A Level H1 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Maths H1 A-Level (Version 2) Answer Key

Subject: Maths H1
Level: A-Level
Total Marks: 60

Section A: Probability and Counting

1. [2]
Choose 2 boys from 7: $\binom{7}{2} = 21$
Choose 2 girls from 5: $\binom{5}{2} = 10$
Total = $21 \times 10 = 210$ ways.
Teaching note: Use combinations since order does not matter. Common mistake: using permutations.

2. [4]
(a) [2] Tree diagram:
First draw: R (6/10), B (4/10).
Second draw without replacement:

• After R: R (5/9), B (4/9)
• After B: R (6/9), B (3/9)
  Marking: Correct branches and probabilities.
  (b) [2] Same colour = RR or BB:
  $P(RR) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$
  $P(BB) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$
  Total = $\frac{42}{90} = \frac{7}{15} \approx 0.467$
  Note: Without replacement reduces denominator to 9.

3. [3]
(a) [1] $P(A \cup B) = 0.55 + 0.40 - 0.20 = 0.75$
(b) [2] Independent if $P(A \cap B) = P(A)P(B)$.
$P(A)P(B) = 0.55 \times 0.40 = 0.22 \neq 0.20$. Not independent.
Teaching: Compare product to intersection.

4. [2]
$X \sim B(15, 0.30)$. $P(X \geq 6) = 1 - P(X \leq 5)$.
Using GC/binomial CDF: $P(X \leq 5) \approx 0.7216$, so answer $\approx 0.278$.
Note: Do not use normal approximation (excluded).

5. [2]
Red cards = 26, hearts = 13. Given red, $P(\text{heart}) = 13/26 = 0.5$.
Formula: $P(H|R) = P(H \cap R)/P(R) = (13/52)/(26/52) = 0.5$.

6. [2]
3 distinct letters from 5, order matters: $^5P_3 = 5 \times 4 \times 3 = 60$.
Note: Password order matters → permutation.

Section A Total: 15 marks

Section B: Distributions and Sampling

7. [4]
(a) [2] $X \sim N(120, 15^2)$.
$Z_1 = (105-120)/15 = -1$, $Z_2 = (135-120)/15 = 1$.
$P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826$.
(b) [2] 90th percentile: $Z_{0.90} = 1.282$, $x = 120 + 1.282(15) = 139.23$ g.

8. [2]
Mean = $np = 20 \times 0.25 = 5$.
Variance = $np(1-p) = 20 \times 0.25 \times 0.75 = 3.75$.

9. [3]
$n=12$, $\sum x = 14+16+15+18+12+17+15+19+13+16+15+14 = 184$.
$\bar{x} = 184/12 = 15.33$ h.
$\sum x^2 = 196+256+225+324+144+289+225+361+169+256+225+196 = 2866$.
$s^2 = \frac{2866 - 12(15.333)^2}{11} = \frac{2866 - 2823.11}{11} = 3.90$ h².
Marking: 1 for mean, 2 for variance with unbiased denominator.

10. [3]
By CLT, $\bar{X} \sim N(62, 7^2/36) = N(62, 1.361)$.
$P(\bar{X} > 64) = P(Z > (64-62)/(7/6)) = P(Z > 1.714) = 0.0433$.
Note: CLT applies as n=36 ≥ 30.

11. [2]
$Z = (40-50)/8 = -1.25$, $P(Z < -1.25) = 0.1056$.

12. [2]
$X \sim B(50, 0.04)$. $P(X \leq 2) = P(0)+P(1)+P(2) \approx 0.1299 + 0.2706 + 0.2762 = 0.6767$.
(GC CDF: 0.6767)

13. [2]
$E(Y) = 3E(X) - 10 = 3(200) - 10 = 590$.
$Var(Y) = 3^2 Var(X) = 9 \times 64 = 576$.

Section B Total: 22 marks

Section C: Hypothesis Testing, Correlation and Regression

14. [5]
$H_0: \mu = 5.0$, $H_1: \mu > 5.0$ (1 mark)
Test stat: $Z = \frac{5.4 - 5.0}{1.2/\sqrt{35}} = \frac{0.4}{0.2029} = 1.97$ (2 marks)
Critical 5% one-tail: 1.645 (1 mark)
Since 1.97 > 1.645, reject $H_0$ (0.5 mark)
There is evidence at 5% that mean daily screen time > 5.0 h (0.5 mark).

15. [4]
(a) [2] Scatter diagram: x-axis 160–185, y-axis 50–70, points plotted as per placeholder, upward trend.
(b) [2] $r = 0.98$ → strong positive linear correlation between height and weight.

16. [2]
$y = 0.75(172) - 72 = 129 - 72 = 57$ kg.

17. [2]
Correlation ≠ causation; confounding variables (diet, sleep) may affect both; or reverse causality possible. (2 points)

18. [3]
$n=40$, $\bar{x} = 820/40 = 20.5$.
$s^2 = \frac{17200 - 40(20.5)^2}{39} = \frac{17200 - 16810}{39} = 9.97$.

19. [3]
$E(2X - Y) = 2(10) - 20 = 0$.
$Var(2X - Y) = 4(4) + 9 = 25$ (independent → add variances with coefficients squared).

20. [3]
$H_0: \mu = 70$, $H_1: \mu > 70$.
$Z = (73-70)/(8/\sqrt{32}) = 3/1.414 = 2.12$.
1% crit = 2.326. Not reject; insufficient evidence mean > 70.

Section C Total: 23 marks
Paper Total: 60 marks