A Level H2 Mathematics Statistics Probability Quiz

A-Level Maths H2 Quiz - Statistics Probability (Answer Key)

1.
(a) Total ways to choose 3 balls from 10: $\binom{10}{3} = 120$.
Ways to choose 1 Red, 1 Blue, 1 Green: $\binom{5}{1}\binom{3}{1}\binom{2}{1} = 5 \times 3 \times 2 = 30$.
$P(\text{different colours}) = \frac{30}{120} = \frac{1}{4} = 0.25$.
[2]

(b) Let $R$ be the number of red balls.
$P(R \ge 2) = P(R=2) + P(R=3)$.
$P(R=2) = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{10 \times 5}{120} = \frac{50}{120}$.
$P(R=3) = \frac{\binom{5}{3}\binom{5}{0}}{\binom{10}{3}} = \frac{10 \times 1}{120} = \frac{10}{120}$.
$P(R \ge 2) = \frac{60}{120} = 0.5$.
[3]

2.
(a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$0.85 = 0.4 + 0.7 - P(A \cap B) \implies P(A \cap B) = 1.1 - 0.85 = 0.25$.
[1]

(b) If independent, $P(A \cap B) = P(A)P(B)$.
$P(A)P(B) = 0.4 \times 0.7 = 0.28$.
Since $0.25 \neq 0.28$, $A$ and $B$ are not independent.
[2]

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$.
$P(B') = 1 - 0.7 = 0.3$.
$P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.25 = 0.15$.
$P(A | B') = \frac{0.15}{0.3} = 0.5$.
[2]

3.
(a) $\sum P(X=x) = 1 \implies 0.1 + k + 0.3 + 0.2 = 1 \implies k = 0.4$.
[1]

(b) $E(X) = 1(0.1) + 2(0.4) + 3(0.3) + 4(0.2) = 0.1 + 0.8 + 0.9 + 0.8 = 2.6$.
$E(X^2) = 1^2(0.1) + 2^2(0.4) + 3^2(0.3) + 4^2(0.2) = 0.1 + 1.6 + 2.7 + 3.2 = 7.6$.
$Var(X) = E(X^2) - [E(X)]^2 = 7.6 - 2.6^2 = 7.6 - 6.76 = 0.84$.
[3]

(c) $E(Y) = E(3X - 2) = 3E(X) - 2 = 3(2.6) - 2 = 7.8 - 2 = 5.8$.
$Var(Y) = Var(3X - 2) = 3^2 Var(X) = 9(0.84) = 7.56$.
[2]

4.
(a) Let $D$ be the number of defective items. $D \sim B(20, 0.05)$.
[1]

(b) $P(D=2) = \binom{20}{2}(0.05)^2(0.95)^{18} \approx 0.1887$.
[2]

(c) $P(D \ge 1) = 1 - P(D=0) = 1 - \binom{20}{0}(0.05)^0(0.95)^{20} = 1 - 0.95^{20} \approx 1 - 0.3585 = 0.6415$.
[2]

5.
(a) $N$ follows a Geometric distribution. $P(N=n) = (5/6)^{n-1}(1/6)$ for $n=1, 2, \dots$
[1]

(b) $P(N > 3) = P(\text{first 3 are not 6}) = (5/6)^3 = \frac{125}{216} \approx 0.579$.
[2]

6.
Let $X$ be the number of emails. $X \sim Po(4)$.
(a) $P(X=3) = \frac{e^{-4} 4^3}{3!} = \frac{64 e^{-4}}{6} \approx 0.1954$.
[2]

(b) $P(X < 2) = P(X=0) + P(X=1) = e^{-4} + 4e^{-4} = 5e^{-4} \approx 0.0916$.
[2]

7.
$Var(2X - 3Y + 5) = 2^2 Var(X) + (-3)^2 Var(Y)$ (since $X, Y$ independent and constant 5 has variance 0).
$= 4(4) + 9(9) = 16 + 81 = 97$.
[3]

8.
Let $H$ be height. $H \sim N(175, 8^2)$.
(a) $P(170 < H < 185) = P(\frac{170-175}{8} < Z < \frac{185-175}{8}) = P(-0.625 < Z < 1.25)$.
Using GC: normalcdf(-0.625, 1.25, 0, 1) $\approx 0.8891 - 0.2660 = 0.6231$.
[3]

(b) $P(H > h) = 0.1 \implies P(H < h) = 0.9$.
$z_{0.9} \approx 1.2816$.
$h = 175 + 1.2816(8) \approx 185.25$ cm.
[3]

9.
Let $M$ be mass. $M \sim N(5.0, \sigma^2)$.
(a) $P(M < 4.8) = 0.05$.
$Z = \frac{4.8 - 5.0}{\sigma}$. From tables, $P(Z < -1.645) = 0.05$.
$\frac{-0.2}{\sigma} = -1.645 \implies \sigma = \frac{0.2}{1.645} \approx 0.1216$.
[3]

(b) $P(4.9 < M < 5.2)$.
$Z_1 = \frac{4.9 - 5.0}{0.1216} \approx -0.822$.
$Z_2 = \frac{5.2 - 5.0}{0.1216} \approx 1.645$.
$P(-0.822 < Z < 1.645) \approx 0.9500 - 0.2055 = 0.7445$.
[3]

10.
Let $L_i$ be lifetime of battery $i$. $L_i \sim N(20, 4^2)$.
(a) Let $S = \sum_{i=1}^4 L_i$. $E(S) = 4(20) = 80$. $Var(S) = 4(4^2) = 64$. $SD(S) = 8$.
$S \sim N(80, 64)$.
$P(S < 70) = P(Z < \frac{70-80}{8}) = P(Z < -1.25) \approx 0.1056$.
[4]

(b) Let $\bar{L} = \frac{S}{4}$. $E(\bar{L}) = 20$. $Var(\bar{L}) = \frac{16}{4} = 4$. $SD(\bar{L}) = 2$.
$\bar{L} \sim N(20, 4)$.
$P(\bar{L} > 22) = P(Z > \frac{22-20}{2}) = P(Z > 1) \approx 0.1587$.
[3]

11.
(a) $E(\bar{X}) = \mu$. $Var(\bar{X}) = \frac{25}{50} = 0.5$.
[2]

(b) If the population is normal, the sample mean $\bar{X}$ is exactly normally distributed for any sample size $n$. The CLT is an approximation for large $n$ when the population distribution is unknown or non-normal.
[1]

12.
Let $W$ be weight. $W \sim N(150, 20^2)$.
(a) $P(W > 160) = P(Z > \frac{160-150}{20}) = P(Z > 0.5) \approx 0.3085$.
[2]

(b) $\bar{W} \sim N(150, \frac{20^2}{10}) = N(150, 40)$. $SD(\bar{W}) = \sqrt{40} \approx 6.325$.
$P(145 < \bar{W} < 155) = P(\frac{145-150}{6.325} < Z < \frac{155-150}{6.325}) = P(-0.79 < Z < 0.79)$.
$\approx 0.7852 - 0.2148 = 0.5704$.
[3]

13.
Let $T$ be time. $T \sim N(45, 10^2)$.
(a) $P(T > 60) = P(Z > \frac{60-45}{10}) = P(Z > 1.5) \approx 0.0668$.
[2]

(b) Expected number $= 100 \times 0.0668 \approx 7$ students.
[1]

14.
$P(X < 10) = 0.2 \implies \frac{10-\mu}{\sigma} = -0.8416$ (1).
$P(X < 20) = 0.9 \implies \frac{20-\mu}{\sigma} = 1.2816$ (2).
From (1): $10 - \mu = -0.8416 \sigma \implies \mu = 10 + 0.8416 \sigma$.
Sub into (2): $\frac{20 - (10 + 0.8416 \sigma)}{\sigma} = 1.2816$.
$\frac{10 - 0.8416 \sigma}{\sigma} = 1.2816 \implies 10 = 2.1232 \sigma \implies \sigma \approx 4.71$.
$\mu = 10 + 0.8416(4.71) \approx 13.96$.
[4]

15.
(a) $H_0: \mu = 500$. $H_1: \mu < 500$.
[2]

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{498 - 500}{12/\sqrt{36}} = \frac{-2}{2} = -1$.
Critical value for 1-tail 5%: $-1.645$.
Since $-1 > -1.645$, we do not reject $H_0$.
Conclusion: There is insufficient evidence at the 5% level to suggest the mean weight is less than 500 g.
[4]

16.
(a) $H_0: \mu = 70$. $H_1: \mu \neq 70$.
[2]

(b) Since $\sigma$ is unknown and $n < 30$ (though $n=25$ is borderline, t-test is appropriate), use t-test.
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{72 - 70}{10/\sqrt{25}} = \frac{2}{2} = 1$.
Degrees of freedom $= 24$.
Critical value for 2-tail 5%: $t_{0.025, 24} \approx 2.064$.
Since $1 < 2.064$, we do not reject $H_0$.
Conclusion: There is insufficient evidence to suggest the mean score has changed.
[4]

17.
(a) Strong negative linear correlation between $x$ and $y$.
[2]

(b) Correlation measures association, not cause. A third variable could influence both, or the relationship could be coincidental.
[1]

18.
(a) $y = 2.5 + 0.8(10) = 10.5$.
[1]

(b) $x=100$ is far outside the range of the data (extrapolation). The linear relationship may not hold outside the observed range.
[1]

19.
(a) p-value $= 2 \times P(Z > 2.1) = 2 \times (1 - 0.9821) = 2 \times 0.0179 = 0.0358$.
[2]

(b) Since $0.0358 < 0.05$, reject $H_0$.
[1]

20.
(a) 95% CI: $\bar{x} \pm z_{0.025} \frac{s}{\sqrt{n}}$.
$50 \pm 1.96 \frac{5}{\sqrt{100}} = 50 \pm 1.96(0.5) = 50 \pm 0.98$.
Interval: $(49.02, 50.98)$.
[3]

(b) We are 95% confident that the true population mean lies within this interval.
[2]