A Level H1 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper - Maths H1 A-Level (Answer Key)

Version 3

Section A: Probability and Distributions

1. (a) At least 3 women means 3 women and 2 men, 4 women and 1 man, or 5 women and 0 men. Number of ways = $\binom{6}{3}\binom{8}{2} + \binom{6}{4}\binom{8}{1} + \binom{6}{5}\binom{8}{0}$ $= 20 \times 28 + 15 \times 8 + 6 \times 1$ $= 560 + 120 + 6 = 686$ [3] (b) Specific man and specific woman are included. We need to choose 3 more members from the remaining $8+6-2=12$ people. Number of ways = $\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$ [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 if $P(A \cap B) = P(A)P(B)$. $P(A)P(B) = 0.4 \times 0.5 = 0.2$. Since $P(A \cap B) = 0.2$, events $A$ and $B$ are independent. [2] (c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$. $P(B') = 1 - 0.5 = 0.5$. $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$. $P(A | B') = \frac{0.2}{0.5} = 0.4$ [2]

3. $X \sim B(12, 0.3)$ (a) $P(X=4) = \binom{12}{4}(0.3)^4(0.7)^8 \approx 0.231$ [1] (b) $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ Using GC: binomcdf(12, 0.3, 2) $\approx 0.168$ [1] (c) $P(X > 8) = 1 - P(X \le 8)$ Using GC: 1 - binomcdf(12, 0.3, 8) $\approx 0.0002$ (or $2.16 \times 10^{-4}$) [2]

4. $H \sim N(175, 8^2)$ (a) $P(170 < H < 185) = \text{normalcdf}(170, 185, 175, 8) \approx 0.628$ [2] (b) $P(H > h) = 0.1 \Rightarrow P(H < h) = 0.9$. Using GC: invNorm(0.9, 175, 8) $\approx 185.25$ cm. $h \approx 185$ cm (3 s.f.) [2]

5. $X \sim N(50, 16)$, $Y \sim N(30, 9)$. Independent. $W = 2X - Y$ (a) $E(W) = 2E(X) - E(Y) = 2(50) - 30 = 70$. $Var(W) = 2^2 Var(X) + (-1)^2 Var(Y) = 4(16) + 1(9) = 64 + 9 = 73$. [3] (b) $W \sim N(70, 73)$. SD $= \sqrt{73} \approx 8.544$. $P(W > 75) = \text{normalcdf}(75, 1E99, 70, \sqrt{73}) \approx 0.279$ [2]

Section B: Sampling and Estimation

6. $n=80, \sum x = 4200, \sum x^2 = 225000$. (a) Unbiased estimate of mean $\bar{x} = \frac{4200}{80} = 52.5$ [1] (b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$ $s^2 = \frac{1}{79} \left( 225000 - \frac{4200^2}{80} \right)$ $s^2 = \frac{1}{79} (225000 - 220500) = \frac{4500}{79} \approx 56.96$ [3]

7. $M \sim N(5.0, 0.1^2)$. Sample size $n=10$. Let $\bar{M}$ be the sample mean. $\bar{M} \sim N(5.0, \frac{0.1^2}{10}) = N(5.0, 0.001)$. (a) $P(\bar{M} < 4.95) = \text{normalcdf}(-1E99, 4.95, 5.0, \sqrt{0.001}) \approx 0.0569$ [3] (b) The Central Limit Theorem is not required because the underlying population distribution is already stated to be normal. The sampling distribution of the mean is normal for any sample size $n$ when the population is normal. [1]

8. (a) $H_0: \mu = 100$ $H_1: \mu < 100$ [2] (b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{98 - 100}{8/\sqrt{50}} = \frac{-2}{1.131} \approx -1.768$. P-value $= P(Z < -1.768) \approx 0.0385$. Since $0.0385 < 0.05$, we reject $H_0$. Conclusion: There is sufficient evidence at the 5% level to suggest that the mean lifetime of the batteries is less than 100 hours. [4]

9. $p = 0.2, n = 200$. (a) Since $n$ is large ($np = 40 > 5, n(1-p) = 160 > 5$), the sample proportion $\hat{P}$ is approximately normally distributed. $\hat{P} \sim N\left(p, \frac{p(1-p)}{n}\right) = N\left(0.2, \frac{0.2(0.8)}{200}\right) = N(0.2, 0.0008)$. [2] (b) $P(\hat{P} > 0.25) = \text{normalcdf}(0.25, 1E99, 0.2, \sqrt{0.0008}) \approx 0.0385$ [3]

Section C: Correlation and Regression

10. (a) Using GC with lists: $r = 1$ (Perfect positive linear correlation). [1] (b) Regression line: $y = 80 + x$ (or $y = 1x + 80$). $a = 80, b = 1$. [2] (c) For every additional year of age, the systolic blood pressure increases by 1 mmHg on average. [1] (d) Estimate: $y = 80 + 70 = 150$ mmHg. Reliability: This is extrapolation (70 is outside the data range 30-65). It may not be reliable as the linear trend might not continue. [2]

11. (a) $S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 1300 - \frac{120^2}{12} = 1300 - 1200 = 100$. $S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 5000 - \frac{240^2}{12} = 5000 - 4800 = 200$. $S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 2500 - \frac{120 \times 240}{12} = 2500 - 2400 = 100$. [3] (b) $b = \frac{S_{xy}}{S_{xx}} = \frac{100}{100} = 1$. $\bar{x} = 10, \bar{y} = 20$. $a = \bar{y} - b\bar{x} = 20 - 1(10) = 10$. Equation: $y = 10 + x$. [3] (c) Predicted $y$ for $x=15$: $\hat{y} = 10 + 15 = 25$. Residual $= y - \hat{y} = 22 - 25 = -3$. [2]

12. (a) Strong positive linear relationship. [1] (b) Correlation does not imply causation. There may be lurking variables (e.g., intelligence, prior knowledge) that affect both study time and scores. [2] (c) $r$ would remain unchanged. The correlation coefficient is invariant under linear scaling (change of units). [2]

13. (a) Substitute $y = 2x + 5$ into $x = 0.4y + 1$: $x = 0.4(2x + 5) + 1$ $x = 0.8x + 2 + 1$ $0.2x = 3 \Rightarrow x = 15$. $y = 2(15) + 5 = 35$. Intersection point: $(15, 35)$. [3] (b) The regression lines always intersect at the point of means $(\bar{x}, \bar{y})$. Thus, $\bar{x}=15, \bar{y}=35$. [1]

14. (a) $P = -2p^2 + 40p - 100$. $\frac{dP}{dp} = -4p + 40$. Set $\frac{dP}{dp} = 0 \Rightarrow 4p = 40 \Rightarrow p = 10$. Check second derivative: $\frac{d^2P}{dp^2} = -4 < 0$, so it is a maximum. Price p = \10$. [2] (b) Max Profit$P(10) = -2(100) + 40(10) - 100 = -200 + 400 - 100 = $100 (i.e., \100,000). [1] (c) Profit usually increases with price up to a point, then decreases as demand drops. This non-monotonic behavior is quadratic (curved), not linear. A linear model cannot capture the turning point (maximum). [1]