A Level H1 Mathematics Statistics Probability Quiz

A-Level Maths H1 Quiz - Statistics Probability: Answer Key

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Question 1 [3 marks]

(a) Unbiased estimate of population mean:

$\bar{x} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Answer: $\bar{x} = 13.6$ hours (3 s.f.) [1]

(b) Unbiased estimate of population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$

Deviations from mean: $(12-13.625)^2 = 2.6406$, $(15-13.625)^2 = 1.8906$, $(10-13.625)^2 = 13.1406$, $(18-13.625)^2 = 19.1406$, $(14-13.625)^2 = 0.1406$, $(11-13.625)^2 = 6.8906$, $(16-13.625)^2 = 5.6406$, $(13-13.625)^2 = 0.3906$

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{8-1} = \frac{49.875}{7} = 7.125$

Answer: $s^2 = 7.13$ hours² (3 s.f.) [2]

Common mistake: Using denominator $n = 8$ instead of $n - 1 = 7$. This gives the biased sample variance, not the unbiased estimate of the population variance.

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Question 2 [3 marks]

(a) Sample mean:

$\bar{x} = \frac{\sum x}{n} = \frac{348.0}{10} = 34.80 \text{ cm}$

Answer: $\bar{x} = 34.8$ cm [1]

(b) Unbiased estimate of population variance:

$s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} = \frac{12\,174.0 - \frac{(348.0)^2}{10}}{9}$

$= \frac{12\,174.0 - \frac{121\,104}{10}}{9} = \frac{12\,174.0 - 12\,110.4}{9} = \frac{63.6}{9} = 7.0666...$

Answer: $s^2 = 7.07$ cm² (3 s.f.) [2]

Teaching note: The computational formula $s^2 = \frac{\sum x^2 - (\sum x)^2/n}{n-1}$ is equivalent to the definition formula but often easier to use with summary statistics. Always use $n-1$ in the denominator for the unbiased estimate.

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Question 3 [4 marks]

(a) Sample mean:

$\bar{x} = \frac{25 + 32 + 28 + 40 + 35 + 30}{6} = \frac{190}{6} = 31.666... \approx 31.7 \text{ minutes}$

Biased variance (denominator $n = 6$):

$\sum(x_i - \bar{x})^2$:
$(25-31.667)^2 = 44.444$, $(32-31.667)^2 = 0.111$, $(28-31.667)^2 = 13.444$, $(40-31.667)^2 = 69.444$, $(35-31.667)^2 = 11.111$, $(30-31.667)^2 = 2.778$

$\sum(x_i - \bar{x})^2 = 141.333$

$\text{Biased variance} = \frac{141.333}{6} = 23.556$

$\text{Sample standard deviation} = \sqrt{23.556} = 4.853...$

Answer: $\bar{x} = 31.7$ min, $s_{\text{biased}} = 4.85$ min (3 s.f.) [2]

(b) The unbiased estimate uses $n-1$ because the sample mean $\bar{x}$ is itself calculated from the same data, which constrains the deviations $\sum(x_i - \bar{x})^2 = 0$. This means only $n-1$ of the deviations are "free to vary" — the last one is determined by the others. Using $n$ systematically underestimates the true population variance (it is biased). Dividing by $n-1$ corrects this bias, giving an unbiased estimator. This is known as Bessel's correction. [2]

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Question 4 [4 marks]

(a) The unbiased estimate of the population mean is simply the sample mean.

Answer: 5.2 hours [1]

(b) Standard error of the mean:

$\text{SE} = \frac{s}{\sqrt{n}} = \frac{1.8}{\sqrt{50}} = \frac{1.8}{7.0711} = 0.2546...$

Answer: SE = 0.255 hours (3 s.f.) [1]

(c) The standard error of the mean measures the variability of sample means from repeated samples of the same size ($n = 50$) drawn from the same population. It tells us how much the sample mean (5.2 hours) is likely to vary from one sample to another. A smaller SE indicates that the sample mean is a more precise estimate of the true population mean. In this context, if we were to repeatedly survey 50 teenagers, the sample mean daily screen time would typically vary by about 0.255 hours from the true population mean. [2]

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Question 5 [4 marks]

(a) Unbiased estimate of population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{80}{5-1} = \frac{80}{4} = 20$

Answer: $s^2 = 20$ [1]

(b) The new value $x_6 = 26$ is 6 units above the original mean of 20. Adding this value will:

• Pull the new mean upward (new mean = $\frac{5 \times 20 + 26}{6} = 21$)
• The new data point is far from the original mean, adding a large squared deviation
• Although the denominator increases from 4 to 5, the sum of squared deviations increases substantially because $x_6 = 26$ is distant from both the old and new means

The new sum of squared deviations: Old sum = 80. Using the identity for updating: $\sum_{\text{new}}(x_i - \bar{x}_{\text{new}})^2 = \sum_{\text{old}}(x_i - \bar{x}_{\text{old}})^2 + \frac{n}{n+1}(x_{n+1} - \bar{x}_{\text{old}})^2$ $= 80 + \frac{5}{6}(26 - 20)^2 = 80 + \frac{5}{6}(36) = 80 + 30 = 110$

New unbiased variance: $s^2_{\text{new}} = \frac{110}{5} = 22$

Since $22 > 20$, the unbiased estimate of the population variance increases. [3]

Answer: The variance increases because the new value is far from the original mean, adding more spread to the data.

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Question 6 [5 marks]

Sample space for two dice rolls: $6 \times 6 = 36$ equally likely outcomes.

(a) Sum = 7: outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — 6 outcomes.

$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$ [1]

(b) First roll = 4 (1 way), second roll > 4, i.e., 5 or 6 (2 ways).

$P(\text{first} = 4 \text{ and second} > 4) = \frac{1}{6} \times \frac{2}{6} = \frac{2}{36} = \frac{1}{18}$

Answer: $\frac{1}{18}$ [2]

(c) $P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - \frac{5}{6} \times \frac{5}{6} = 1 - \frac{25}{36} = \frac{11}{36}$

Answer: $\frac{11}{36}$ [2]

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Question 7 [5 marks]

Let $F$ = female, $M$ = male, $L$ = owns a laptop.
$P(F) = 0.6$, $P(M) = 0.4$, $P(L|F) = 0.3$, $P(L|M) = 0.5$.

(a) By the law of total probability:

$P(L) = P(L|F)P(F) + P(L|M)P(M) = (0.3)(0.6) + (0.5)(0.4) = 0.18 + 0.20 = 0.38$

Answer: $P(L) = 0.38$ [2]

(b) By Bayes' theorem:

$P(F|L) = \frac{P(L|F)P(F)}{P(L)} = \frac{(0.3)(0.6)}{0.38} = \frac{0.18}{0.38} = \frac{9}{19} \approx 0.4737$

Answer: $P(F|L) = \frac{9}{19} \approx 0.474$ (3 s.f.) [3]

Teaching note: This is a classic Bayes' theorem problem. The key insight is that even though there are more female students, male students are more likely to own laptops, so a laptop owner is almost equally likely to be male or female.

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Question 8 [5 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Drawing 3 without replacement.

(a) $P(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22}$

Alternatively: $\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$

Answer: $\frac{1}{22}$ [2]

(b) All different colours means 1 red, 1 blue, 1 green:

$P(\text{1 of each colour}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11}$

Answer: $\frac{3}{11}$ [3]

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Question 9 [5 marks]

Let $X \sim B(7, 0.25)$ where $X$ = number of rainy days in a week.

(a) $P(X = 3) = \binom{7}{3}(0.25)^3(0.75)^4 = 35 \times 0.015625 \times 0.31640625 = 0.1730...$

Answer: $P(X = 3) = 0.173$ (3 s.f.) [2]

(b) $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

$P(X = 0) = (0.75)^7 = 0.13348...$

$P(X = 1) = \binom{7}{1}(0.25)^1(0.75)^6 = 7 \times 0.25 \times 0.17798 = 0.31146...$

$P(X \geq 2) = 1 - 0.13348 - 0.31146 = 0.55505...$

Answer: $P(X \geq 2) = 0.555$ (3 s.f.) [3]

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Question 10 [5 marks]

(a) Using the addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.7 = 0.4 + 0.6 - P(A \cap B)$ $P(A \cap B) = 1.0 - 0.7 = 0.3$

Answer: $P(A \cap B) = 0.3$ [1]

(b) For independence, check if $P(A \cap B) = P(A) \times P(B)$:

$P(A) \times P(B) = 0.4 \times 0.6 = 0.24$

Since $0.3 \neq 0.24$, events $A$ and $B$ are not independent. [2]

(c) $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.7 = 0.3$

Answer: $P(A' \cap B') = 0.3$ [2]

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Question 11 [5 marks]

(a) Sum of probabilities = 1:

$0.1 + 0.2 + 0.3 + a + 0.1 = 1$ $0.7 + a = 1 \implies a = 0.3$

Answer: $a = 0.3$ [1]

(b) $E(X) = \sum x \cdot P(X=x)$:

$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.3) + 5(0.1)$ $= 0.1 + 0.4 + 0.9 + 1.2 + 0.5 = 3.1$

Answer: $E(X) = 3.1$ [2]

(c) $E(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.3) + 5^2(0.1)$ $= 0.1 + 0.8 + 2.7 + 4.8 + 2.5 = 10.9$

$Var(X) = E(X^2) - [E(X)]^2 = 10.9 - (3.1)^2 = 10.9 - 9.61 = 1.29$

Answer: $Var(X) = 1.29$ [2]

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Question 12 [5 marks]

$X \sim \text{Po}(2.5)$

(a) $P(X = 3) = \frac{e^{-2.5}(2.5)^3}{3!} = \frac{e^{-2.5} \times 15.625}{6} = e^{-2.5} \times 2.6042 = 0.2138...$

Answer: $P(X = 3) = 0.214$ (3 s.f.) [2]

(b) $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

$P(X = 0) = e^{-2.5} = 0.082085$

$P(X = 1) = e^{-2.5} \times 2.5 = 0.20521$

$P(X \geq 2) = 1 - 0.082085 - 0.20521 = 0.71270...$

Answer: $P(X \geq 2) = 0.713$ (3 s.f.) [2]

(c) $P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = 2) = \frac{e^{-2.5}(2.5)^2}{2!} = e^{-2.5} \times 3.125 = 0.25652$

$P(X < 4) = 0.082085 + 0.20521 + 0.25652 + 0.21376 = 0.75758...$

Answer: $P(X < 4) = 0.758$ (3 s.f.) [1]

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Question 13 [5 marks]

$X \sim B(5, 0.6)$

(a) $P(X = 3) = \binom{5}{3}(0.6)^3(0.4)^2 = 10 \times 0.216 \times 0.16 = 0.3456$

Answer: $P(X = 3) = 0.346$ (3 s.f.) [2]

(b) $P(X \geq 4) = P(X = 4) + P(X = 5)$

$P(X = 4) = \binom{5}{4}(0.6)^4(0.4)^1 = 5 \times 0.1296 \times 0.4 = 0.2592$

$P(X = 5) = (0.6)^5 = 0.07776$

$P(X \geq 4) = 0.2592 + 0.07776 = 0.33696$

Answer: $P(X \geq 4) = 0.337$ (3 s.f.) [1]

(c) $E(X) = np = 5 \times 0.6 = 3$

$Var(X) = np(1-p) = 5 \times 0.6 \times 0.4 = 1.2$

Answer: $E(X) = 3$, $Var(X) = 1.2$ [2]

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Question 14 [5 marks]

$X \sim B(20, 0.3)$

(a) $P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14}$

Using calculator: $\binom{20}{6} = 38\,760$

$P(X = 6) = 38\,760 \times 0.000729 \times 0.067822 = 0.1916...$

Answer: $P(X = 6) = 0.192$ (3 s.f.) [2]

(b) $P(X \leq 4) = \sum_{k=0}^{4}\binom{20}{k}(0.3)^k(0.7)^{20-k}$

Using calculator or cumulative binomial tables:

$P(X = 0) = 0.000798$, $P(X = 1) = 0.006839$, $P(X = 2) = 0.027846$, $P(X = 3) = 0.071604$, $P(X = 4) = 0.130421$

$P(X \leq 4) = 0.23751...$

Answer: $P(X \leq 4) = 0.238$ (3 s.f.) [2]

(c) We need the smallest $k$ such that $P(X \leq k) \geq 0.95$.

$P(X \leq 8) \approx 0.8867$, $P(X \leq 9) \approx 0.9520$

Answer: $k = 9$ [1]

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Question 15 [5 marks]

(a) $X \sim \text{Po}(4)$ for a given minute.

$P(X = 5) = \frac{e^{-4}(4)^5}{5!} = \frac{e^{-4} \times 1024}{120} = e^{-4} \times 8.5333 = 0.1563...$

Answer: $P(X = 5) = 0.156$ (3 s.f.) [2]

(b) $P(X \geq 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)$

$P(X = 0) = e^{-4} = 0.018316$

$P(X = 1) = 4e^{-4} = 0.073263$

$P(X = 2) = \frac{16e^{-4}}{2} = 0.146525$

$P(X \geq 3) = 1 - 0.018316 - 0.073263 - 0.146525 = 0.76190...$

Answer: $P(X \geq 3) = 0.762$ (3 s.f.) [2]

(c) Over 30 minutes, the total number of calls $Y \sim \text{Po}(4 \times 30) = \text{Po}(120)$.

Assumption: Calls occur independently and the rate is constant over the 30-minute period.

$P(Y = 120) = \frac{e^{-120}(120)^{120}}{120!}$

Using Stirling's approximation or a calculator, this is approximately:

$P(Y = 120) \approx \frac{1}{\sqrt{2\pi \times 120}} = \frac{1}{\sqrt{753.98}} \approx 0.0364$

(More precisely, using a calculator: $P(Y = 120) \approx 0.0365$)

Answer: $P(Y = 120) \approx 0.0365$ (3 s.f.) [1]

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Question 16 [5 marks]

$X \sim N(50, 25)$, so $\mu = 50$, $\sigma = 5$.

(a) $P(X > 55) = P\left(Z > \frac{55 - 50}{5}\right) = P(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$

Answer: $P(X > 55) = 0.159$ (3 s.f.) [1]

(b) $P(45 < X < 58) = P\left(\frac{45-50}{5} < Z < \frac{58-50}{5}\right) = P(-1 < Z < 1.6)$

$= \Phi(1.6) - \Phi(-1) = 0.9452 - 0.1587 = 0.7865$

Answer: $P(45 < X < 58) = 0.787$ (3 s.f.) [2]

(c) $P(X < k) = 0.9 \implies P\left(Z < \frac{k-50}{5}\right) = 0.9$

$\frac{k-50}{5} = 1.2816$ (from standard normal tables)

$k = 50 + 5(1.2816) = 56.408$

Answer: $k = 56.4$ (3 s.f.) [2]

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Question 17 [5 marks]

$X \sim N(150, 144)$, so $\mu = 150$, $\sigma = 12$.

(a) $P(140 < X < 165) = P\left(\frac{140-150}{12} < Z < \frac{165-150}{12}\right) = P(-0.8333 < Z < 1.25)$

$= \Phi(1.25) - \Phi(-0.8333) = 0.8944 - 0.2023 = 0.6921$

Answer: $P(140 < X < 165) = 0.692$ (3 s.f.) [2]

(b) $P(X > m) = 0.15 \implies P(X < m) = 0.85$

$P\left(Z < \frac{m-150}{12}\right) = 0.85$

$\frac{m-150}{12} = 1.0364$

$m = 150 + 12(1.0364) = 162.433$

Answer: $m = 162$ g (3 s.f.) [3]

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Question 18 [6 marks]

$X \sim N(172, 64)$, so $\mu = 172$, $\sigma = 8$.

(a) $P(165 < X < 180) = P\left(\frac{165-172}{8} < Z < \frac{180-172}{8}\right) = P(-0.875 < Z < 1)$

$= \Phi(1) - \Phi(-0.875) = 0.8413 - 0.1908 = 0.6505$

Answer: $P(165 < X < 180) = 0.651$ (3 s.f.) [2]

(b) Let $p = 0.6505$ be the probability from part (a). Let $Y \sim B(4, 0.6505)$ be the number of males (out of 4) with heights in the range.

$P(Y = 2) = \binom{4}{2}(0.6505)^2(1-0.6505)^2 = 6 \times 0.4232 \times 0.1222 = 0.3103...$

Answer: $P(Y = 2) = 0.310$ (3 s.f.) [2]

(c) By the Central Limit Theorem, the sample mean $\bar{X} \sim N\left(172, \frac{64}{25}\right) = N(172, 2.56)$, so $\sigma_{\bar{X}} = \frac{8}{5} = 1.6$.

$P(170 < \bar{X} < 174) = P\left(\frac{170-172}{1.6} < Z < \frac{174-172}{1.6}\right) = P(-1.25 < Z < 1.25)$

$= \Phi(1.25) - \Phi(-1.25) = 0.8944 - 0.1056 = 0.7888$

Answer: $P(170 < \bar{X} < 174) = 0.789$ (3 s.f.) [2]

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Question 19 [6 marks]

$X \sim N(45, 36)$, so $\mu = 45$, $\sigma = 6$.

(a) $P(X > 50) = P\left(Z > \frac{50-45}{6}\right) = P(Z > 0.8333) = 1 - \Phi(0.8333) = 1 - 0.7977 = 0.2023$

Answer: $P(X > 50) = 0.202$ (3 s.f.) [2]

(b) $P(X < t) = 0.80 \implies \frac{t-45}{6} = 0.8416$

$t = 45 + 6(0.8416) = 50.050$

Answer: $t = 50.1$ minutes (3 s.f.) [2]

(c) From part (a), $p = 0.2023$. Let $W \sim B(30, 0.2023)$.

$P(W \geq 5) = 1 - P(W \leq 4)$

$E(W) = 30 \times 0.2023 = 6.069$, $Var(W) = 30 \times 0.2023 \times 0.7977 = 4.841$

Using binomial calculation (calculator):

$P(W = 0) = 0.00227$, $P(W = 1) = 0.01378$, $P(W = 2) = 0.04106$, $P(W = 3) = 0.07993$, $P(W = 4) = 0.11634$

$P(W \leq 4) = 0.25338$

$P(W \geq 5) = 1 - 0.25338 = 0.74662$

Answer: $P(W \geq 5) = 0.747$ (3 s.f.) [2]

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Question 20 [7 marks]

$X \sim N(\mu, 100)$, so $\sigma = 10$.

(a) $P(X < 480) = 0.05$

$P\left(Z < \frac{480 - \mu}{10}\right) = 0.05$

$\frac{480 - \mu}{10} = -1.6449$ (the 5th percentile of standard normal)

$480 - \mu = -16.449$

$\mu = 480 + 16.449 = 496.449$

Answer: $\mu \approx 496.45$ g [2]

(b) $P(490 < X < 510) = P\left(\frac{490-496.45}{10} < Z < \frac{510-496.45}{10}\right) = P(-0.645 < Z < 1.355)$

$= \Phi(1.355) - \Phi(-0.645) = 0.9123 - 0.2595 = 0.6528$

Answer: $P(490 < X < 510) = 0.653$ (3 s.f.) [2]

(c) $\bar{X} \sim N\left(496.45, \frac{100}{16}\right) = N(496.45, 6.25)$, so $\sigma_{\bar{X}} = 2.5$.

$P(492 < \bar{X} < 500) = P\left(\frac{492-496.45}{2.5} < Z < \frac{500-496.45}{2.5}\right) = P(-1.78 < Z < 1.42)$

$= \Phi(1.42) - \Phi(-1.78) = 0.9222 - 0.0375 = 0.8847$

Answer: $P(492 < \bar{X} < 500) = 0.885$ (3 s.f.) [1]

(d) We want $P(X < 480) = 0.01$ with new $\mu'$.

$\frac{480 - \mu'}{10} = -2.3263$ (the 1st percentile)

$480 - \mu' = -23.263$

$\mu' = 480 + 23.263 = 503.263$

Answer: New $\mu = 503$ g (3 s.f.) [2]

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Mark Summary:

Q	Marks		Q	Marks		Q	Marks		Q	Marks
1	3		6	5		11	5		16	5
2	3		7	5		12	5		17	5
3	4		8	5		13	5		18	6
4	4		9	5		14	5		19	6
5	4		10	5		15	5		20	7

Total: 50 marks ✓