A Level H1 Mathematics Practice Paper 2

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

Version 2 of 5

Section A: Probability and Distributions

1. (a) $X \sim B(25, 0.04)$
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(b) $P(X=2) = \binom{25}{2} (0.04)^2 (0.96)^{23} \approx 0.187$
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(1 mark for correct substitution, 1 mark for answer)

(c) $P(X \ge 1) = 1 - P(X=0) = 1 - (0.96)^{25} \approx 1 - 0.360 = 0.640$
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(1 mark for method $1-P(X=0)$, 1 mark for answer)

2. (a) Let $Y \sim B(10, 0.3)$.
$P(Y > 7) = P(Y=8) + P(Y=9) + P(Y=10)$
$= \binom{10}{8}(0.3)^8(0.7)^2 + \binom{10}{9}(0.3)^9(0.7)^1 + \binom{10}{10}(0.3)^{10}$
$\approx 0.00145 + 0.00014 + 0.00001 = 0.00160$
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(1 mark for correct sum setup, 1 mark for answer)

(b) $E(Y) = np = 10 \times 0.3 = 3$
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3. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.2 = 0.9$
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(b) Check if $P(A \cap B) = P(A)P(B)$.
$P(A)P(B) = 0.6 \times 0.5 = 0.3$.
Since $0.2 \neq 0.3$, $A$ and $B$ are not independent.
[2]
(1 mark for calculation of product, 1 mark for conclusion)

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$.
$P(B') = 1 - 0.5 = 0.5$.
$P(A \cap B') = P(A) - P(A \cap B) = 0.6 - 0.2 = 0.4$.
$P(A | B') = \frac{0.4}{0.5} = 0.8$.
[2]
(1 mark for numerator/denominator logic, 1 mark for answer)

4. Let $M \sim N(150, 12^2)$.

(a) $P(140 < M < 165)$.
Using GC: normalcdf(140, 165, 150, 12) $\approx 0.691$.
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(b) $P(M > m) = 0.10 \Rightarrow P(M < m) = 0.90$.
Using GC: invNorm(0.90, 150, 12) $\approx 165.38$.
$m \approx 165$ g (3 s.f.).
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5. $X \sim N(20, 9)$, $Y \sim N(15, 16)$. Independent.

(a) $E(W) = E(2X - Y) = 2E(X) - E(Y) = 2(20) - 15 = 40 - 15 = 25$.
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(b) $Var(W) = Var(2X - Y) = 2^2 Var(X) + (-1)^2 Var(Y) = 4(9) + 1(16) = 36 + 16 = 52$.
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(1 mark for formula $4Var(X) + Var(Y)$, 1 mark for answer)

Section B: Sampling and Estimation

6. $n=50, \sum h = 8250, \sum h^2 = 1,361,500$.

(a) Unbiased estimate of mean $\bar{x} = \frac{8250}{50} = 165$ cm.
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(b) Unbiased estimate of variance $s^2 = \frac{n}{n-1} \left( \frac{\sum h^2}{n} - \bar{x}^2 \right)$.
$s^2 = \frac{50}{49} \left( \frac{1,361,500}{50} - 165^2 \right)$
$s^2 = \frac{50}{49} (27,230 - 27,225) = \frac{50}{49} (5) \approx 5.10$.
[3]
(1 mark for formula, 1 mark for substitution, 1 mark for answer)

7. $\sigma = 15, n=36, \bar{x} = 120$. 95% CI.

(a) Formula: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$.
$z_{0.025} = 1.96$.
Margin of error $= 1.96 \times \frac{15}{\sqrt{36}} = 1.96 \times 2.5 = 4.9$.
CI: $120 \pm 4.9 \Rightarrow (115.1, 124.9)$.
[3]
(1 mark for standard error, 1 mark for z-value/margin, 1 mark for interval)

(b) Yes, 115 is not in the interval $(115.1, 124.9)$, so it is not a plausible value at the 95% confidence level.
(Note: 115 is very close, but strictly outside. If student says "No" with correct reasoning, accept. If student calculates test stat, also accept.)
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8. $V \sim N(500, 5^2)$. Sample $n=16$.

(a) Distribution of sample mean $\bar{V} \sim N(500, \frac{5^2}{16}) = N(500, 1.5625)$.
$P(\bar{V} < 498)$.
$Z = \frac{498 - 500}{5/\sqrt{16}} = \frac{-2}{1.25} = -1.6$.
$P(Z < -1.6) \approx 0.0548$.
[3]
(1 mark for dist of mean, 1 mark for standardizing, 1 mark for prob)

(b) The Central Limit Theorem is not required because the population distribution is already normal. The sample mean of a normal population is always normal, regardless of sample size.
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9. $W \sim N(25, 9)$. $\sigma = 3$.

(a) $P(W > 28)$.
$Z = \frac{28-25}{3} = 1$.
$P(Z > 1) = 1 - 0.8413 = 0.1587$.
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(b) Sample $n=9$. $\bar{W} \sim N(25, \frac{9}{9}) = N(25, 1)$.
$P(\bar{W} > 28)$.
$Z = \frac{28-25}{1} = 3$.
$P(Z > 3) \approx 0.00135$.
[3]
(1 mark for new variance, 1 mark for Z, 1 mark for prob)

10. (a) Assign each of the 200 residents a unique number from 1 to 200. Use a random number generator to produce 20 distinct integers between 1 and 200. Select the residents corresponding to these numbers.
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(1 mark for numbering, 1 mark for random selection of distinct numbers)

(b) Simple random sampling ensures every resident has an equal chance of being selected, reducing selection bias. Convenience sampling may over-represent certain groups (e.g., those home during the day).
[1]

Section C: Hypothesis Testing and Regression

11. $\mu_0 = 1200, \sigma = 100, n=50, \bar{x} = 1180$. $\alpha = 0.05$.

(a) $H_0: \mu = 1200$
$H_1: \mu < 1200$
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(b) Test statistic $Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{1180 - 1200}{100/\sqrt{50}} = \frac{-20}{14.14} \approx -1.414$.
Critical value for one-tail 5%: $-1.645$.
Since $-1.414 > -1.645$ (or P-value $0.0786 > 0.05$), we do not reject $H_0$.
Conclusion: There is insufficient evidence at the 5% level to suggest the mean lifetime is less than 1200 hours.
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(1 mark for Z calc, 1 mark for critical value/p-value, 1 mark for comparison, 1 mark for context conclusion)

12. (a) Using GC: $r \approx 0.986$.
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(b) There is a strong, positive, linear correlation between age and reaction time. As age increases, reaction time tends to increase.
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13. (a) Using GC: $y = 196.67 + 3.57x$ (values approx).
Exact: $b = \frac{S_{xy}}{S_{xx}}$, $a = \bar{y} - b\bar{x}$.
$\bar{x} = 45, \bar{y} = 340$.
$S_{xx} = 1750, S_{xy} = 6250$.
$b = 6250/1750 = 3.57$.
$a = 340 - 3.57(45) = 179.35$ (Check: GC gives $a \approx 179.3, b \approx 3.57$).
Equation: $y = 179.3 + 3.57x$.
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(b) $y = 179.3 + 3.57(45) = 340$ ms.
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(c) Age 90 is outside the range of the data (20-70). This is extrapolation, and the linear relationship may not hold for older ages.
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14. Two-tail, $\alpha = 0.10$.

(a) Critical values are $\pm z_{0.05} = \pm 1.645$.
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(b) Test statistic $Z = 1.85$.
Since $1.85 > 1.645$, the result falls in the critical region.
Reject $H_0$.
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15. $\mu_0 = 10, \sigma = 3, n=25$. One-tail (increase), $\alpha = 0.05$.

(a) Critical region for $\bar{X}$.
Critical $Z = 1.645$.
$\bar{X}_{crit} = \mu_0 + 1.645 \frac{\sigma}{\sqrt{n}} = 10 + 1.645 \frac{3}{5} = 10 + 0.987 = 10.987$.
Critical region: $\bar{X} > 10.99$ (2 d.p.).
[3]
(1 mark for SE, 1 mark for Z, 1 mark for boundary)

(b) $11.2 > 10.99$, so reject $H_0$. There is evidence the mean time has increased.
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16. $p = 0.4, n=100$.

(a) $\hat{P} \sim N(p, \frac{p(1-p)}{n}) = N(0.4, \frac{0.4(0.6)}{100}) = N(0.4, 0.0024)$.
[2]
(1 mark for mean, 1 mark for variance)

(b) $P(\hat{P} > 0.45)$.
$Z = \frac{0.45 - 0.4}{\sqrt{0.0024}} = \frac{0.05}{0.04899} \approx 1.02$.
$P(Z > 1.02) = 1 - 0.8461 = 0.1539$.
[3]

17. $M \sim N(175, 7^2)$, $W \sim N(162, 6^2)$.

(a) Let $D = M - W$.
$E(D) = 175 - 162 = 13$.
$Var(D) = 7^2 + 6^2 = 49 + 36 = 85$.
$D \sim N(13, 85)$.
$P(D > 0)$.
$Z = \frac{0 - 13}{\sqrt{85}} = \frac{-13}{9.22} \approx -1.41$.
$P(Z > -1.41) = P(Z < 1.41) \approx 0.9207$.
[3]

(b) Let $T = M_1 + M_2$.
$E(T) = 175 + 175 = 350$.
$Var(T) = 7^2 + 7^2 = 98$.
$T \sim N(350, 98)$.
$P(T > 360)$.
$Z = \frac{360 - 350}{\sqrt{98}} = \frac{10}{9.90} \approx 1.01$.
$P(Z > 1.01) = 1 - 0.8438 = 0.1562$.
[3]

18. (a) $E(X) = \sum x P(x) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7$.
[2]

(b) $E(X^2) = 1^2(0.1) + 2^2(0.3) + 3^2(0.4) + 4^2(0.2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1$.
$Var(X) = E(X^2) - [E(X)]^2 = 8.1 - (2.7)^2 = 8.1 - 7.29 = 0.81$.
[3]

19. (a) Tree Diagram:
First branch: Type A (0.6), Type B (0.4).
Second branch from A: Defective (0.02), Not Def (0.98).
Second branch from B: Defective (0.05), Not Def (0.95).
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(b) $P(D) = P(A \cap D) + P(B \cap D) = (0.6)(0.02) + (0.4)(0.05) = 0.012 + 0.020 = 0.032$.
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(c) $P(A | D) = \frac{P(A \cap D)}{P(D)} = \frac{0.012}{0.032} = \frac{12}{32} = 0.375$.
[2]

20. $E \sim N(150, 30^2)$.

(a) $P(E > 200)$.
$Z = \frac{200 - 150}{30} = \frac{50}{30} = 1.67$.
$P(Z > 1.67) = 1 - 0.9525 = 0.0475$.
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(b) $P(E < k) = 0.25$.
Using GC: invNorm(0.25, 150, 30) $\approx 129.77$.
k \approx \129.77$.
[2]