A Level H1 Mathematics Statistics Probability Quiz

A-Level Maths H1 Quiz - Statistics Probability (Answer Key)

1. (a) Total people = 11. Choose 4. $\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$. [1] (b) At least 2 women means: 2W 2M, 3W 1M, or 4W 0M. 2W 2M: $\binom{5}{2}\binom{6}{2} = 10 \times 15 = 150$ 3W 1M: $\binom{5}{3}\binom{6}{1} = 10 \times 6 = 60$ 4W 0M: $\binom{5}{4}\binom{6}{0} = 5 \times 1 = 5$ Total = $150 + 60 + 5 = 215$. [2]

2. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.7 = 0.4 + 0.5 - P(A \cap B)$ $P(A \cap B) = 0.9 - 0.7 = 0.2$. [1] (b) Check independence: Is $P(A \cap B) = P(A)P(B)$? $P(A)P(B) = 0.4 \times 0.5 = 0.2$. Since $0.2 = 0.2$, events $A$ and $B$ are independent. [2]

3. Let $M$ = Math, $P$ = Physics. $P(M) = 0.6, P(P) = 0.4, P(M \cap P) = 0.2$. (a) $P(\text{Neither}) = 1 - P(M \cup P) = 1 - [P(M) + P(P) - P(M \cap P)]$ $= 1 - [0.6 + 0.4 - 0.2] = 1 - 0.8 = 0.2$. [1] (b) $P(M | P) = \frac{P(M \cap P)}{P(P)} = \frac{0.2}{0.4} = 0.5$. [2]

4. (a) Tree Diagram: Start -> Red (4/10) -> Red (3/9), Blue (6/9) Start -> Blue (6/10) -> Red (4/9), Blue (5/9) [2] (b) Different colors: (Red then Blue) or (Blue then Red). $P(RB) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}$ $P(BR) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}$ Total = $\frac{48}{90} = \frac{8}{15} \approx 0.533$. [1]

5. Let $X$ be number of rainy days. $X \sim B(5, 0.3)$. (a) $P(X=2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10 \times 0.09 \times 0.343 = 0.3087$. [2] (b) $P(X \ge 1) = 1 - P(X=0) = 1 - (0.7)^5 = 1 - 0.16807 = 0.83193 \approx 0.832$. [1]

6. $X \sim B(12, 0.25)$. (a) $P(X=3) = \binom{12}{3}(0.25)^3(0.75)^9 \approx 0.258$. [1] (b) $P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$. $P(X=0) = (0.75)^{12} \approx 0.0317$ $P(X=1) = 12(0.25)(0.75)^{11} \approx 0.1267$ $P(X \ge 2) = 1 - (0.0317 + 0.1267) = 1 - 0.1584 = 0.8416 \approx 0.842$. [2]

7. (a) Conditions:

1. Fixed number of trials ($n=20$).
2. Constant probability of success ($p=0.05$).
3. Trials are independent.
4. Two outcomes (defective/not defective). (Any two). [2] (b) Let $D$ be number of defective bulbs. $D \sim B(20, 0.05)$. $P(D > 2) = 1 - P(D \le 2) = 1 - \text{binomcdf}(20, 0.05, 2)$. Using calculator: $P(D \le 2) \approx 0.9245$. $P(D > 2) = 1 - 0.9245 = 0.0755$. [2]

8. $H \sim N(175, 8^2)$. (a) $P(H < 165) = \text{normalcdf}(-\infty, 165, 175, 8) \approx 0.1056$. [1] (b) $P(H > h) = 0.1 \Rightarrow P(H < h) = 0.9$. $h = \text{invNorm}(0.9, 175, 8) \approx 185.25$ cm. [2]

9. $W \sim N(\mu, 0.5^2)$. (a) $P(W < 4.36) = 0.1$. Standardizing: $Z = \frac{4.36 - \mu}{0.5}$. From tables/calc, $P(Z < z) = 0.1 \Rightarrow z \approx -1.2816$. $\frac{4.36 - \mu}{0.5} = -1.2816 \Rightarrow 4.36 - \mu = -0.6408 \Rightarrow \mu = 5.0008 \approx 5.00$ kg. [3] (b) $P(4.5 < W < 5.5)$ with $\mu=5, \sigma=0.5$. $= \text{normalcdf}(4.5, 5.5, 5, 0.5) \approx 0.6827$. [2]

10. (a) $W = X - Y$. $E(W) = E(X) - E(Y) = 50 - 30 = 20$. Since independent, $Var(W) = Var(X) + Var(Y) = 16 + 9 = 25$. $W \sim N(20, 25)$. [2] (b) $P(W > 22) = \text{normalcdf}(22, \infty, 20, 5)$. $Z = \frac{22-20}{5} = 0.4$. $P(Z > 0.4) = 1 - 0.6554 = 0.3446$. [2]

11. $S \sim N(200, 20^2)$. (a) $P(S > 230) = \text{normalcdf}(230, \infty, 200, 20) \approx 0.0668$. [1] (b) Find $k$ such that $P(S < k) = 0.95$. $k = \text{invNorm}(0.95, 200, 20) \approx 232.9$. Stock 233 copies. [2]

12. $T \sim N(\mu, \sigma^2)$. $P(T < 30) = 0.2 \Rightarrow \frac{30-\mu}{\sigma} = z_1 \approx -0.8416$. $P(T < 50) = 0.9 \Rightarrow \frac{50-\mu}{\sigma} = z_2 \approx 1.2816$. Eq 1: $30 - \mu = -0.8416\sigma$ Eq 2: $50 - \mu = 1.2816\sigma$ Subtract Eq 1 from Eq 2: $20 = 2.1232\sigma \Rightarrow \sigma \approx 9.42$. Substitute $\sigma$: $\mu = 30 + 0.8416(9.42) \approx 37.93$. Mean $\approx 37.9$, SD $\approx 9.42$. [4]

13. (a) Unbiased estimate of mean = sample mean = 168 cm. [1] (b) Unbiased estimate of variance $s^2 = \frac{n}{n-1} \times \text{sample variance}$? Note: Question says "sample variance is 36". Usually, if "sample variance" is given as $s^2_{biased} = \frac{\sum(x-\bar{x})^2}{n}$, then unbiased is $\frac{n}{n-1}s^2_{biased}$. However, standard convention in many texts: if "variance of the sample" is stated, it often implies $s^2 = \frac{\sum(x-\bar{x})^2}{n-1}$ is already the unbiased estimator if calculated from data. Clarification for H1: If 36 is $\frac{\sum x^2}{n} - \bar{x}^2$ (biased), then Unbiased $= \frac{100}{99} \times 36 = 36.36$. If 36 is already $s^2$ (unbiased), then answer is 36. Given "sample variance" usually refers to biased estimator in raw data contexts unless specified "unbiased", we assume biased. Unbiased Estimate $= \frac{100}{99}(36) \approx 36.4$. [1] (c) Variance of sample mean $\bar{X} = \frac{\sigma^2}{n}$. We estimate $\sigma^2$ with unbiased estimate 36.36. $Var(\bar{X}) = \frac{36.36}{100} = 0.3636$. [1]

14. (a) $\sigma = 15, n=25, \bar{x}=140$. 90% CI. $z_{0.05} = 1.645$. CI $= \bar{x} \pm z \frac{\sigma}{\sqrt{n}} = 140 \pm 1.645 \frac{15}{5} = 140 \pm 1.645(3) = 140 \pm 4.935$. $(135.065, 144.935)$. [3] (b) If we were to take many random samples of size 25 and construct a 90% CI for each, 90% of these intervals would contain the true population mean. [1]

15. (a) $H_0: \mu = 500$. $H_1: \mu < 500$. [2] (b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{485 - 500}{40/\sqrt{40}} = \frac{-15}{6.3246} \approx -2.37$. Critical value for 1-tail 5%: $-1.645$. Since $-2.37 < -1.645$, we reject $H_0$. Alternatively, p-value $= P(Z < -2.37) \approx 0.0089$. Since $0.0089 < 0.05$, reject $H_0$. Conclusion: There is sufficient evidence at the 5% level to suggest the mean lifetime is less than 500 hours. [4]

16. (a) $H_0: \mu = 65$. $H_1: \mu \ne 65$. $n=50$ (large), so use Z-test approx. $s=10$. $Z = \frac{68 - 65}{10/\sqrt{50}} = \frac{3}{1.414} \approx 2.12$. Critical values for 2-tail 1%: $\pm 2.576$. Since $2.12 < 2.576$, we do not reject $H_0$. Conclusion: There is insufficient evidence to suggest the mean score has changed. [4] (b) The sample size $n=50$ is large ($>30$), so by the Central Limit Theorem, the sampling distribution of the mean is approximately normal, regardless of the population distribution. [1]

17. (a) Using GC: $r \approx 0.998$. [1] (b) Regression line $y$ on $x$: $a \approx 66.57, b \approx 1.257$. $y = 66.6 + 1.26x$ (to 3 sf). [2] (c) $x=62 \Rightarrow y = 66.57 + 1.257(62) \approx 144.5$ mmHg. [1] (d) Unreliable. 90 is outside the range of the data (40-75), so this is extrapolation. The linear relationship may not hold for older ages. [1]

18. (a) $b_{yx} = 2.5$, $b_{xy} = 0.3$. $r^2 = b_{yx} \times b_{xy} = 2.5 \times 0.3 = 0.75$. $r = \sqrt{0.75} \approx 0.866$. [2] (b) Positive, because both regression coefficients ($2.5$ and $0.3$) are positive. [1]

19. (a) Convenience sampling. [1] (b) People at a mall on Monday morning may not represent the whole population (e.g., working people are excluded, elderly/retired over-represented). [1] (c) Assign each of the 10,000 residents a unique number from 1 to 10,000. Use a random number generator to select 50 distinct numbers. Select the residents corresponding to these numbers. [2]

20. $T \sim N(12, 2^2)$. (a) Sample mean $\bar{T}$ for $n=16$. $\bar{T} \sim N(12, \frac{2^2}{16}) = N(12, 0.25)$. SD = 0.5. $P(\bar{T} < 11) = \text{normalcdf}(-\infty, 11, 12, 0.5)$. $Z = \frac{11-12}{0.5} = -2$. $P(Z < -2) \approx 0.0228$. [3] (b) Total time $S = \sum_{i=1}^{16} T_i$. $E(S) = 16 \times 12 = 192$. $Var(S) = 16 \times 2^2 = 64$. SD = 8. $S \sim N(192, 64)$. $P(S > 200) = \text{normalcdf}(200, \infty, 192, 8)$. $Z = \frac{200-192}{8} = 1$. $P(Z > 1) \approx 0.1587$. [2]