Statistics Overview – A Level Maths Year 2

1 The Big Idea

Year 1 statistics was mostly about describing data and working with one discrete model, the binomial distribution. Year 2 statistics is about modelling: fitting lines and curves to bivariate data, handling probabilities that depend on extra information, and using a continuous model — the normal distribution — for measurements.

This page covers three DfE content areas: L · Data presentation and interpretation (regression and exponential models), M · Probability (conditional probability and modelling), and N · Statistical distributions (the normal distribution and choosing between models).

In every topic the same question returns: is the model appropriate? A regression line is only reliable inside the data; a probability calculation only holds if its assumptions hold; a normal model is only sensible for roughly symmetrical, bell-shaped data. Expect at least one mark per question for a comment in context.

What is new compared with Year 1

The correlation coefficient $r$ and regression lines $y = a + bx$
Linearising $y = kb^x$ and $y = ax^n$ with logarithms
Conditional probability $\mathrm{P}(A \mid B)$
The normal distribution $\mathrm{N}(\mu,\sigma^2)$ and the normal approximation to the binomial

Calculators do the arithmetic. You are expected to use your calculator's statistics functions for $r$, regression coefficients and normal probabilities. Marks go for setting the problem up correctly — stating the distribution, the parameters, the region — and for interpreting the answer.

2 The Language You Need

Term	Meaning	Example
PMCC	Product moment correlation coefficient $r$; strength of linear association, $-1 \leqslant r \leqslant 1$.	$r = 0.94$: strong positive
Explanatory variable	The independent variable, plotted on the $x$-axis.	Hours of revision
Response variable	The dependent variable, predicted from the explanatory one.	Test score
Extrapolation	Predicting outside the range of the data — unreliable.	Using $1$–$8$ year data for a $20$-year-old car
Conditional probability	$\mathrm{P}(A \mid B)$: probability of $A$ given that $B$ has occurred.	$\mathrm{P}(\text{disease} \mid \text{positive test})$
Independent	$\mathrm{P}(A \mid B) = \mathrm{P}(A)$.	Two separate dice
Continuous	Can take any value in a range; $\mathrm{P}(X = a) = 0$.	Height, time, mass
$\mathrm{N}(\mu,\sigma^2)$	Normal distribution, mean $\mu$, variance $\sigma^2$.	$\mathrm{N}(170,\ 64)$ has $\sigma = 8$
$\Phi(z)$	$\mathrm{P}(Z < z)$ for the standard normal $Z \sim \mathrm{N}(0,1)$.	$\Phi(1.96) = 0.975$
Continuity correction	Replacing integer $k$ by $k \pm 0.5$ when approximating a discrete distribution by a continuous one.	$\mathrm{P}(X \leqslant 20) \approx \mathrm{P}(Y < 20.5)$

3 The Core Techniques

Four results that do most of the work

$\log y = \log k + x\log b$ (for $y = kb^x$)
$\mathrm{P}(A \mid B) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$
$Z = \dfrac{X - \mu}{\sigma}$
$\mathrm{B}(n,p) \approx \mathrm{N}\big(np,\ np(1-p)\big)$

Worked Example 1 — Conditional probability with a normal distribution

The lengths of a species of fish are $X \sim \mathrm{N}(64,\ 5^2)$ cm. A fish caught is longer than the legal minimum of $60$ cm. Find the probability that it is longer than $70$ cm.

①We need $\mathrm{P}(X > 70 \mid X > 60) = \dfrac{\mathrm{P}(X > 70 \text{ and } X > 60)}{\mathrm{P}(X > 60)}$.

②Any fish longer than $70$ is longer than $60$, so the numerator is just $\mathrm{P}(X > 70) = 1 - \Phi(1.2) = 0.11507$.

③$\mathrm{P}(X > 60) = 1 - \Phi(-0.8) = 0.78814$

④$\mathrm{P}(X > 70 \mid X > 60) = \dfrac{0.11507}{0.78814} = 0.146$ (3 s.f.)

This single question uses M (conditional probability) and N (the normal distribution). Mixing content areas like this is common in exams.

Worked Example 2 — Choosing and approximating

$40\%$ of voters support a proposal. In a random sample of $150$ voters, use a suitable approximation to find the probability that fewer than $55$ support it.

①$X \sim \mathrm{B}(150,\ 0.4)$: a count of successes in a fixed number of independent trials.

②$np = 60$ and $nq = 90$, both $> 5$, so use $Y \sim \mathrm{N}(60,\ 36)$, $\sigma = 6$.

③Fewer than $55$ means $X \leqslant 54$, so use $\mathrm{P}(Y < 54.5)$.

④$\Phi\!\left(\dfrac{54.5 - 60}{6}\right) = \Phi(-0.917) = 0.180$ (3 s.f.)

4 The Statistics Subtopics

Every Year 2 statistics topic outside hypothesis testing, and the page that covers it. The labels give the DfE content area.

LRegression and Exponential ModelsInterpreting $r$, regression lines $y = a + bx$, interpolation and extrapolation, and linearising $y = kb^x$ and $y = ax^n$.
MConditional Probability$\mathrm{P}(A \mid B)$ from two-way tables, Venn diagrams and tree diagrams; independence; modelling assumptions.
NThe Normal DistributionProperties, standardising, probabilities, inverse normal, and finding unknown $\mu$ and $\sigma$.
NChoosing a DistributionBinomial or normal? The normal approximation with continuity correction, and checking a model against data.

Testing whether a correlation or a mean is significant is covered in Hypothesis Testing.

5 Common Mistakes to Avoid

Using the variance as the standard deviation. $\mathrm{N}(50,\ 16)$ has $\sigma = 4$. Check what the second parameter is every time you open the calculator's normal function.

Confusing $\mathrm{P}(A \mid B)$ with $\mathrm{P}(B \mid A)$. Divide by the probability of the event after the bar.

Mixing up independent and mutually exclusive. Independent: $\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$. Mutually exclusive: $\mathrm{P}(A \cap B) = 0$.

Forgetting or misapplying the continuity correction. List the integers you want first, then go half a unit beyond them.

Claiming causation. A high $r$ shows association, not cause and effect.

Undoing logarithms wrongly. If $\log_{10} y = 0.6 + 0.2x$ then $k = 10^{0.6}$, not $0.6$; for a log–log line the gradient is the power $n$.

6 Quick Reference

PMCC

$-1 \leqslant r \leqslant 1$; linear association only.

Regression

$y = a + bx$; $b$ is change in $y$ per unit $x$.

Exponential model

$y = kb^x$: plot $\log y$ against $x$.

Power model

$y = ax^n$: plot $\log y$ against $\log x$.

Conditional

$\mathrm{P}(A \mid B) = \mathrm{P}(A \cap B) / \mathrm{P}(B)$

Independence

$\mathrm{P}(A \mid B) = \mathrm{P}(A)$

Normal

Symmetric; inflection at $\mu \pm \sigma$; $Z = (X - \mu)/\sigma$.

Approximation

$\mathrm{B}(n,p) \approx \mathrm{N}(np,\ npq)$ if $np, nq > 5$.

Continuity

Integer $k$ becomes $k \pm 0.5$.

7 Practice Questions

Ten questions drawing on the whole Year 2 Statistics section. Work each one through before opening the solution.

Question 1

For $40$ adults, the correlation coefficient between shoe size and score on a memory test is $r = 0.08$. Interpret this value.

▶ Show solution

$r$ is very close to $0$: there is little or no linear correlation between shoe size and memory score.

(It does not rule out some non-linear relationship, but there is no reason to expect one here.)

Question 2

The regression line of $\log_{10} y$ on $x$ is $\log_{10} y = 1.1 + 0.3x$. Express $y$ in the form $kb^x$.

▶ Show solution

$k = 10^{1.1} = 12.6$ and $b = 10^{0.3} = 2.00$ (3 s.f.)

$y = 12.6 \times 2.00^{x}$ — roughly, $y$ doubles each time $x$ increases by $1$.

Question 3

$\mathrm{P}(A) = 0.7$, $\mathrm{P}(B) = 0.4$ and $\mathrm{P}(A \cap B) = 0.28$. Determine whether $A$ and $B$ are independent, and find $\mathrm{P}(B \mid A)$.

▶ Show solution

$\mathrm{P}(A)\,\mathrm{P}(B) = 0.7 \times 0.4 = 0.28 = \mathrm{P}(A \cap B)$, so independent.

$\mathrm{P}(B \mid A) = \dfrac{0.28}{0.7} = 0.4$ — equal to $\mathrm{P}(B)$, as expected.

Question 4

Machine $A$ makes $60\%$ of a factory's components and $3\%$ of its output is defective. Machine $B$ makes the rest, and $5\%$ of its output is defective. A component is found to be defective. Find the probability it was made by machine $A$.

▶ Show solution

$\mathrm{P}(D) = 0.6 \times 0.03 + 0.4 \times 0.05 = 0.018 + 0.02 = 0.038$

$\mathrm{P}(A \mid D) = \dfrac{0.018}{0.038} = \dfrac{9}{19} = 0.474$ (3 s.f.)

Question 5

$X \sim \mathrm{N}(80,\ 25)$. Find $\mathrm{P}(X < 72)$.

▶ Show solution

$\sigma = 5$, $z = \dfrac{72 - 80}{5} = -1.6$.

$\Phi(-1.6) = 0.0548$ (3 s.f.)

Question 6

$X \sim \mathrm{N}(\mu,\ 36)$ and $\mathrm{P}(X > 50) = 0.2$. Find $\mu$.

▶ Show solution

$\mathrm{P}(X < 50) = 0.8$, so $z = 0.8416$.

$\dfrac{50 - \mu}{6} = 0.8416 \Rightarrow \mu = 50 - 5.050 = 45.0$ (3 s.f.)

Question 7

$\mathrm{P}(A) = 0.5$, $\mathrm{P}(B \mid A) = 0.4$ and $\mathrm{P}(A \cup B) = 0.75$. Find $\mathrm{P}(B)$ and $\mathrm{P}(A \mid B)$.

▶ Show solution

$\mathrm{P}(A \cap B) = 0.5 \times 0.4 = 0.2$

$0.75 = 0.5 + \mathrm{P}(B) - 0.2 \Rightarrow \mathrm{P}(B) = 0.45$

$\mathrm{P}(A \mid B) = \dfrac{0.2}{0.45} = \dfrac{4}{9}$

Question 8

A student models the number of girls in families with $4$ children by $\mathrm{B}(4,\ 0.5)$. State two assumptions this model makes and comment on whether each is reasonable.

▶ Show solution

Constant probability: each child is a girl with probability $0.5$. Roughly reasonable, though slightly more boys than girls are born (about $0.49$ for a girl).

Independence: the sex of each child does not depend on the others. Largely reasonable, but identical twins are always the same sex, which breaks independence.

Question 9

$X \sim \mathrm{B}(150,\ 0.4)$. Use a normal approximation to find $\mathrm{P}(X > 65)$.

▶ Show solution

$Y \sim \mathrm{N}(60,\ 36)$. $X > 65$ means $X \geqslant 66$, so use $\mathrm{P}(Y > 65.5)$.

$1 - \Phi\!\left(\dfrac{5.5}{6}\right) = 1 - \Phi(0.917) = 0.180$ (3 s.f.)

(This equals the answer to Worked Example 2 by symmetry: $65.5$ and $54.5$ are each $5.5$ from the mean.)

Question 10

Marks in an exam are modelled by $\mathrm{N}(58,\ 12^2)$.

(a) Find the probability that a candidate scores more than $75$. (b) The top $10\%$ of candidates receive the top grade. Find the minimum mark needed. (c) Given that a candidate scored above the mean, find the probability that they scored more than $75$. (d) Find the probability that, in a random sample of $30$ candidates, at least $3$ score more than $75$.

▶ Show solution

(a) $z = \dfrac{17}{12} = 1.417$; $1 - \Phi(1.417) = 0.0783$ (3 s.f.)

(b) $\Phi(z) = 0.9 \Rightarrow z = 1.2816$; $58 + 1.2816 \times 12 = 73.4$, so a mark of $74$ (marks are whole numbers).

(c) $\mathrm{P}(X > 75 \mid X > 58) = \dfrac{0.07829}{0.5} = 0.157$ (3 s.f.)

(d) Let $W \sim \mathrm{B}(30,\ 0.07829)$. $\mathrm{P}(W \geqslant 3) = 1 - \mathrm{P}(W \leqslant 2) = 0.421$ (3 s.f.)

Statistics Overview · A Level Maths Year 2 · Created with MathJax