A Level H1 Mathematics Practice Paper 5

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

Version 5 - Marking Scheme

Section A: Probability and Distributions

1. (a) Let $X$ be the number of defective components. $X \sim B(25, 0.04)$ [1]

(b) $P(X=2) = \binom{25}{2} (0.04)^2 (0.96)^{23}$ $= 300 \times 0.0016 \times 0.3905...$ $\approx 0.1876$ [2] (Accept 0.188)

(c) $P(X \ge 1) = 1 - P(X=0)$ $P(X=0) = (0.96)^{25} \approx 0.3604$ $P(X \ge 1) = 1 - 0.3604 = 0.6396$ [2] (Accept 0.640)

2. Let $W$ be the weight of a bag of rice. $W \sim N(5.0, 0.15^2)$.

(a) $P(W < 4.8)$ Using GC: normalcdf(-1E99, 4.8, 5.0, 0.15) $\approx 0.0912$ [2]

(b) We want $P(W > w) = 0.95$, which implies $P(W < w) = 0.05$. Using GC: invNorm(0.05, 5.0, 0.15) $w \approx 4.753$ kg [2] (Accept 4.75)

3. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.8 = 0.6 + 0.5 - P(A \cap B)$ $P(A \cap B) = 1.1 - 0.8 = 0.3$ [1]

(b) Check if $P(A \cap B) = P(A)P(B)$. $P(A)P(B) = 0.6 \times 0.5 = 0.3$. Since $P(A \cap B) = 0.3$, $P(A \cap B) = P(A)P(B)$. Therefore, $A$ and $B$ are independent. [2] (1 mark for calculation, 1 mark for conclusion with reason)

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ $P(B') = 1 - 0.5 = 0.5$. $P(A \cap B') = P(A) - P(A \cap B) = 0.6 - 0.3 = 0.3$. $P(A | B') = \frac{0.3}{0.5} = 0.6$ [2]

4. (a) Sum of probabilities must be 1. $k + 2k + 3k + 4k = 1$ $10k = 1 \Rightarrow k = 0.1$ [1]

(b) $E(X) = \sum x P(X=x)$ $= 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)$ $= 0.1 + 0.4 + 0.9 + 1.6$ $= 3.0$ [2]

Section B: Sampling and Estimation

5. $n = 80, \sum t = 1200, \sum t^2 = 18500$.

(a) Unbiased estimate of mean $\bar{t} = \frac{1200}{80} = 15$ hours. [1]

(b) Unbiased estimate of variance $s^2 = \frac{n}{n-1} \left( \frac{\sum t^2}{n} - \bar{t}^2 \right)$ $s^2 = \frac{80}{79} \left( \frac{18500}{80} - 15^2 \right)$ $s^2 = \frac{80}{79} (231.25 - 225)$ $s^2 = \frac{80}{79} (6.25)$ $s^2 \approx 6.329$ [3] (1 mark for formula/setup, 1 mark for substitution, 1 mark for answer)

6. (a) Since the population is normal, the sample mean $\bar{M}$ is also normally distributed. $\bar{M} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$ $\bar{M} \sim N\left(0.25, \frac{0.05^2}{100}\right)$ or $N(0.25, 0.000025)$ [2]

(b) We want $P(\bar{M} > 0.26)$. Standard error $SE = \frac{0.05}{\sqrt{100}} = 0.005$. $Z = \frac{0.26 - 0.25}{0.005} = \frac{0.01}{0.005} = 2$. $P(Z > 2) = 1 - P(Z < 2) \approx 1 - 0.9772 = 0.0228$. Using GC: normalcdf(0.26, 1E99, 0.25, 0.005) $\approx 0.0228$. [3]

7. $n=50, \bar{x}=12.5, s^2=4.0 \Rightarrow s=2.0$. Since $n$ is large ($>30$), we use the Z-distribution (or t-distribution approximated by Z in H1 context often, but strictly t if $\sigma$ unknown. H1 syllabus allows Z for large samples using $s$ as estimate for $\sigma$). Standard Error $SE = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{50}} \approx 0.2828$. Critical value for 95% confidence ($z_{0.025}$) is 1.96.

(a) Confidence Interval: $\bar{x} \pm z \frac{s}{\sqrt{n}}$ $12.5 \pm 1.96(0.2828)$ $12.5 \pm 0.554$ $(11.946, 13.054)$ Answer: $(11.9, 13.1)$ m (to 3 s.f.) [4] (1 mark for SE, 1 mark for critical value, 1 mark for margin of error, 1 mark for interval)

(b) "95% confidence" means that if we were to take many random samples of size 50 and construct a confidence interval for each, 95% of those intervals would contain the true population mean height. It does not mean there is a 95% probability that this specific interval contains the mean. [2]

8. (a) The Central Limit Theorem states that for a large sample size ($n > 30$), the sampling distribution of the sample mean $\bar{X}$ will be approximately normally distributed, regardless of the shape of the population distribution, with mean $\mu$ and variance $\frac{\sigma^2}{n}$. [2]

(b) If the population is already normally distributed, the sample mean $\bar{X}$ is exactly normally distributed for any sample size $n$. Therefore, the approximation provided by the CLT is not needed; the exact distribution is known. [1]

Section C: Hypothesis Testing and Regression

9. (a) $H_0: \mu = 1200$ $H_1: \mu < 1200$ [2]

(b) Test Statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ $Z = \frac{1180 - 1200}{100/\sqrt{64}} = \frac{-20}{100/8} = \frac{-20}{12.5} = -1.6$. P-value $= P(Z < -1.6)$. Using GC/Tables: $P(Z < -1.6) \approx 0.0548$. [3] (1 mark for Z formula, 1 mark for Z value, 1 mark for p-value)

(c) Since $p\text{-value} (0.0548) > 0.05$, we do not reject $H_0$. There is insufficient evidence at the 5% level to support the claim that the mean lifetime is less than 1200 hours. [1]

10. Using GC Statistics Mode:

(a) $r \approx -0.996$ (to 3 s.f.) [2]

(b) Regression line $y = a + bx$. $a \approx 20964.28$ $b \approx -1328.57$ Equation: $y = 20964 - 1329x$ (coefficients to 4 s.f. or integers as appropriate for context, usually 3-4 s.f. required). Let's use 3 s.f.: $y = 21000 - 1330x$. Better precision for calculation: $y = 20964.3 - 1328.6x$. [3] (1 mark for a, 1 mark for b, 1 mark for equation)

(c) For $x=6$: $y = 20964.28 - 1328.57(6) = 20964.28 - 7971.42 = 12992.86$. Estimated price \approx \12,993$. [1]

(d) $x=15$ is outside the range of the data ($2 \le x \le 10$). This is extrapolation. The linear relationship may not hold for older cars (e.g., price might plateau at scrap value). Therefore, the estimate is unreliable. [2]

11. (a) Yes, because $0.03 < 0.05$. [1]

(b) No, because $0.03 > 0.01$. [1]

(c) The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. A small p-value indicates strong evidence against the null hypothesis. [2]

12. (a) For a two-tail test at 10% significance, $\alpha = 0.10$. Each tail has area $0.05$. Critical values are $z$ such that $P(Z < -z) = 0.05$ and $P(Z > z) = 0.05$. $z \approx 1.645$. Critical Region: $Z < -1.645$ or $Z > 1.645$. [2]

(b) $Z = 1.85$. Since $1.85 > 1.645$, the test statistic falls in the critical region. Reject $H_0$. [1]