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A-Level Maths – Statistics

Sampling, data presentation, probability, distributions, hypothesis testing, regression.

12 questions  ·  69 marks  ·  answers included  ·  A-Level (Edexcel/AQA/OCR)
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This worksheet covers:

  • Sampling
  • Measures of location and spread (including coding)
  • Probability and Venn diagrams
  • The binomial and normal distributions
  • Hypothesis testing (binomial, normal mean and correlation)
  • Regression

Calculator with statistical functions allowed throughout; give probabilities to 4 decimal places. Allow about 80 minutes.

Questions

Section A – Skills

Short questions.
1

A school has $1200$ students: $500$ in Year 12 and $700$ in Year 13. A sample of $60$ students is to be taken.(a) Describe how a stratified sample could be taken, stating how many students should be chosen from each year group.(b) State one advantage of stratified sampling over simple random sampling in this context.

Show answer
(a) $25$ from Year 12 and $35$ from Year 13, chosen by simple random sampling (e.g. numbered lists and random numbers) within each year group(b) The sample reflects the proportion of each year group in the population, so is more representative
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2

The times, in minutes, taken by $8$ students to complete a puzzle are$12,\ 15,\ 11,\ 18,\ 20,\ 14,\ 16,\ 13$.(a) Find the median and the interquartile range.(b) Calculate the mean and the standard deviation.

Show answer
(a) Median $= 14.5$; $Q_1 = 12.5$, $Q_3 = 17$, IQR $= 4.5$ (accept $Q_1 = 12.75$, $Q_3 = 16.5$, IQR $= 3.75$ by interpolation)(b) Mean $= 14.875$; standard deviation $= 2.85$ (population) or $3.04$ (sample)
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3

The masses $x$ grams of $50$ apples are coded using $y = \dfrac{x – 100}{5}$. The coded data have mean $3.2$ and standard deviation $1.5$. Find the mean and standard deviation of the masses of the apples.

Show answer
Mean $= 100 + 5(3.2) = 116$ g; standard deviation $= 5 \times 1.5 = 7.5$ g
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4

Events $A$ and $B$ are such that $P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cap B) = 0.15$.(a) Find $P(A \cup B)$.(b) Find $P(A \mid B)$.(c) Determine whether $A$ and $B$ are independent.

Show answer
(a) $0.4 + 0.5 – 0.15 = 0.75$(b) $\frac{0.15}{0.5} = 0.3$(c) $P(A)P(B) = 0.2 e 0.15$, so not independent
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5

The random variable $X \sim B(20,\ 0.3)$. Find(a) $P(X = 6)$(b) $P(X \le 4)$(c) $P(X > 7)$.

Show answer
(a) $0.1916$(b) $0.2375$(c) $1 – P(X \le 7) = 0.2277$
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Section B – Exam style

6

The lengths of steel rods are normally distributed with mean $50$ cm and standard deviation $8$ cm. Find the probability that a randomly chosen rod(a) is shorter than $60$ cm,(b) is longer than $45$ cm,(c) has a length between $40$ cm and $58$ cm.(d) Find the length $l$ such that $10$ percent of rods are longer than $l$.

Show answer
(a) $0.8944$(b) $0.7340$(c) $0.7357$(d) $l = 50 + 1.2816 \times 8 = 60.3$ cm
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7

The random variable $Y \sim N(\mu,\ \sigma^2)$. It is given that $P(Y < 20) = 0.1$ and $P(Y > 32) = 0.2$. Find the values of $\mu$ and $\sigma$.

Show answer
$\frac{20-\mu}{\sigma} = -1.2816$, $\frac{32-\mu}{\sigma} = 0.8416$; subtracting gives $\sigma = \frac{12}{2.1232} = 5.65$, $\mu = 27.2$
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8

A spinner is claimed to land on red with probability $0.3$. Jo thinks the probability is greater than $0.3$. She spins it $20$ times and it lands on red $10$ times.(a) Write down suitable hypotheses.(b) Test, at the $5$ percent significance level, whether there is evidence to support Jo’s claim.(c) Find the critical region for a two-tailed test at the $5$ percent level, using the same model, and state the actual significance level.

Show answer
(a) $H_0: p = 0.3$, $H_1: p > 0.3$(b) $P(X \ge 10) = 0.0480 < 0.05$, so reject $H_0$; there is evidence at the $5$ percent level that the probability is greater than $0.3$(c) Critical region $X \le 1$ or $X \ge 11$; actual significance level $= 0.0076 + 0.0171 = 0.0248$
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9

A machine fills bags of flour. The mass of flour in a bag is normally distributed with standard deviation $15$ g. The mean is supposed to be $500$ g. A random sample of $25$ bags has mean mass $506$ g. Test, at the $5$ percent level, whether the mean mass has increased.

Show answer
$H_0: \mu = 500$, $H_1: \mu > 500$. $\bar{X} \sim N(500, 3^2)$; $z = \frac{506-500}{3} = 2 > 1.6449$ (or $P(\bar{X} \ge 506) = 0.0228 < 0.05$). Reject $H_0$: evidence the mean mass has increased.
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10

A bag contains $5$ red and $3$ blue counters. Two counters are taken at random without replacement.(a) Draw a tree diagram (or list the outcomes) and find the probability that the counters are different colours.(b) Given that the second counter is blue, find the probability that the first was red.

Show answer
(a) $\frac{5}{8}\cdot\frac{3}{7} + \frac{3}{8}\cdot\frac{5}{7} = \frac{30}{56} = \frac{15}{28}$(b) $P(\text{2nd blue}) = \frac{3}{8}$; $P(\text{1st red} \mid \text{2nd blue}) = \frac{15/56}{3/8} = \frac{5}{7}$
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Section C – Data response

11

A researcher records the daily maximum temperature $t$ (in degrees C) and the number of ice creams sold $s$ at a kiosk on $10$ days. The regression line of $s$ on $t$ is $s = -48 + 12.5t$ and the product moment correlation coefficient is $r = 0.62$.(a) Interpret the value $12.5$ in context.(b) Estimate the sales on a day when $t = 24$, and comment on the reliability of this estimate if the recorded temperatures ranged from $15$ to $27$.(c) Explain why the equation should not be used to predict sales when $t = 40$.(d) Test, at the $5$ percent significance level, whether there is evidence of positive correlation between temperature and sales. (Critical value for $n = 10$, one-tailed $5$ percent: $0.5494$.)

Show answer
(a) For each $1$ degree rise in temperature, sales increase by about $12.5$ ice creams(b) $s = -48 + 12.5(24) = 252$; reliable-ish since $24$ is within the data range (interpolation)(c) $40$ is outside the range of the data (extrapolation), so the linear model may not hold(d) $H_0: \rho = 0$, $H_1: \rho > 0$; $0.62 > 0.5494$ so reject $H_0$: evidence of positive correlation
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12

A large data set records the daily mean windspeed (knots) at a weather station. The values for one month are summarised: $n = 31$, $\sum x = 279$, $\sum x^2 = 2821$.(a) Calculate the mean and standard deviation.(b) An outlier is defined as a value more than $2$ standard deviations from the mean. Show that a reading of $19$ knots would be an outlier.(c) Give one reason why the data may not be representative of the whole year.

Show answer
(a) Mean $= 9$; $\sigma = \sqrt{\frac{2821}{31} – 81} = \sqrt{10} = 3.16$(b) $9 + 2(3.16) = 15.3 < 19$, so $19$ is an outlier(c) Only one month is used; windspeed varies seasonally
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