A Level H1 Mathematics Practice Paper 4

TuitionGoWhere Practice Paper - Maths H1 A-Level (Answers)

Version 4 - Statistics & Probability Focus

Section A: Probability and Distributions

1. (a) Total people = $6 + 5 = 11$. Choose 4. Number of ways = $\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$. [1]

(b) "At least 2 women" means 2 women, 3 women, or 4 women.

• 2 Women, 2 Men: $\binom{5}{2}\binom{6}{2} = 10 \times 15 = 150$
• 3 Women, 1 Man: $\binom{5}{3}\binom{6}{1} = 10 \times 6 = 60$
• 4 Women, 0 Men: $\binom{5}{4}\binom{6}{0} = 5 \times 1 = 5$ Total = $150 + 60 + 5 = 215$. [3]

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 $P(A \cap B) = 0.2$, the events are independent. [2]

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$. $P(B') = 1 - P(B) = 1 - 0.5 = 0.5$. Since $A$ and $B$ are independent, $A$ and $B'$ are also independent. $P(A \cap B') = P(A)P(B') = 0.4 \times 0.5 = 0.2$. Alternatively: $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$. $P(A | B') = \frac{0.2}{0.5} = 0.4$. [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) $P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$. $P(X=0) = (0.7)^{12} \approx 0.0138$. $P(X=1) = \binom{12}{1}(0.3)^1(0.7)^{11} \approx 0.0712$. $P(X \ge 2) = 1 - (0.0138 + 0.0712) = 1 - 0.0850 = 0.915$. [2]

4. $H \sim N(175, 8^2)$. (a) $P(170 < H < 185)$. Using GC: normalcdf(170, 185, 175, 8) $\approx 0.628$. [2]

(b) $P(H > h) = 0.10 \Rightarrow P(H < h) = 0.90$. Using GC: invNorm(0.90, 175, 8) $\approx 185.25$. $h \approx 185$ cm (3 s.f.). [2]

5. (a) $W = X - Y$. $E(W) = E(X) - E(Y) = 50 - 30 = 20$. Since $X, Y$ independent, $Var(W) = Var(X) + Var(Y) = 4^2 + 3^2 = 16 + 9 = 25$. $W \sim N(20, 25)$. (Or $N(20, 5^2)$). [2]

(b) $P(W > 25)$. Using GC: normalcdf(25, 1E99, 20, 5) $\approx 0.159$. [2]

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Section B: Sampling and Estimation

6. $n = 80, \sum x = 4200, \sum x^2 = 225500$. (a) Unbiased estimate of mean $\bar{x} = \frac{4200}{80} = 52.5$. [1]

(b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$. $s^2 = \frac{1}{79} \left( 225500 - \frac{4200^2}{80} \right)$ $s^2 = \frac{1}{79} (225500 - 220500) = \frac{5000}{79} \approx 63.3$. [3]

7. $X \sim N(500, 10^2)$. Sample size $n=16$. Let $\bar{X}$ be the sample mean. $\bar{X} \sim N(500, \frac{10^2}{16}) = N(500, 6.25)$. SD of $\bar{X} = \sqrt{6.25} = 2.5$.

(a) $P(\bar{X} < 495)$. Using GC: normalcdf(-1E99, 495, 500, 2.5) $\approx 0.0228$. [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 $n$. [1]

8. Population $\sigma = 12$. Sample size $n$. $\bar{X} \sim N(\mu, \frac{144}{n})$. We are given $P(\bar{X} > 502) = 0.1587$. This implies $P(\bar{X} < 502) = 1 - 0.1587 = 0.8413$. From standard normal tables/GC, the Z-score corresponding to cumulative probability 0.8413 is approximately $1.0$ (since $\Phi(1) \approx 0.8413$). $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$. Here, we assume the test is centered around the true mean $\mu$. However, the problem implies finding $n$ given the probability bound relative to the mean. Wait, usually such questions give $\mu$. Let's assume the question implies the probability that the sample mean deviates from the population mean $\mu$ by more than 2 units? Re-reading: "It is given that $P(\bar{x} > 502) = 0.1587$." This statement is incomplete without knowing $\mu$. Correction based on standard H1 patterns: Usually, this implies testing against a hypothesized mean or finding $n$ such that the margin of error is specific. Let's assume the population mean $\mu = 500$ (standard context from Q7, though Q8 says "different machine"). If $\mu$ is not given, we cannot solve. Assumption for Solution: Let us assume the population mean $\mu = 500$ g (consistent with typical problem structures where 502 is the observed value or boundary). If $\mu = 500$: $Z = \frac{502 - 500}{12/\sqrt{n}} = \frac{2\sqrt{n}}{12} = \frac{\sqrt{n}}{6}$. We know $P(Z > z) = 0.1587 \Rightarrow z \approx 1.0$. $\frac{\sqrt{n}}{6} = 1 \Rightarrow \sqrt{n} = 6 \Rightarrow n = 36$. [4] (Note: If $\mu$ was different, the Z-score would change. Given 0.1587 is exactly $1 - \Phi(1)$, the Z-score is 1.)

9. An unbiased estimator is a statistic where the expected value (or mean of the sampling distribution) is equal to the true population parameter it is estimating. For example, $E(\bar{X}) = \mu$ and $E(S^2) = \sigma^2$. [2]

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Section C: Hypothesis Testing

10. (a) $H_0: \mu = 120$ $H_1: \mu < 120$ [2]

(b) Test Statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$. $\bar{x} = 118, \mu = 120, \sigma = 10, n = 50$. $Z = \frac{118 - 120}{10/\sqrt{50}} = \frac{-2}{1.414} \approx -1.414$. P-value $= P(Z < -1.414)$. Using GC: normalcdf(-1E99, -1.414, 0, 1) $\approx 0.0787$. Since $0.0787 > 0.05$, we do not reject $H_0$. Conclusion: There is insufficient evidence at the 5% level to suggest that the mean lifetime of the batteries is less than 120 hours. [5]

11. (a) $H_0: \mu = 65$ $H_1: \mu > 65$ Test Statistic $Z = \frac{69 - 65}{15/\sqrt{36}} = \frac{4}{15/6} = \frac{4}{2.5} = 1.6$. P-value $= P(Z > 1.6)$. Using GC: normalcdf(1.6, 1E99, 0, 1) $\approx 0.0548$. Significance level $\alpha = 0.01$. Since $0.0548 > 0.01$, we do not reject $H_0$. Conclusion: There is insufficient evidence at the 1% level to support the school's claim that their students perform better than the national average. [5]

(b) A Type I error occurs when we reject the null hypothesis when it is actually true. In this context, it would mean concluding that the school's students perform better than the national average when, in reality, their mean score is equal to the national average (65). [3]

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Section D: Correlation and Regression

12. (a) Scatter diagram:

• Axes labeled "Age (years)" and "Pressure (mmHg)".
• Points plotted correctly: (35,110), (40,115)... (70,152).
• Positive linear trend visible. [2]

(b) Using GC (Linear Reg): $r \approx 0.997$ (to 3 d.p.). [2]

(c) Regression line $y = a + bx$. Using GC: $b \approx 1.057$ $a \approx 72.5$ Equation: $y = 72.5 + 1.06x$ (coefficients to 3 s.f.). [3]

(d) Estimate for $x = 62$: $y = 72.5 + 1.057(62) \approx 138.0$ mmHg. Reliability: This is an interpolation because 62 is within the range of the data (35 to 70). Therefore, the estimate is likely reliable. [3]