \dfrac{3x^2}{125}$ for $0 \le x \le 5$, and $f(x) = 0$ otherwise.

[1] for correct differentiation, [1] for stating the domain.

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Question 13(c) [2]

The median $m$ satisfies $F(m) = 0.5$:

$\frac{m^3}{125} = 0.5$ $m^3 = 62.5$ $m = \sqrt[3]{62.5} \approx 3.97$

Answer: $m \approx 3.97$

[1] for setting $F(m) = 0.5$, [1] for correct answer.

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Question 14(a) [2]

$\mathrm{P}(\text{Online} \mid \text{Arts}) = \frac{62}{100} = 0.62$

Answer: $0.62$

[1] for correct conditional probability setup, [1] for correct answer.

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Question 14(b) [3]

$\mathrm{P}(\text{Science or In-Person}) = \mathrm{P}(\text{Science}) + \mathrm{P}(\text{In-Person}) - \mathrm{P}(\text{Science and In-Person})$

$= \frac{100}{200} + \frac{93}{200} - \frac{55}{200} = \frac{138}{200} = 0.69$

Answer: $0.69$

[1] for using inclusion-exclusion, [1] for correct values, [1] for correct answer.

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Question 14(c) [2]

$\mathrm{P}(\text{both online}) = \frac{107}{200} \times \frac{106}{199} = \frac{11342}{39800} \approx 0.285$

Answer: $0.285$

[1] for correct multiplication of conditional probabilities, [1] for correct answer.

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Question 15(a) [3]

$X \sim \mathrm{N}(25, \sigma^2)$. Given $\mathrm{P}(X > 30) = 0.20$.

$\mathrm{P}\!\left(Z > \frac{30 - 25}{\sigma}\right) = 0.20$

From tables, $\mathrm{P}(Z > 0.8416) = 0.20$.

$\frac{5}{\sigma} = 0.8416$ $\sigma = \frac{5}{0.8416} \approx 5.94$

Answer: $\sigma \approx 5.94$

[1] for correct standardisation, [1] for correct $z$-value, [1] for correct $\sigma$.

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Question 15(b) [3]

From part (a), $p = \mathrm{P}(\text{wait} > 30) = 0.20$.

Let $Y \sim \mathrm{B}(8, 0.20)$.

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

$\mathrm{P}(Y=0) = (0.8)^8 = 0.1678$

$\mathrm{P}(Y=1) = 8(0.2)(0.8)^7 = 8 \times 0.2 \times 0.2097 = 0.3355$

$\mathrm{P}(Y=2) = \binom{8}{2}(0.2)^2(0.8)^6 = 28 \times 0.04 \times 0.2621 = 0.2936$

$\mathrm{P}(Y \le 2) = 0.1678 + 0.3355 + 0.2936 = 0.797$

Answer: $0.797$

[1] for identifying $\mathrm{B}(8, 0.2)$, [1] for correct individual terms, [1] for correct sum.

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Question 15(c) [3]

For $n = 100$, $p = 0.20$: $\mu = np = 20$, $\sigma^2 = np(1-p) = 16$.

Using normal approximation $Y \sim \mathrm{N}(20, 16)$ with continuity correction:

$\mathrm{P}(Y > 25) = \mathrm{P}(Y \ge 25.5) = \mathrm{P}\!\left(Z > \frac{25.5 - 20}{4}\right) = \mathrm{P}(Z > 1.375)$

$= 1 - \Phi(1.375) = 1 - 0.9154 = 0.0846$

Answer: $0.0846$

[1] for correct normal approximation parameters, [1] for continuity correction, [1] for correct answer.

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Question 16(a) [3]

The possible sums range from 2 to 12. There are $6 \times 6 = 36$ equally likely outcomes.

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

[1] for correct range of $S$, [1] for correct enumeration of outcomes, [1] for correct probabilities.

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Question 16(b) [3]

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

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

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

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

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

$\mathrm{Var}(S) = 54.833 - 49 = 5.833 = \frac{35}{6}$

Answer: $\mathrm{E}(S) = 7$, $\mathrm{Var}(S) = \dfrac{35}{6} \approx 5.83$

[1] for correct $\mathrm{E}(S)$, [1] for correct $\mathrm{E}(S^2)$, [1] for correct $\mathrm{Var}(S)$.

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Question 16(c) [3]

$\mathrm{P}(S = 7) = \frac{6}{36} = \frac{1}{6}$. Let $W \sim \mathrm{B}\!\left(5, \frac{1}{6}\right)$.

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

$\mathrm{P}(W=0) = \left(\frac{5}{6}\right)^5 = \frac{3125}{7776} \approx 0.4019$

$\mathrm{P}(W=1) = 5 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^4 = 5 \times \frac{1}{6} \times \frac{625}{1296} = \frac{3125}{7776} \approx 0.4019$

$\mathrm{P}(W \ge 2) = 1 - 0.4019 - 0.4019 = 0.196$

Answer: $0.196$

[1] for correct $p = \frac{1}{6}$, [1] for correct binomial calculation, [1] for correct answer.

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Question 17(a) [2]

$\mathrm{E}(\bar{X}) = \mu = 12$

$\mathrm{Var}(\bar{X}) = \frac{\sigma^2}{n} = \frac{9}{36} = 0.25$

Answer: Mean $= 12$, Variance $= 0.25$

[1] for each correct value.

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Question 17(b) [3]

By CLT, $\bar{X} \sim \mathrm{N}(12, 0.25)$ approximately.

$\mathrm{P}(\bar{X} > 12.5) = \mathrm{P}\!\left(Z > \frac{12.5 - 12}{\sqrt{0.25}}\right) = \mathrm{P}(Z > 1.0) = 1 - \Phi(1.0) = 1 - 0.8413 = 0.159$

Answer: $0.159$

[1] for correct normal distribution for $\bar{X}$, [1] for correct standardisation, [1] for correct answer.

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Question 17(c) [1]

Answer: The sample size $n = 36$ is sufficiently large ($n \ge 30$) for the Central Limit Theorem to apply, so the distribution of $\bar{X}$ is approximately normal regardless of the underlying distribution of $X$.

[1] for mentioning large sample size / $n \ge 30$.

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Question 18(a) [3]

$n = 50$, $\bar{x} = 18.2$, $s^2 = 16.0$, so $s = 4.0$.

For a 95% confidence interval, $z_{0.025} = 1.96$.

$\text{CI} = \bar{x} \pm z_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 18.2 \pm 1.96 \times \frac{4.0}{\sqrt{50}}$

$= 18.2 \pm 1.96 \times 0.5657 = 18.2 \pm 1.109$

$= (17.09,\ 19.31)$

Answer: 95% CI is $(17.09,\ 19.31)$ hours

[1] for correct critical value, [1] for correct standard error, [1] for correct interval.

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Question 18(b) [1]

Answer: Since 20 hours lies outside the 95% confidence interval $(17.09,\ 19.31)$, the company's claim that the mean battery life is 20 hours is not supported by the data. It is unlikely that the true population mean is 20 hours.

[1] for correct comparison and conclusion.

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Question 18(c) [2]

Answer: Increasing the sample size $n$ decreases the standard error $\frac{\sigma}{\sqrt{n}}$, which makes the confidence interval narrower. A larger sample provides more information about the population, leading to a more precise estimate.

[1] for stating the interval becomes narrower, [1] for correct justification involving standard error.

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Question 19(a) [2]

$X \sim \mathrm{Po}(2.5)$

$\mathrm{P}(X = 3) = \frac{e^{-2.5}(2.5)^3}{3!} = \frac{0.082085 \times 15.625}{6} = \frac{1.2826}{6} \approx 0.214$

Answer: $0.214$

[1] for correct Poisson formula, [1] for correct answer.

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Question 19(b) [3]

$\mathrm{P}(\text{at least 2 in one week}) = 1 - \mathrm{P}(X=0) - \mathrm{P}(X=1)$

$\mathrm{P}(X=0) = e^{-2.5} \approx 0.082085$

$\mathrm{P}(X=1) = 2.5 \times e^{-2.5} \approx 0.205212$

$\mathrm{P}(X \ge 2) = 1 - 0.082085 - 0.205212 = 0.712703$

For two consecutive weeks (independent):

$\mathrm{P}(\text{both weeks have at least 2}) = (0.712703)^2 \approx 0.508$

Answer: $0.508$

[1] for $\mathrm{P}(X \ge 2)$ in one week, [1] for squaring (independence), [1] for correct answer.

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Question 19(c) [3]

For a two-week period, $Y \sim \mathrm{Po}(5.0)$.

$\mathrm{P}(Y < 4) = \mathrm{P}(Y=0) + \mathrm{P}(Y=1) + \mathrm{P}(Y=2) + \mathrm{P}(Y=3)$

$\mathrm{P}(Y=0) = e^{-5} = 0.006738$

$\mathrm{P}(Y=1) = 5e^{-5} = 0.033690$

$\mathrm{P}(Y=2) = \frac{25e^{-5}}{2} = 0.084224$

$\mathrm{P}(Y=3) = \frac{125e^{-5}}{6} = 0.140374$

$\mathrm{P}(Y < 4) = 0.006738 + 0.033690 + 0.084224 + 0.140374 = 0.265$

Answer: $0.265$

[1] for correct $\lambda = 5$, [1] for correct individual terms, [1] for correct sum.

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Question 20(a) [2]

$\int_0^3 \frac{2}{9}x\,dx = \frac{2}{9} \cdot \left[\frac{x^2}{2}\right]_0^3 = \frac{2}{9} \cdot \frac{9}{2} = 1$

Also $f(x) = \frac{2}{9}x \ge 0$ for $0 \le x \le 3$. Hence $f(x)$ is a valid PDF. $\quad \blacksquare$

[1] for showing the integral equals 1, [1] for noting non-negativity.

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Question 20(b) [3]

$\mathrm{E}(X) = \int_0^3 x \cdot \frac{2}{9}x\,dx = \frac{2}{9}\int_0^3 x^2\,dx = \frac{2}{9} \cdot \left[\frac{x^3}{3}\right]_0^3 = \frac{2}{9} \cdot 9 = 2$

$\mathrm{E}(X^2) = \int_0^3 x^2 \cdot \frac{2}{9}x\,dx = \frac{2}{9}\int_0^3 x^3\,dx = \frac{2}{9} \cdot \left[\frac{x^4}{4}\right]_0^3 = \frac{2}{9} \cdot \frac{81}{4} = \frac{162}{36} = 4.5$

$\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = 4.5 - 4 = 0.5$

Answer: $\mathrm{E}(X) = 2$, $\mathrm{Var}(X) = 0.5$

[1] for correct $\mathrm{E}(X)$, [1] for correct $\mathrm{E}(X^2)$, [1] for correct $\mathrm{Var}(X)$.

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Question 20(c) [2]

From part (b), $\mathrm{E}(X) = 2$.

$\mathrm{P}(X > 2) = \int_2^3 \frac{2}{9}x\,dx = \frac{2}{9} \cdot \left[\frac{x^2}{2}\right]_2^3 = \frac{2}{9} \cdot \left(\frac{9}{2} - \frac{4}{2}\right) = \frac{2}{9} \cdot \frac{5}{2} = \frac{10}{18} = \frac{5}{9}$

Answer: $\dfrac{5}{9}$ or $0.556$

[1] for correct integral setup, [1] for correct answer.

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End of Answer Key