A Level H1 Mathematics Statistics Probability Quiz

A-Level Maths H1 Quiz - Statistics Probability (Answers)

1. Solution:

   • Total ways = $\binom{7}{2}\binom{6}{3} + \binom{7}{1}\binom{6}{4} + \binom{7}{0}\binom{6}{5}$
   • $= (21 \times 20) + (7 \times 15) + (1 \times 6) = 420 + 105 + 6 = 531$.
   • Marks: 1 for each combination set, 1 for final sum. [3]

2. Solution:

   • Sample space $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$
   • $B = \{HHH, HHT, HTH, HTT\}$, $n(B) = 4$.
   • $A \cap B = \{HHH, HHT, HTH\}$, $n(A \cap B) = 3$.
   • $P(A|B) = \frac{3}{4} = 0.75$.
   • Marks: 1 for identifying $B$, 1 for conditional probability. [2]

3. Solution:

   • $P(X \cap Y) = P(X) + P(Y) - P(X \cup Y) = 0.6 + 0.4 - 0.8 = 0.2$.
   • Check independence: $P(X)P(Y) = 0.6 \times 0.4 = 0.24$.
   • Since $0.2 \neq 0.24$, events $X$ and $Y$ are NOT independent.
   • Marks: 1 for $P(X \cap Y)$, 1 for $P(X)P(Y)$, 1 for conclusion. [3]

4. Solution:

   • Tree: Root $\rightarrow$ Red (5/9), Blue (4/9).
   • From Red $\rightarrow$ Red (4/8), Blue (4/8).
   • From Blue $\rightarrow$ Red (5/8), Blue (3/8).
   • $P(\text{Same}) = P(RR) + P(BB) = (\frac{5}{9} \times \frac{4}{8}) + (\frac{4}{9} \times \frac{3}{8}) = \frac{20}{72} + \frac{12}{72} = \frac{32}{72} = \frac{4}{9} \approx 0.444$.
   • Marks: 1 for tree, 2 for calculation. [3]

5. Solution:

   • Total arrangements of STATISTICS (10 letters: 3S, 3T, 2I): $\frac{10!}{3!3!2!} = 50,400$.
   • Arrangements where 3 S's are together (treat SSS as one block): $\frac{8!}{3!2!} = 3,360$.
   • Not all together = $50,400 - 3,360 = 47,040$.
   • Marks: 1 for total, 1 for "together" case, 1 for subtraction. [3]

6. Solution:

   • $X \sim B(12, 0.35)$. $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
   • $P(X=0) = \binom{12}{0}(0.35)^0(0.65)^{12} \approx 0.0057$
   • $P(X=1) = \binom{12}{1}(0.35)^1(0.65)^{11} \approx 0.0368$
   • $P(X=2) = \binom{12}{2}(0.35)^2(0.65)^{10} \approx 0.1088$
   • $P(X \geq 3) = 1 - 0.1513 = 0.849$.
   • Marks: 1 for complement method, 1 for final answer. [2]

7. Solution:

   • $X \sim B(15, 0.2)$. $P(X=4) = \binom{15}{4}(0.2)^4(0.8)^{11}$.
   • $P(X=4) = 1365 \times 0.0016 \times 0.0859 \approx 0.188$.
   • Marks: 1 for formula, 1 for answer. [2]

8. Solution:

   • $Y \sim N(60, 25)$. $\sigma = 5$.
   • $z_1 = \frac{52-60}{5} = -1.6$; $z_2 = \frac{68-60}{5} = 1.6$.
   • $P(-1.6 < Z < 1.6) = \Phi(1.6) - \Phi(-1.6) = 0.9452 - 0.0548 = 0.890$.
   • Marks: 1 for $z$-scores, 2 for probability. [3]

9. Solution:

   • $P(X < 40) = 0.15 \rightarrow z_1 = -1.036 \rightarrow 40 = \mu - 1.036\sigma$
   • $P(X > 70) = 0.10 \rightarrow z_2 = 1.282 \rightarrow 70 = \mu + 1.282\sigma$
   • Subtracting: $30 = 2.318\sigma \rightarrow \sigma \approx 12.9$.
   • $\mu = 40 + 1.036(12.9) \approx 53.4$.
   • Marks: 1 per $z$-equation, 2 for solving. [4]

10. Solution:

    • $E(W) = 2E(X) - 3E(Y) = 2(10) - 3(20) = 20 - 60 = -40$.
    • $Var(W) = 2^2 Var(X) + (-3)^2 Var(Y) = 4(4) + 9(9) = 16 + 81 = 97$.
    • Marks: 1 for mean, 2 for variance. [3]

11. Solution:

    • $\bar{x} = \frac{162+170+165+178+160+172}{6} = \frac{1007}{6} \approx 167.8$ cm.
    • Marks: 2 for correct sum/division. [2]

12. Solution:

    • $s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$
    • $\sum(x_i - \bar{x})^2 = (162-167.8)^2 + \dots + (172-167.8)^2 \approx 33.64 + 4.84 + 7.84 + 104.04 + 60.84 + 17.64 = 228.84$.
    • $s^2 = \frac{228.84}{5} \approx 45.8$.
    • Marks: 2 for sum of squares, 1 for dividing by $n-1$. [3]

13. Solution:

    • $\bar{X} \sim N(100, \frac{15^2}{36}) = N(100, 6.25)$. $\sigma_{\bar{x}} = 2.5$.
    • $z_1 = \frac{97-100}{2.5} = -1.2$; $z_2 = \frac{103-100}{2.5} = 1.2$.
    • $P(-1.2 < Z < 1.2) = 0.8849 - 0.1151 = 0.770$.
    • Marks: 1 for $\sigma_{\bar{x}}$, 1 for $z$-scores, 2 for probability. [4]

14. Solution:

    • 1. Assign a unique number from 1 to 2,000 to each household.
    • 2. Calculate interval $k = 2000/50 = 40$.
    • 3. Pick a random starting number $r$ between 1 and 40.
    • 4. Select households $r, r+40, r+80 \dots$
    • Marks: 1 for interval, 1 for random start/selection process. [2]

15. Solution:

    • $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \rightarrow 2 = \frac{12}{\sqrt{n}}$
    • $\sqrt{n} = 6 \rightarrow n = 36$.
    • Marks: 2 for formula setup, 1 for answer. [3]

16. Solution:

    • $H_0: \mu = 1200, H_1: \mu < 1200$.
    • $z = \frac{1170 - 1200}{80/\sqrt{40}} = \frac{-30}{12.65} \approx -2.37$.
    • Critical value for 5% (one-tail) is $z = -1.645$.
    • Since $-2.37 < -1.645$, reject $H_0$.
    • Conclusion: There is sufficient evidence at the 5% level to suggest the mean life is less than 1,200 hours.
    • Marks: 1 for hypotheses, 2 for $z$-stat, 1 for critical value, 1 for conclusion. [5]

17. Solution:

    • Two-tailed test at 1% significance.
    • $\alpha/2 = 0.005$. $z = \pm 2.576$.
    • Marks: 2 for correct values. [2]

18. Solution:

    • For every additional hour studied, the test score is predicted to increase by 2.5 marks on average.
    • Marks: 2 for clear contextual interpretation. [2]

19. Solution:

    • $\bar{x} = 6, \bar{y} = 15$.
    • $\sum(x-\bar{x})(y-\bar{y}) = (-4)(-10) + (-2)(-4) + (0)(0) + (2)(6) + (4)(10) = 40+8+0+12+40 = 100$.
    • $\sum(x-\bar{x})^2 = 16+4+0+4+16 = 40$.
    • $\sum(y-\bar{y})^2 = 100+16+0+36+100 = 252$.
    • $r = \frac{100}{\sqrt{40 \times 252}} = \frac{100}{100.4} \approx 0.996$.
    • Marks: 1 for means, 1 for numerator, 1 for denominator, 1 for final $r$. [4]

20. Solution:

    • $Y = 2.5(7) + 12 = 17.5 + 12 = 29.5$.
    • Since $7$ is within the range $[2, 10]$, this is interpolation.
    • Marks: 2 for calculation, 1 for interpolation. [3]