Notes · MathematicsUK · A-Levels

Statistical distributions

This topic introduces discrete probability distributions as models, the binomial distribution B(n, p) for the number of successes in fixed independent trials, and the normal distribution N(mu, sigma-squared) as a model for continuous data. It covers calculating probabilities from each and choosing an appropriate distribution for a situation.

4 sections~9 min reading time3 competenciesLevel Standard 2 · Advanced 2

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Exam profile

AO1 · Calculate probabilities using the binomial and normal distributionsAO2 · Justify the modelling assumptions of each distributionAO3 · Select and apply an appropriate distribution to a context

Operators:findcalculatemodeldeterminestateshow that

basic level

AS-Level covers discrete probability distributions and the binomial distribution as a model.

higher level

The full A-Level adds the normal distribution, standardising, and using the normal as a model (including the choice of distribution).

Depth

Text

Contents · 4 sections

Statistical distributions

Discrete random variables#

StandardDfE GCE Mathematics — distributionsAQA 7357 (Statistics)

A discrete probability distribution

Fig. 1The probabilities sum to 1; any event probability is the sum of the relevant bars.

A discrete random variable

X

takes a countable set of values, each with a probability. Its probability distribution lists the possible values against their probabilities, often as a table or a formula

P(X = x)

. The single defining property is that the probabilities must be valid — each between 0 and 1 — and must sum to 1 over all possible values:

\sum P(X = x) = 1

. This condition is used constantly to find an unknown probability.

A distribution can be given by a formula, such as

P(X = x) = kx

for

x = 1, 2, 3, 4

; here the constant

k

is found by imposing

\sum P(X = x) = 1

, giving

k(1 + 2 + 3 + 4) = 1

k = \tfrac{1}{10}

. Setting up and solving this normalisation equation is the standard first step whenever a distribution contains an unknown constant.

Once the full distribution is known, the probability of any event is found by adding the probabilities of the values in it:

P(X \geq 3)

is the sum of

P(X = 3)

and any larger values. Cumulative probabilities such as

P(X \leq x)

are handled the same way, by summing from the bottom up.

In A-Level Mathematics, the emphasis is on discrete distributions as models and on calculating probabilities from them; calculating the mean and variance of a general discrete random variable is not part of this specification (it belongs to Further Mathematics). The named model you must know in detail is the binomial distribution, treated next.

\sum_{\text{all } x} P(X = x) = 1, \qquad 0 \leq P(X = x) \leq 1

A valid probability distribution

The probabilities must be non-negative and sum to 1; this fixes any unknown constant.

Worked example

Finding an unknown probability

The distribution of $X$ is $P(X = x) = cx^2$ for $x = 1, 2, 3$ . Find $c$ and $P(X \geq 2)$ .

Sum to 1
$c(1^2 + 2^2 + 3^2) = c(1 + 4 + 9) = 14c = 1$ , so $c = \tfrac{1}{14}$ .
Compute the probabilities
$P(X = 2) = \tfrac{4}{14}$ , $P(X = 3) = \tfrac{9}{14}$ .
$P(X \geq 2) = \frac{4}{14} + \frac{9}{14} = \frac{13}{14}$
Add
$P(X \geq 2) = \tfrac{13}{14}$ .

Result: $c = \tfrac{1}{14}$ and $P(X \geq 2) = \tfrac{13}{14}$ .

Typical mistakes

Not using the fact that probabilities sum to 1 when a constant is unknown.
Including or excluding an endpoint incorrectly in $P(X < x)$ versus $P(X \leq x)$ .

Active revision

A discrete random variable has $P(X = x) = k(5 - x)$ for $x = 1, 2, 3, 4$ . Find $k$ and $P(X \geq 3)$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The binomial distribution#

StandardDfE GCE Mathematics — distributionsAQA 7357 (Statistics)

A binomial distribution

Fig. 2

B(6, 0.5)

is symmetric about its mean

np = 3

; the bars give

P(X = r)

The binomial distribution models the number of successes

X

in a fixed number

n

of independent trials, each with the same probability

p

of success. The conditions are strict and examinable: a fixed number of trials, only two outcomes per trial (success or failure), a constant probability of success, and independence between trials. When these hold we write

X \sim B(n, p)

The probability of exactly

r

successes is

P(X = r) = \binom{n}{r} p^{r}(1 - p)^{n - r}

: the binomial coefficient counts the ways of choosing which

r

trials succeed,

p^r

is the probability those succeed, and

(1 - p)^{n - r}

that the rest fail. The distribution is symmetric when

p = 0.5

and skewed otherwise; the mean number of successes is

np

In practice most binomial probabilities are found using the cumulative function on a calculator, which gives

P(X \leq r)

. From it you obtain other probabilities by subtraction:

P(X = r) = P(X \leq r) - P(X \leq r - 1)

P(X \geq r) = 1 - P(X \leq r - 1)

, and

P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)

. Translating a worded phrase ('at least', 'more than', 'fewer than') into the correct inequality is where marks are won or lost.

Checking that the binomial conditions actually apply is part of modelling. If trials are not independent (drawing without replacement from a small population) or the probability changes between trials, the binomial is not appropriate, and saying so — with a reason — is an examined AO3 skill.

X \sim B(n, p): \quad P(X = r) = \binom{n}{r} p^{r}(1 - p)^{n - r}

The binomial probability

Choose which $r$ of $n$ trials succeed; the mean is $np$ .

Worked example

A binomial calculation

$X \sim B(10, 0.2)$ . Find $P(X = 3)$ and $P(X \geq 1)$ .

Exact probability
$P(X = 3) = \binom{10}{3}(0.2)^3(0.8)^7 = 120 \times 0.008 \times 0.2097$ .
$P(X = 3) = \binom{10}{3}(0.2)^3(0.8)^7 \approx 0.2013$
Evaluate
$= 0.2013$ (4 d.p.).
At least one
$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.8)^{10} = 1 - 0.1074 = 0.8926$ .

Result: $P(X = 3) \approx 0.2013$ and $P(X \geq 1) \approx 0.8926$ .

Typical mistakes

Off-by-one errors, e.g. using $P(X \leq r)$ for $P(X < r)$ .
Applying the binomial when trials are not independent or $p$ is not constant.

Active revision

A biased coin has $P(\text{head}) = 0.3$ . It is tossed 12 times. Find the probability of (a) exactly 4 heads, and (b) at least 5 heads.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

The normal distribution#

AdvancedDfE GCE Mathematics — distributionsAQA 7357 (Statistics)

The standard normal curve

Fig. 3About 68% of the distribution lies within one standard deviation of the mean (shaded).

The normal distribution models a continuous variable that clusters symmetrically around a mean, such as heights, masses or measurement errors. It is written

X \sim N(\mu, \sigma^2)

, where

\mu

is the mean and

\sigma^2

the variance; its graph is the familiar symmetric bell curve, centred on

\mu

, with its spread controlled by

\sigma

. The total area under the curve is 1, and probabilities are areas under it.

Because the curve is symmetric, useful facts follow immediately: the mean, median and mode coincide at

\mu

, and about 68% of the data lie within one standard deviation of the mean, 95% within two, and 99.7% within three (the 68-95-99.7 rule). These proportions give quick estimates and a sense-check on any calculated probability.

Probabilities are found by standardising: the transformation

Z = \dfrac{X - \mu}{\sigma}

converts any normal variable to the standard normal

Z \sim N(0, 1)

, whose probabilities come from tables or a calculator. Standardising measures how many standard deviations a value is from the mean, and it is the bridge between a specific problem and the universal standard normal curve.

Two directions of problem occur. Forward problems give a value and ask for a probability (an area): standardise and read off. Inverse problems give a probability and ask for the value: find the

z

-value with that area, then reverse the standardisation as

x = \mu + z\sigma

. Sketching the curve and shading the required area first prevents almost all sign and tail errors.

X \sim N(\mu, \sigma^2): \quad Z = \frac{X - \mu}{\sigma} \sim N(0, 1)

Standardising

Convert to the standard normal to read probabilities; reverse with $x = \mu + z\sigma$ .

Worked example

A normal probability

$X \sim N(50, 4^2)$ . Find $P(X < 56)$ .

Standardise
$Z = \dfrac{56 - 50}{4} = 1.5$ .
Read the probability
$P(X < 56) = P(Z < 1.5) = 0.9332$ (from the standard normal).
$P(Z < 1.5) = 0.9332$
State
About 93.3% of values are below 56.

Result: $P(X < 56) \approx 0.9332$ .

Typical mistakes

Confusing the variance $\sigma^2$ with the standard deviation $\sigma$ when standardising.
Shading or reading the wrong tail, giving $1 -$ the intended probability.

Active revision

The masses of apples are modelled by $N(150, 20^2)$ grams. Find the probability that an apple has mass over 180 g, and the mass exceeded by only 10% of apples.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Modelling and choosing a distribution#

AdvancedDfE GCE Mathematics — distributionsAQA 7357 (Statistics)

Choosing between the two models

Fig. 4The binomial models discrete counts of successes; the normal models symmetric continuous data.

Choosing the right distribution is a modelling decision driven by the structure of the situation. The binomial fits a count of successes in a fixed number of independent, identical trials — the number of sixes in 20 rolls, or of defective items in a batch of fixed size. The normal fits a continuous quantity that is symmetric and clustered about a mean — heights, weights, or errors — and can also approximate other distributions in suitable circumstances.

The distinction between discrete and continuous is a first filter: a count of occurrences is discrete (points binomial), whereas a measurement on a continuous scale is continuous (points normal). Whether the binomial conditions genuinely hold — especially independence and constant probability — must then be checked; sampling without replacement from a small population, for instance, breaks independence.

Every model is an approximation, and part of the skill is stating its limitations. A normal model assigns non-zero probability to impossible values (a negative height), and a binomial model assumes trials really are independent; acknowledging such imperfections, and judging whether they matter for the question, is the honest modelling the specification rewards.

Finally, the distributions connect to the wider course. The binomial underlies hypothesis tests for a proportion, and the normal underlies tests for a mean and the interpretation of standardised scores. Recognising which distribution a problem calls for is therefore not just a labelling exercise but the gateway to the inference of the next topic.

Worked example

Selecting a distribution

A quality inspector checks a random sample of 30 items, each independently having a 5% chance of being faulty. State a distribution for the number of faulty items and justify it.

Identify the structure
There is a fixed number of trials ( $n = 30$ ), two outcomes (faulty or not), constant probability ( $p = 0.05$ ) and independence.
Name the distribution
The number of faulty items $X \sim B(30, 0.05)$ .
$X \sim B(30, 0.05)$
Comment
The binomial is appropriate provided items are genuinely independent and the fault rate is constant across the sample.

Result: $X \sim B(30, 0.05)$ , valid under independence and a constant 5% fault rate.

Typical mistakes

Using a binomial model when trials are not independent, or a normal model for clearly skewed data.
Failing to state the limitations of the chosen model.

Active revision

For each situation, state a suitable distribution with reasons: (a) the number of left-handed people in a random sample of 50; (b) the time taken to run 100 m by trained athletes.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Statistical distributions

4 sections~9 min reading time3 competenciesLevel Standard 2 · Advanced 2

Statistical distributions

Key points

Finding an unknown probability

Key points

A binomial calculation

Key points

A normal probability

Key points

Selecting a distribution

Sources

Statistical distributions

Key points

Finding an unknown probability

Key points

A binomial calculation

Key points

A normal probability

Key points

Selecting a distribution

Sources