A Level H1 Mathematics Practice Paper 4

TuitionGoWhere Exam Practice (AI) - Answer Key

A-Level H1 Mathematics (8865) - Practice Paper (Version 4)

Topic: Statistics & Probability

Total Marks: 60

Section A: Probability and Counting Principles

1. (a) Total people = $8 + 6 = 14$. Select 5. $^{14}C_5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$ [1]

(b) Select 3 men from 8 and 2 women from 6. $^8C_3 \times ^6C_2 = 56 \times 15 = 840$ [2]

(c) At least 4 women means 4 women and 1 man, OR 5 women and 0 men. Case 1 (4W, 1M): $^6C_4 \times ^8C_1 = 15 \times 8 = 120$ Case 2 (5W, 0M): $^6C_5 \times ^8C_0 = 6 \times 1 = 6$ Total = $120 + 6 = 126$ [2]

2. (a) Let $X$ be the number of defective items in the sample. $X \sim B(15, 0.1)$ [1]

(b) $P(X = 2) = ^{15}C_2 (0.1)^2 (0.9)^{13}$ $= 105 \times 0.01 \times 0.25418... \approx 0.2669$ Answer: 0.267 (3 s.f.) [2]

(c) $P(X > 1) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$ $P(X=0) = (0.9)^{15} \approx 0.2059$ $P(X=1) = 15(0.1)(0.9)^{14} \approx 0.3432$ $P(X > 1) = 1 - (0.2059 + 0.3432) = 1 - 0.5491 = 0.4509$ Answer: 0.451 (3 s.f.) [2]

3. (a) Check if $P(A \cap B) = P(A)P(B)$. $P(A)P(B) = 0.4 \times 0.5 = 0.2$ Given $P(A \cap B) = 0.2$. Since $P(A \cap B) = P(A)P(B)$, events $A$ and $B$ are independent. [2] (1 for calculation, 1 for conclusion with reason)

(b) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $= 0.4 + 0.5 - 0.2 = 0.7$ [1]

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ $P(B') = 1 - P(B) = 1 - 0.5 = 0.5$ $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$ (Since A and B are independent, $A$ and $B'$ are also independent, so $0.4 \times 0.5 = 0.2$) $P(A | B') = \frac{0.2}{0.5} = 0.4$ [2]

Section B: Descriptive Statistics and Estimation

4. (a) Mean $\bar{t} = \frac{\sum t}{n} = \frac{120}{10} = 12$ minutes. [1]

(b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left[ \sum t^2 - \frac{(\sum t)^2}{n} \right]$ $s^2 = \frac{1}{9} \left[ 1550 - \frac{120^2}{10} \right]$ $s^2 = \frac{1}{9} [ 1550 - 1440 ] = \frac{110}{9} \approx 12.22$ Answer: 12.2 (3 s.f.) [3] (1 for formula/setup, 1 for substitution, 1 for answer)

5. (a) Mean of coded data $\bar{h} = \frac{\sum h}{n} = \frac{40}{80} = 0.5$. Since $h = \frac{x - 170}{5} \implies x = 5h + 170$. $\bar{x} = 5\bar{h} + 170 = 5(0.5) + 170 = 2.5 + 170 = 172.5$ cm. [3]

(b) Variance of coded data $s_h^2 = \frac{1}{n-1} \left[ \sum h^2 - \frac{(\sum h)^2}{n} \right]$ $s_h^2 = \frac{1}{79} \left[ 120 - \frac{40^2}{80} \right] = \frac{1}{79} [ 120 - 20 ] = \frac{100}{79}$ Variance is invariant under change of origin but scales by square of multiplier for change of scale. $s_x^2 = 5^2 \times s_h^2 = 25 \times \frac{100}{79} = \frac{2500}{79} \approx 31.645$ Answer: 31.6 (3 s.f.) [3]

6. (a) $E(\bar{X}) = \mu$ $Var(\bar{X}) = \frac{\sigma^2}{n}$ [2]

(b) Dividing by $n-1$ makes the estimator unbiased. Dividing by $n$ tends to underestimate the population variance. [1]

Section C: Normal Distribution and Sampling

7. Let $M$ be the mass of a bag. $M \sim N(5.0, 0.1^2)$. (a) $P(M < 4.9)$. Using GC: normalcdf(-E99, 4.9, 5.0, 0.1) $P(M < 4.9) \approx 0.1587$ Answer: 0.159 (3 s.f.) [2]

(b) We want $P(M > m) = 0.95$, which implies $P(M < m) = 0.05$. Using GC: invNorm(0.05, 5.0, 0.1) $m \approx 4.8355$ Answer: 4.84 kg (3 s.f.) [3]

8. Let $X$ be expenditure. $X \sim N(150, 40^2)$. Sample size $n=16$. Sample mean $\bar{X} \sim N(150, \frac{40^2}{16}) = N(150, 100)$. Standard deviation of $\bar{X}$ is $\sqrt{100} = 10$. (a) $P(\bar{X} < 140)$. Using GC: normalcdf(-E99, 140, 150, 10) $P(\bar{X} < 140) \approx 0.1587$ Answer: 0.159 (3 s.f.) [3]

(b) The Central Limit Theorem is not required because the population distribution is already stated to be normal. Therefore, the sampling distribution of the mean is normal for any sample size. [1]

(c) Standard error $SE = \frac{\sigma}{\sqrt{n}}$. If $n$ increases from 16 to 50, the denominator $\sqrt{n}$ increases. Therefore, the standard error decreases. Specifically, it changes from $\frac{40}{\sqrt{16}}=10$ to $\frac{40}{\sqrt{50}} \approx 5.66$. [2]

9. $X \sim N(10, 4) \implies E(X)=10, Var(X)=4$. $Y \sim N(5, 9) \implies E(Y)=5, Var(Y)=9$. (a) $E(2X - Y) = 2E(X) - E(Y) = 2(10) - 5 = 15$. [1]

(b) Since $X$ and $Y$ are independent: $Var(2X - Y) = 2^2 Var(X) + (-1)^2 Var(Y) = 4(4) + 1(9) = 16 + 9 = 25$. [2]

(c) Linear combination of independent normal variables is normal. $2X - Y \sim N(15, 25)$. [1]

Section D: Hypothesis Testing and Regression

10. (a) $H_0: \mu = 1000$ $H_1: \mu < 1000$ [2]

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ $Z = \frac{980 - 1000}{100/\sqrt{50}} = \frac{-20}{14.142} \approx -1.414$ Critical value for one-tail test at 5%: $z_{0.05} = -1.645$. Since $-1.414 > -1.645$ (or $P(Z < -1.414) \approx 0.0787 > 0.05$), we do not reject $H_0$. Conclusion: There is insufficient evidence at the 5% significance level to support the claim that the mean lifetime is less than 1000 hours. [4] (1 for Z calc, 1 for critical value/p-value, 1 for comparison, 1 for conclusion in context)

11. (a) Scatter diagram:

• Axes labeled "Age (years)" and "Price ($100s)".
• Points plotted correctly: (2,15), (3,14), (5,11), (7,9), (8,8), (10,5).
• Reasonable scale. [2]

(b) Using GC: $r \approx -0.986$ Answer: -0.986 (3 s.f.) [2]

(c) Using GC for regression $y$ on $x$: $a \approx 17.57$, $b \approx -1.29$ Equation: $y = 17.6 - 1.29x$ (coefficients to 3 s.f.) [3]

(d) For every additional year of age, the selling price of the car decreases by approximately \129$(since$y$ is in hundreds). [1]

(e) Age 20 is outside the range of the observed data (2 to 10 years). This is extrapolation, and the linear relationship may not hold for older cars (e.g., price cannot go below zero, or classic car value might increase). [1]