Notes · Further MathematicsUK · A-Levels

Optional application: Statistics

Statistics is one of the three optional applied routes in AQA Further Mathematics 7367, examined only for students entered for it. It extends the statistics of A-Level Mathematics with discrete and continuous random variables, the Poisson and exponential distributions, Type I and Type II errors, chi-squared tests and inference using the t-distribution. This note presents the route in full for students who choose it.

5 sections~15 min reading time3 competenciesLevel Standard 1 · Advanced 4

T·121212 / 13

Exam profile

AO1 · Compute probabilities, expectations and test statistics accuratelyAO2 · Choose an appropriate distribution or test and justify the choiceAO3 · Interpret the outcome of a test or a confidence interval in context

Operators:findtestshow thathencecalculateinterpret

basic level

As an optional applied route, Statistics is examined only for students entered for it; AS-level content covers discrete random variables and the Poisson distribution.

higher level

The full A-Level route adds continuous distributions, Type I/II errors and power, chi-squared tests, and t-distribution inference with confidence intervals.

Depth

Text

Contents · 5 sections

Optional application: Statistics

Discrete random variables and the Poisson distribution#

StandardAQA 7367 Statistics optionDfE Further Mathematics (optional applied)

The Poisson(3) distribution

Fig. 1For

\lambda = 3

the probabilities peak near

x = 3

and tail off; the distribution is right-skewed.

A discrete random variable

X

takes a countable set of values with a probability distribution

P(X = x)

summing to

1

. Its expectation (mean) is

E(X) = \sum x\,P(X = x)

and its variance is

\operatorname{Var}(X) = E(X^2) - [E(X)]^2

, where

E(X^2) = \sum x^2\,P(X=x)

. These weighted sums are the basic descriptive measures, and the identity for variance — 'mean of the square minus the square of the mean' — is the practical way to compute it.

The Poisson distribution models the number of independent events occurring in a fixed interval when they happen at a constant average rate — telephone calls per minute, flaws per metre of wire, particles per second. Its probability function is

P(X = x) = \dfrac{\mathrm{e}^{-\lambda}\lambda^x}{x!}

for

x = 0, 1, 2, \dots

, where

\lambda

is the mean number of events. A distinctive feature is that its mean and variance are equal, both being

\lambda

; a data set whose sample mean and variance are close is a hint that a Poisson model may fit.

A key property is additivity: if

X \sim \text{Poisson}(\lambda_1)

and

Y \sim \text{Poisson}(\lambda_2)

are independent, then

X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2)

. This lets you rescale the interval — the number of events in two hours, when the hourly rate is

\lambda

, is Poisson with mean

2\lambda

— and combine independent Poisson sources. Recognising when to add the means is central to Poisson problem-solving.

The distribution is right-skewed for small

\lambda

and becomes more symmetric as

\lambda

grows, as the bar chart of

\text{Poisson}(3)

illustrates: the probabilities rise to a peak near

x = \lambda

and then tail off. In practice, cumulative probabilities such as

P(X \geq 2)

are found via the complement

1 - P(X = 0) - P(X = 1)

, and the model's assumptions — independence and a constant rate — should be stated when it is applied.

P(X = x) = \frac{\mathrm{e}^{-\lambda}\lambda^x}{x!}, \qquad E(X) = \operatorname{Var}(X) = \lambda

Poisson distribution

Mean equals variance; models independent events at a constant rate.

\operatorname{Var}(X) = E(X^2) - [E(X)]^2

Variance of a discrete variable

Mean of the square minus the square of the mean.

Worked example

Poisson probabilities

Calls arrive at mean rate $3$ per minute (Poisson). Find $P(X \geq 2)$ in one minute and $P(Y = 5)$ in two minutes.

Set up the one-minute variable
$X \sim \text{Poisson}(3)$ . $P(X = 0) = \mathrm{e}^{-3} = 0.0498$ ; $P(X = 1) = \mathrm{e}^{-3}\cdot 3 = 0.1494$ .
Use the complement
$P(X \geq 2) = 1 - P(X=0) - P(X=1) = 1 - 0.0498 - 0.1494 = 0.8008$ .
$P(X \geq 2) = 1 - 0.0498 - 0.1494 = 0.8008$
Rescale for two minutes
Over two minutes the mean is $2\times 3 = 6$ , so $Y \sim \text{Poisson}(6)$ .
Evaluate P(Y = 5)
$P(Y = 5) = \dfrac{\mathrm{e}^{-6}6^5}{5!} = \dfrac{0.002479 \times 7776}{120} = 0.1606$ .

Result: $P(X \geq 2) = 0.801$ (3 s.f.) and $P(Y = 5) = 0.161$ (3 s.f.).

Typical mistakes

Forgetting to add the means when combining independent Poisson variables or changing the interval length.
Confusing $P(X \geq 2)$ with $P(X > 2)$ , or omitting a term when using the complement.

Active revision

Calls arrive at a switchboard at a mean rate of $3$ per minute, modelled as a Poisson process. Find the probability of at least $2$ calls in a given minute, and the probability of exactly $5$ calls in a two-minute period.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Continuous random variables and the exponential distribution#

AdvancedAQA 7367 Statistics optionDfE Further Mathematics (optional applied)

The exponential probability density

Fig. 2The density

f(x) = 0.5\,\mathrm{e}^{-0.5x}

starts at

\lambda = 0.5

and decays; the area beyond any point is

\mathrm{e}^{-0.5x}

A continuous random variable is described by a probability density function

f(x)

satisfying

f(x) \geq 0

and

\displaystyle\int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 1

. Probabilities are areas under the density,

P(a \leq X \leq b) = \displaystyle\int_a^b f(x)\,\mathrm{d}x

, and the probability of any single exact value is zero. The cumulative distribution function

F(x) = P(X \leq x) = \displaystyle\int_{-\infty}^{x} f(t)\,\mathrm{d}t

accumulates this area from the left.

The mean and variance are the continuous analogues of the discrete sums:

E(X) = \displaystyle\int x\,f(x)\,\mathrm{d}x

and

\operatorname{Var}(X) = \displaystyle\int x^2 f(x)\,\mathrm{d}x - [E(X)]^2

. The median

m

is the value with

F(m) = \tfrac12

, splitting the area in half. Setting up and evaluating these integrals — and using the cdf to find probabilities and the median — is the core skill.

The exponential distribution, with density

f(x) = \lambda\mathrm{e}^{-\lambda x}

for

x \geq 0

, models waiting times between the events of a Poisson process. Its cdf is

F(x) = 1 - \mathrm{e}^{-\lambda x}

, its mean is

\dfrac{1}{\lambda}

and its variance

\dfrac{1}{\lambda^2}

. Because the mean is the reciprocal of the Poisson rate, a Poisson process with

3

events per minute has exponential inter-event times with mean

\tfrac13

minute — the two distributions are two views of the same process.

A striking feature of the exponential distribution is that it is memoryless: the probability of waiting a further time

t

is the same regardless of how long one has already waited,

P(X > s + t \mid X > s) = P(X > t)

. This models genuinely random arrivals with no ageing, and its density is a simple decreasing exponential, as the figure shows. Recognising when the memoryless, constant-rate assumptions are appropriate is part of choosing the model.

f(x) \geq 0, \quad \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 1, \quad F(x) = \int_{-\infty}^{x} f(t)\,\mathrm{d}t

Continuous random variable

Probabilities are areas; the cdf accumulates them.

f(x) = \lambda\mathrm{e}^{-\lambda x}\ (x \geq 0), \quad F(x) = 1 - \mathrm{e}^{-\lambda x}, \quad E(X) = \frac{1}{\lambda}

Exponential distribution

Waiting times of a Poisson process; mean is the reciprocal of the rate.

Worked example

Exponential lifetime

A component's lifetime (hours) is exponential with mean $200$ . Find $P(X > 300)$ and the median lifetime.

Find the rate
Mean $= \dfrac{1}{\lambda} = 200$ , so $\lambda = \dfrac{1}{200} = 0.005$ , and $F(x) = 1 - \mathrm{e}^{-0.005x}$ .
Compute P(X > 300)
$P(X > 300) = 1 - F(300) = \mathrm{e}^{-0.005 \times 300} = \mathrm{e}^{-1.5} = 0.2231$ .
$P(X > 300) = \mathrm{e}^{-1.5} = 0.2231$
Median
Solve $F(m) = \tfrac12$ : $1 - \mathrm{e}^{-0.005m} = 0.5 \Rightarrow \mathrm{e}^{-0.005m} = 0.5 \Rightarrow m = \dfrac{\ln 2}{0.005}$ .
Evaluate
$m = \dfrac{0.6931}{0.005} = 138.6$ hours; the median is less than the mean, reflecting the right skew.

Result: $P(X > 300) = 0.223$ and the median lifetime is $138.6$ hours.

Typical mistakes

Treating $f(x)$ itself as a probability (it is a density; probabilities are areas).
Forgetting the domain $x \geq 0$ for the exponential, or confusing the mean $\tfrac{1}{\lambda}$ with $\lambda$ .

Active revision

The lifetime (in hours) of a component follows an exponential distribution with mean $200$ . Find the probability that a component lasts more than $300$ hours, and find the median lifetime.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Errors in hypothesis testing#

AdvancedAQA 7367 Statistics optionDfE Further Mathematics (optional applied)

A hypothesis test decides between a null hypothesis

H_0

and an alternative

H_1

using sample data, and because the data are random the decision can be wrong in two distinct ways. A Type I error is rejecting

H_0

when it is in fact true — a false alarm. A Type II error is failing to reject

H_0

when it is in fact false — a missed detection. The two errors are of different kinds and cannot be reduced simultaneously without more data.

The probability of a Type I error is the significance level

\alpha

of the test: by choosing the critical region so that, under

H_0

, the test statistic falls in it with probability

\alpha

, you fix the Type I error rate. For a discrete test statistic the actual size of the critical region may be a little below the nominal

\alpha

, and stating the true probability of a Type I error is examinable.

The probability of a Type II error, denoted

\beta

, depends on the specific true value of the parameter under

H_1

, because 'how likely we are to miss a false null' depends on how false it is. It is computed as the probability that the statistic falls outside the critical region when the parameter takes a particular alternative value. The power of the test is

1 - \beta

, the probability of correctly rejecting a false null; a more powerful test is better at detecting a real effect.

There is an unavoidable trade-off: shrinking the critical region lowers

\alpha

(fewer false alarms) but raises

\beta

(more missed detections), and vice versa. Only increasing the sample size can reduce both. Understanding this trade-off, computing

\alpha

and

\beta

for a given test, and interpreting the power in context is the substance of this section — it deepens the mechanical hypothesis testing of A-Level Mathematics into an appreciation of what a test can and cannot guarantee.

P(\text{Type I}) = \alpha, \qquad P(\text{Type II}) = \beta, \qquad \text{power} = 1 - \beta

Error probabilities and power

$\alpha$ is fixed by the critical region; $\beta$ depends on the true alternative value.

Worked example

Type I and Type II errors for a coin test

$X \sim B(10, p)$ ; test $H_0: p = 0.5$ by rejecting if $X \in \{0,1,9,10\}$ . Find $P(\text{Type I})$ and $P(\text{Type II})$ when $p = 0.8$ .

Type I error
Under $H_0$ ( $p=0.5$ ), reject when $X \in \{0,1,9,10\}$ . By symmetry $P = 2\big(P(X=0) + P(X=1)\big) = 2(0.5^{10} + 10\cdot 0.5^{10}) = 2(11)(0.000977) = 0.0215$ .
$P(\text{Type I}) = 2\,(0.5)^{10}(1 + 10) = 0.0215$
Set up the Type II error
A Type II error is failing to reject $H_0$ when $p = 0.8$ , i.e. $X \in \{2,3,\dots,8\}$ under $B(10, 0.8)$ .
Compute the complement
$P(\text{Type II}) = 1 - P(X \in \{0,1,9,10\})$ under $B(10,0.8)$ . Here $P(X=9) = 10(0.8)^9(0.2) = 0.2684$ , $P(X=10) = 0.8^{10} = 0.1074$ , and $P(X=0), P(X=1)$ are negligible.
Evaluate
$P(\text{reject} \mid p=0.8) \approx 0 + 0 + 0.2684 + 0.1074 = 0.3758$ , so $P(\text{Type II}) = 1 - 0.3758 = 0.6242$ .

Result: $P(\text{Type I}) = 0.0215$ ; $P(\text{Type II}) = 0.624$ (so the power at $p=0.8$ is only $0.376$ ).

Typical mistakes

Swapping the definitions of Type I and Type II errors.
Computing a Type II error probability without a specific alternative value, or using the null value by mistake.

Active revision

A coin is tested for bias by tossing it $10$ times; $H_0: p = 0.5$ is rejected if the number of heads is $0, 1, 9$ or $10$ . Find the probability of a Type I error, and the probability of a Type II error if in fact $p = 0.8$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Chi-squared tests#

AdvancedAQA 7367 Statistics optionDfE Further Mathematics (optional applied)

The chi-squared distribution and its critical region

Fig. 3For

4

degrees of freedom the

5\%

critical value is

9.49

; a statistic beyond it (shaded tail) rejects the null hypothesis.

The chi-squared (

\chi^2

) test compares observed frequencies with those expected under a hypothesis. The test statistic is

\chi^2 = \displaystyle\sum \frac{(O - E)^2}{E}

, summed over all classes, where

O

is the observed frequency and

E

the expected frequency of each class. A small value means the observations agree well with the model; a large value indicates a poor fit and leads to rejecting the null hypothesis.

In a goodness-of-fit test the expected frequencies come from a proposed distribution (uniform, binomial, Poisson, or a given ratio). The statistic is compared with a critical value from

\chi^2

tables at the chosen significance level, using the degrees of freedom

\nu = (\text{number of classes}) - 1 - (\text{number of parameters estimated from the data})

. Each parameter estimated from the sample (such as a mean used to fit a Poisson) costs one degree of freedom.

For a contingency table testing whether two categorical variables are independent, the expected frequency of a cell is

\dfrac{(\text{row total}) \times (\text{column total})}{\text{grand total}}

, and the degrees of freedom are

(r-1)(c-1)

for an

r \times c

table. The same statistic

\sum \dfrac{(O-E)^2}{E}

is formed and compared with the critical value; a significant result means the variables are associated, not independent.

Two practical rules protect the test's validity: expected frequencies should generally be at least

5

, and where they are too small, adjacent classes are pooled (which reduces the degrees of freedom accordingly). The conclusion must be stated in context — 'there is (in)sufficient evidence at the

5\%

level to reject the hypothesis that ...' — and never as a bald 'significant', since the whole point is to interpret the fit or the association meaningfully.

\chi^2 = \sum \frac{(O - E)^2}{E}, \qquad \nu = (\text{classes}) - 1 - (\text{estimated parameters})

Chi-squared statistic and degrees of freedom

Large $\chi^2$ means a poor fit; each estimated parameter costs a degree of freedom.

E_{ij} = \frac{(\text{row}_i\ \text{total})(\text{col}_j\ \text{total})}{\text{grand total}}, \qquad \nu = (r-1)(c-1)

Contingency table

Expected cell frequencies for a test of independence.

Worked example

Goodness-of-fit test for a fair die

A die rolled $120$ times gives $14, 22, 19, 25, 18, 22$ . Test at $5\%$ whether it is fair.

Hypotheses and expected frequencies
$H_0$ : the die is fair, so each face is expected $\dfrac{120}{6} = 20$ times; $H_1$ : it is not fair.
Compute the statistic
$\chi^2 = \dfrac{(14-20)^2 + (22-20)^2 + (19-20)^2 + (25-20)^2 + (18-20)^2 + (22-20)^2}{20} = \dfrac{36+4+1+25+4+4}{20} = \dfrac{74}{20} = 3.7$ .
$\chi^2 = \frac{74}{20} = 3.7$
Degrees of freedom and critical value
$\nu = 6 - 1 = 5$ (no parameters estimated); the $5\%$ critical value is $\chi^2_{5} = 11.07$ .
Conclude
Since $3.7 < 11.07$ , do not reject $H_0$ : there is insufficient evidence at the $5\%$ level that the die is unfair.

Result: $\chi^2 = 3.7 < 11.07$ ; the data are consistent with a fair die.

Typical mistakes

Miscounting degrees of freedom by forgetting to subtract for parameters estimated from the data.
Using observed rather than expected frequencies in the denominator, or failing to pool small expected classes.

Active revision

A die is rolled $120$ times, giving the frequencies $14, 22, 19, 25, 18, 22$ for faces $1$ to $6$ . Test at the $5\%$ level whether the die is fair.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

The t-distribution and confidence intervals#

AdvancedAQA 7367 Statistics optionDfE Further Mathematics (optional applied)

When the population variance is unknown and estimated from a small sample, the standardised sample mean does not follow a normal distribution but Student's

t

-distribution. The statistic

t = \dfrac{\bar x - \mu}{s/\sqrt{n}}

, where

s

is the sample standard deviation (with divisor

n-1

), has a

t

-distribution with

n - 1

degrees of freedom. The

t

-distribution is symmetric and bell-shaped like the normal but has heavier tails, reflecting the extra uncertainty from estimating the variance.

As the sample size increases, the

t

-distribution approaches the standard normal, because a large sample estimates the variance reliably. For small samples, however, the heavier tails matter: the critical values are larger than the normal ones, so tests and intervals based on

t

are appropriately more cautious. Using a normal critical value with a small sample would understate the uncertainty.

A confidence interval for the mean, when the variance is unknown, is

\bar x \pm t_{n-1}\,\dfrac{s}{\sqrt{n}}

, where

t_{n-1}

is the critical value from the

t

-tables for the required confidence level and

n-1

degrees of freedom. A

95\%

interval means that if the sampling were repeated many times, about

95\%

of the intervals so constructed would contain the true mean — a statement about the procedure, not a probability that the fixed true mean lies in one particular interval.

The same

t

-statistic underlies a hypothesis test for a mean with unknown variance: compare

t = \dfrac{\bar x - \mu_0}{s/\sqrt{n}}

with the critical value for

n-1

degrees of freedom. The assumptions — a random sample from an approximately normal population — should be stated, and the interval or test result interpreted in context. This completes the route by giving small-sample inference its correct, honest treatment.

t = \frac{\bar x - \mu}{s/\sqrt{n}} \quad \text{with } n-1 \text{ degrees of freedom}

The t-statistic

For small samples with unknown variance; heavier tails than the normal.

\bar x \pm t_{n-1}\,\frac{s}{\sqrt{n}}

Confidence interval for the mean

$t_{n-1}$ is the critical value for the chosen confidence level.

Worked example

A 95% confidence interval

A sample of $n = 10$ has $\bar x = 24.6$ and $s = 2.1$ . Find a $95\%$ confidence interval for $\mu$ (take $t_9 = 2.262$ ).

Standard error
$\dfrac{s}{\sqrt{n}} = \dfrac{2.1}{\sqrt{10}} = \dfrac{2.1}{3.1623} = 0.6641$ .
Margin of error
$t_9 \times \dfrac{s}{\sqrt{n}} = 2.262 \times 0.6641 = 1.502$ .
$\text{margin} = 2.262 \times 0.6641 = 1.502$
Form the interval
$24.6 \pm 1.502$ , i.e. $(23.10,\ 26.10)$ .
Interpret
We are $95\%$ confident, in the procedural sense, that the population mean lies between $23.1$ and $26.1$ .

Result: The $95\%$ confidence interval is $(23.1,\ 26.1)$ .

Typical mistakes

Using a normal (z) critical value instead of a $t$ value for a small sample with unknown variance.
Interpreting a $95\%$ interval as 'a $95\%$ probability that the true mean is in this interval' rather than as a property of the method.

Active revision

A sample of $10$ measurements has mean $\bar x = 24.6$ and sample standard deviation $s = 2.1$ . Construct a $95\%$ confidence interval for the population mean (use $t_{9} = 2.262$ ).

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

AQA

AQA A-level Further Mathematics 7367 specification

Optional application: Statistics

Key points

Poisson probabilities

Key points

Exponential lifetime

Key points

Type I and Type II errors for a coin test

Key points

Goodness-of-fit test for a fair die

Key points

A 95% confidence interval

Sources