Notes · StatisticsUK · A-Levels

Introduction to probability distributions

This topic introduces discrete random variables and their probability distributions, the cumulative distribution function, and the expectation and variance that summarise a distribution. It develops the algebra of linear transformations E(aX + b) and Var(aX + b) and the discrete uniform distribution, providing the language used by every named distribution that follows.

5 sections~11 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

T·0555 / 18

Exam profile

AO1 · Compute probabilities, expectation and variance from a discrete distributionAO2 · Use the expectation and variance algebra for linear transformationsAO3 · Model a discrete situation with a random variable and interpret its mean and spread

Operators:findcalculatedetermineshow thatmodelinterpret

basic level

The AS foundation expects discrete probability distributions and the use of the total-probability-of-one condition to find an unknown constant.

higher level

The full A-Level expects fluent expectation and variance, the linear-transformation rules and the discrete uniform distribution.

Depth

Text

Contents · 5 sections

Introduction to probability distributions

Discrete random variables and probability functions#

FoundationPearson Edexcel 9ST0, Topic 4 (Paper 1)

A discrete probability distribution

Fig. 1The distribution of X: probabilities 0.1, 0.2, 0.3, 0.4 sum to 1.

A discrete random variable

X

takes a countable set of values, each with an associated probability, and its probability distribution lists those values against their probabilities

P(X = x)

, given as a table or a formula. The two defining conditions are that every probability lies between 0 and 1, and that they sum to 1 over all possible values:

\sum P(X = x) = 1

. This normalisation condition is the tool used to find an unknown constant in a distribution.

When a distribution is given by a formula such as

P(X = x) = kx

for

x = 1, 2, 3, 4

, the constant

k

is found by imposing

\sum P(X = x) = 1

: here

k(1 + 2 + 3 + 4) = 10k = 1

, so

k = \tfrac{1}{10}

. Setting up and solving this equation is the standard first step whenever a distribution contains an unknown, and it must be done before any probability or average is computed.

Once the full distribution is known, any event probability is the sum of the probabilities of the values in it. So

P(X \geq 3)

is obtained by adding

P(X = 3)

and any larger values, and

P(1 < X \leq 3)

by adding

P(X = 2)

and

P(X = 3)

. Care with strict versus non-strict inequalities is essential for a discrete variable, because the endpoints carry probability.

The distribution below, with

P(X = x)

equal to 0.1, 0.2, 0.3 and 0.4 for

x = 1, 2, 3, 4

, is used throughout this topic. Its bars sum to 1, and it is deliberately asymmetric so that the mean, variance and their transformations are non-trivial to compute.

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

A valid probability distribution

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

Worked example

Finding an unknown constant

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

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

Typical mistakes

Not imposing that the probabilities sum to 1 when a constant is unknown.
Mishandling strict versus non-strict inequalities for a discrete variable, so including or excluding an endpoint wrongly.

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: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

The cumulative distribution function#

StandardPearson Edexcel 9ST0, Topic 4 (Paper 1)

The cumulative distribution function

Fig. 2F(x) = P(X <= x) rises to 1: values 0.1, 0.3, 0.6, 1.0 at x = 1, 2, 3, 4.

The cumulative distribution function (cdf) of a discrete random variable is

F(x) = P(X \leq x)

, the running total of the probabilities up to and including

x

. It starts at 0 below the smallest value, increases in a step at each value of

X

by that value's probability, and reaches 1 at the largest value. For the distribution 0.1, 0.2, 0.3, 0.4 the cdf takes the values 0.1, 0.3, 0.6 and 1.0 at

x = 1, 2, 3, 4

The cdf makes range probabilities immediate:

P(X \leq x)

is read directly,

P(X > x) = 1 - F(x)

, and

P(a < X \leq b) = F(b) - F(a)

. For a discrete variable these must be handled with care over the endpoints, since

P(X < x)

differs from

P(X \leq x)

by exactly

P(X = x)

Conversely, the individual probabilities can be recovered from the cdf by differencing:

P(X = x) = F(x) - F(x-1)

, where

x-1

denotes the previous value of

X

. This two-way relationship between the probability function and the cdf is exactly the discrete analogue of the relationship between a density and its integral for continuous variables in a later topic.

The cdf is the natural object for the cumulative tables used with the binomial and Poisson distributions, where a calculator or table returns

P(X \leq r)

and every other probability is obtained from it by subtraction. Fluency with

F(x)

here pays off directly in those distributions and in hypothesis testing, where critical regions are defined through cumulative probabilities.

F(x) = P(X \leq x), \qquad P(a < X \leq b) = F(b) - F(a)

The cumulative distribution function

The running total of probabilities; range probabilities come from differences of $F$ .

Worked example

Range probability from the cdf

Given $F(1) = 0.1$ , $F(2) = 0.3$ , $F(3) = 0.6$ , $F(4) = 1.0$ , find $P(2 \leq X \leq 3)$ .

Rewrite the range
$P(2 \leq X \leq 3) = F(3) - F(1)$ , because $F(1)$ removes $X = 1$ but keeps $X = 2, 3$ .
Substitute
$F(3) - F(1) = 0.6 - 0.1 = 0.5$ .
$P(2 \leq X \leq 3) = F(3) - F(1) = 0.6 - 0.1 = 0.5$
Check
Directly, $P(X = 2) + P(X = 3) = 0.2 + 0.3 = 0.5$ , which agrees.

Result: $P(2 \leq X \leq 3) = 0.5$ .

Typical mistakes

Confusing $P(X < x)$ with $P(X \leq x)$ , which differ by $P(X = x)$ .
Forgetting that the cdf must reach exactly 1 at the largest value.

Active revision

For the distribution with $P(X=x)$ equal to 0.1, 0.2, 0.3, 0.4 at $x = 1, 2, 3, 4$ , write down $F(x)$ and find $P(2 \leq X \leq 3)$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Expectation of a discrete random variable#

StandardPearson Edexcel 9ST0, Topic 4 (Paper 1)

The expectation, or mean, of a discrete random variable is

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

, the average value of

X

weighted by its probabilities. It is the long-run average outcome: if the experiment were repeated many times, the mean of the observed values would settle towards

E(X)

. It need not be a value the variable can actually take — the expected score on the distribution below is exactly 3.0, and for a fair die it is 3.5.

Computing

E(X)

is a matter of multiplying each value by its probability and summing. For the distribution with

P(X = x)

equal to 0.1, 0.2, 0.3, 0.4 at

x = 1, 2, 3, 4

E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3.0

. Laying the calculation out in a table of

x

P(X = x)

and

xP(X = x)

keeps it organised and reduces slips.

Expectation extends to functions of

X

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

. The most important case is

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

, needed for the variance in the next section. Note carefully that

E(X^2)

is generally NOT equal to

[E(X)]^2

; the difference between them is exactly the variance.

Expectation is linear:

E(aX + b) = aE(X) + b

for constants

a

and

b

. This follows directly from the definition and means a linear rescaling of the variable rescales its mean in the same way, a fact used constantly when changing units or standardising.

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

Expectation

The probability-weighted average of $X$ (or of a function of $X$ ).

Worked example

Computing E(X) and E(X^2)

For the distribution with $P(X = x)$ equal to 0.1, 0.2, 0.3, 0.4 at $x = 1, 2, 3, 4$ , find $E(X)$ and $E(X^2)$ .

E(X)
$1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$ .
$E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$
E(X²)
$1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0$ .
Note
$[E(X)]^2 = 9$ , which is not $E(X^2) = 10$ ; the gap of 1 is the variance.

Result: $E(X) = 3.0$ and $E(X^2) = 10.0$ .

Typical mistakes

Writing $E(X^2) = [E(X)]^2$ ; these are different and their difference is the variance.
Forgetting to weight each value by its probability, so averaging the values only.

Active revision

For $P(X = x)$ equal to 0.1, 0.2, 0.3, 0.4 at $x = 1, 2, 3, 4$ , find $E(X)$ and $E(X^2)$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Variance and standard deviation#

StandardPearson Edexcel 9ST0, Topic 4 (Paper 1)

The variance of a discrete random variable measures the spread of its distribution about the mean,

\text{Var}(X) = E\big((X - \mu)^2\big)

where

\mu = E(X)

. In practice it is almost always computed by the equivalent form

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

, which needs only the two expectations already found. The standard deviation

\sqrt{\text{Var}(X)}

returns the spread to the original units of

X

For the distribution with

E(X) = 3.0

and

E(X^2) = 10.0

, the variance is

10 - 3^2 = 1

, so the standard deviation is 1. The computational form is preferred because it avoids the awkward table of squared deviations, though the two forms are algebraically identical and give the same answer.

Variance behaves differently from expectation under a linear transformation:

\text{Var}(aX + b) = a^2 \text{Var}(X)

. Adding a constant

b

shifts every value equally and leaves the spread unchanged, so

b

has no effect; multiplying by

a

scales the deviations by

a

and hence the variance by

a^2

. The standard deviation scales by

|a|

These rules combine cleanly: if

Y = 2X + 3

then

E(Y) = 2E(X) + 3

and

\text{Var}(Y) = 4\,\text{Var}(X)

. Recognising that only the multiplier affects the spread — and that it does so as its square — prevents the common error of adding the shift to the variance, and it underlies the standard-error formulae met in estimation.

\text{Var}(X) = E(X^2) - [E(X)]^2, \qquad \text{Var}(aX + b) = a^2\,\text{Var}(X)

Variance and its linear transformation

A shift $b$ leaves the variance unchanged; a scale $a$ multiplies it by $a^2$ .

Worked example

Variance and a linear transformation

For the distribution with $E(X) = 3$ and $E(X^2) = 10$ , find $\text{Var}(X)$ and hence $E(Y)$ and $\text{Var}(Y)$ where $Y = 2X + 3$ .

Variance of X
$\text{Var}(X) = 10 - 3^2 = 1$ .
Mean of Y
$E(Y) = 2(3) + 3 = 9$ .
Variance of Y
$\text{Var}(Y) = 2^2 \times 1 = 4$ ; the $+3$ has no effect.
$\text{Var}(Y) = 2^2 \times \text{Var}(X) = 4$
Standard deviation of Y
$\sqrt{4} = 2$ .

Result: $\text{Var}(X) = 1$ , $E(Y) = 9$ , $\text{Var}(Y) = 4$ and the standard deviation of $Y$ is 2.

Typical mistakes

Adding the shift constant $b$ into the variance; only the scale factor matters.
Scaling the variance by $a$ instead of $a^2$ .

Active revision

If $\text{Var}(X) = 1$ and $Y = 2X + 3$ , find $\text{Var}(Y)$ and the standard deviation of $Y$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Linear transformations and the discrete uniform distribution#

AdvancedPearson Edexcel 9ST0, Topic 4 (Paper 1)

The discrete uniform distribution assigns equal probability to each of

n

equally likely values, the standard model being the score on a fair die, uniform on

1, 2, \ldots, n

with

P(X = x) = \frac{1}{n}

. It is the simplest named discrete distribution and a useful check-case for the expectation and variance machinery, because its mean and variance have closed forms.

For the uniform distribution on

1

n

, summing the arithmetic and square series gives

E(X) = \frac{n + 1}{2}

and

\text{Var}(X) = \frac{n^2 - 1}{12}

. For a fair six-sided die these give

E(X) = 3.5

and

\text{Var}(X) = \frac{35}{12} = 2.92

(3 s.f.), results worth remembering as a benchmark and as a check on general calculations.

The linear-transformation rules bring modelling flexibility. If a game pays out

\pounds(2X + 1)

on a die score

X

, the expected payout is

2E(X) + 1 = 2(3.5) + 1 = \pounds8

and its variance is

4\,\text{Var}(X) = 4 \times \frac{35}{12} = \frac{35}{3}

. This lets the mean and spread of a rescaled quantity be found without rebuilding its distribution.

The value of these results is conceptual as much as computational: they show that a distribution is fully summarised for many purposes by its mean and variance, and that these summaries transform predictably under rescaling. That principle carries directly into the named distributions and into the standard-error and confidence-interval work of the inference topics.

E(X) = \frac{n + 1}{2}, \qquad \text{Var}(X) = \frac{n^2 - 1}{12}

Discrete uniform on 1 to n

For equally likely values $1, 2, \ldots, n$ ; for a fair die ( $n = 6$ ) these give 3.5 and 2.92.

Worked example

Mean and variance of a fair die, rescaled

A fair six-sided die shows score $X$ . A game pays $\pounds(2X + 1)$ . Find the expected payout and its variance.

Mean of X
$E(X) = \frac{6 + 1}{2} = 3.5$ .
Variance of X
$\text{Var}(X) = \frac{6^2 - 1}{12} = \frac{35}{12}$ .
Expected payout
$E(2X + 1) = 2(3.5) + 1 = 8$ , so the expected payout is $\pounds8$ .
Variance of payout
$\text{Var}(2X + 1) = 2^2 \times \frac{35}{12} = \frac{35}{3} = 11.7$ (3 s.f.).
$\text{Var}(2X+1) = 4 \times \frac{35}{12} = \frac{35}{3}$

Result: The expected payout is $\pounds8$ with variance $\frac{35}{3} = 11.7$ (3 s.f.).

Typical mistakes

Using $\frac{n^2 - 1}{12}$ for a uniform distribution that does not start at 1 without adjusting.
Applying $\text{Var}(aX + b) = a^2\text{Var}(X)$ but rescaling the standard deviation by $a^2$ instead of $|a|$ .

Active revision

A fair eight-sided die is scored $1$ to $8$ . Find $E(X)$ and $\text{Var}(X)$ .

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

References & sources

Sources

Pearson Edexcel

Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Department for Education / Ofqual

GCE AS and A level subject content (Statistics)

Introduction to probability distributions

Key points

Finding an unknown constant

Key points

Range probability from the cdf

Key points

Computing E(X) and E(X^2)

Key points

Variance and a linear transformation

Key points

Mean and variance of a fair die, rescaled

Sources