Notes · StatisticsUK · A-Levels

Probability distributions

This topic develops continuous random variables through the probability density function and the cumulative distribution function, the expectation and variance of a continuous variable, and the continuous uniform distribution. It then treats the combination of independent random variables — the algebra of means and variances of sums and differences — which underlies the distribution of the sample mean and all later inference.

5 sections~10 min reading time3 competenciesLevel Standard 3 · Advanced 2

T·0999 / 18

Exam profile

AO1 · Work with probability density and cumulative distribution functions and compute E(X) and Var(X)AO2 · Use the algebra of means and variances of independent random variablesAO3 · Model with continuous distributions and combine random variables in context

Operators:findcalculatedetermineshow thatintegratemodel

basic level

The AS foundation treats discrete distributions; continuous random variables and the combination of variables are principally full A-Level (Paper 2) content.

higher level

The full A-Level expects pdfs and cdfs, continuous expectation and variance, the uniform distribution, and combining independent variables.

Depth

Text

Contents · 5 sections

Probability distributions

Continuous random variables and the pdf#

StandardPearson Edexcel 9ST0, Topic 9 (Paper 2)

A probability density function

Fig. 1The density f(x) = x/8 on [0, 4]; the shaded area P(1 < X < 3) = 0.5.

A continuous random variable takes values across an interval rather than a countable list, and its distribution is described by a probability density function

f(x)

. The density is not a probability itself; instead probabilities are areas under it, so

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

. Because a single point has zero width,

P(X = a) = 0

, and strict and non-strict inequalities coincide.

A valid density satisfies two conditions: it is non-negative everywhere,

f(x) \geq 0

, and the total area beneath it is 1,

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

. When a density is given up to an unknown constant, that constant is found by imposing the total-area condition, exactly as the sum-to-one condition fixed a discrete constant.

The triangular density

f(x) = \tfrac{1}{8}x

0 \leq x \leq 4

(and 0 elsewhere), shown below, has total area

\int_0^4 \tfrac{1}{8}x\,\mathrm{d}x = \tfrac{1}{8}\cdot\tfrac{16}{2} = 1

, confirming it is valid. The shaded strip is

P(1 < X < 3) = \int_1^3 \tfrac{1}{8}x\,\mathrm{d}x = \tfrac{1}{8}\cdot\tfrac{9 - 1}{2} = 0.5

The shape of the density carries the same information as a histogram of a large sample: where

f

is high the variable is more likely to fall, and the balance point of the area is the mean. Reading a density qualitatively — where it peaks, whether it is symmetric — is a useful first step before any integral is computed.

P(a < X < b) = \int_a^b f(x)\,\mathrm{d}x, \qquad \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 1, \qquad f(x) \geq 0

The probability density function

Probabilities are areas; the total area is 1 and the density is non-negative.

Worked example

Finding a density constant and a probability

A continuous variable has $f(x) = kx$ for $0 \leq x \leq 4$ and 0 elsewhere. Find $k$ and $P(1 < X < 3)$ .

Total area to 1
$\int_0^4 kx\,\mathrm{d}x = k\left[\tfrac{x^2}{2}\right]_0^4 = 8k = 1$ , so $k = \tfrac{1}{8}$ .
Interval probability
$P(1 < X < 3) = \int_1^3 \tfrac{1}{8}x\,\mathrm{d}x = \tfrac{1}{8}\left[\tfrac{x^2}{2}\right]_1^3 = \tfrac{1}{8}\cdot\tfrac{9 - 1}{2}$ .
$P(1 < X < 3) = \frac{1}{8}\cdot\frac{8}{2} = 0.5$
Evaluate
$= \tfrac{1}{8} \times 4 = 0.5$ .

Result: $k = \tfrac{1}{8}$ and $P(1 < X < 3) = 0.5$ .

Typical mistakes

Treating $f(x)$ as a probability rather than a density (areas are probabilities).
Integrating over the wrong range, especially where the density is defined piecewise.

Active revision

The density is $f(x) = kx^2$ for $0 \leq x \leq 3$ and 0 elsewhere. Find $k$ and $P(X > 2)$ .

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 9 (Paper 2)

The cumulative distribution function

Fig. 2F(x) = x^2/16 for the triangular density; F(2) = 0.25 and F(4) = 1.

The cumulative distribution function of a continuous variable is

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

, the area under the density up to

x

. It increases from 0 to 1, and interval probabilities come from differences:

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

. For the triangular density

f(x) = \tfrac{1}{8}x

[0, 4]

, integration gives

F(x) = \tfrac{x^2}{16}

on that interval, shown below rising to 1 at

x = 4

The density and the cdf are linked by calculus:

F

is the integral of

f

, and conversely

f(x) = F'(x)

, the derivative of the cdf. This means the cdf is the smooth, always-increasing partner of the density, and either one determines the other — a continuous echo of the differencing relationship between the discrete cdf and probability function.

The cdf is the direct route to medians and quartiles of a continuous distribution. The median

m

solves

F(m) = 0.5

, and the lower and upper quartiles solve

F(x) = 0.25

and

F(x) = 0.75

. For the triangular example,

F(m) = \tfrac{m^2}{16} = 0.5

gives

m = \sqrt{8} = 2.83

(3 s.f.), which lies above the midpoint 2 because the density leans right.

Working with the cdf is often quicker than repeated integration, because once

F

is known every probability and quantile is a substitution or a simple equation. Establishing

F(x)

early, and stating its value at the endpoints of the support (0 below and 1 above), is good practice on any continuous-distribution question.

F(x) = \int_{-\infty}^{x} f(t)\,\mathrm{d}t, \qquad f(x) = F'(x), \qquad F(m) = 0.5 \text{ gives the median}

The continuous cumulative distribution function

Integrate the density for $F$ ; differentiate $F$ for the density; solve $F(m)=0.5$ for the median.

Worked example

Median from the cdf

For the triangular density with $F(x) = \tfrac{x^2}{16}$ on $[0, 4]$ , find the median.

Set F equal to 0.5
$\tfrac{m^2}{16} = 0.5$ , so $m^2 = 8$ .
Solve
$m = \sqrt{8} = 2.83$ (3 s.f.), taking the positive root in $[0, 4]$ .
$m = \sqrt{8} = 2.83$
Interpret
The median 2.83 exceeds the midpoint 2 because the density increases across the interval, placing more probability to the right.

Result: The median is $\sqrt{8} = 2.83$ (3 s.f.).

Typical mistakes

Forgetting the constant of integration / the lower limit, so $F$ does not start at 0.
Solving $f(m) = 0.5$ instead of $F(m) = 0.5$ for the median.

Active revision

For $F(x) = \tfrac{x^2}{16}$ on $[0, 4]$ , find the median and $P(X > 3)$ .

Active recall

Recall the key points — then reveal.

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

Expectation and variance of a continuous variable#

AdvancedPearson Edexcel 9ST0, Topic 9 (Paper 2)

The expectation of a continuous variable replaces the discrete sum with an integral:

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

over the support. It is the balance point of the density — the value about which the area is evenly distributed — and for a symmetric density it sits at the centre of symmetry. The same pattern gives

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

for any function of

X

The variance uses

E(X^2) = \int x^2 f(x)\,\mathrm{d}x

and the familiar identity

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

. For the triangular density

f(x) = \tfrac{1}{8}x

[0, 4]

E(X) = \int_0^4 \tfrac{x^2}{8}\,\mathrm{d}x = \tfrac{1}{8}\cdot\tfrac{64}{3} = \tfrac{8}{3}

and

E(X^2) = \int_0^4 \tfrac{x^3}{8}\,\mathrm{d}x = 8

, so

\text{Var}(X) = 8 - \left(\tfrac{8}{3}\right)^2 = \tfrac{8}{9}

The linear-transformation rules carry over unchanged from the discrete case:

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

and

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

. This consistency reflects that expectation and variance are defined by the same weighting idea whether the weights are probabilities (discrete) or a density times

\mathrm{d}x

(continuous).

These integrals are the definitions behind the closed-form means and variances of the named distributions — the exponential mean

\tfrac{1}{\lambda}

and the uniform mean

\tfrac{a + b}{2}

are both instances. Being able to compute them from first principles, and to check them against a symmetry argument where one applies, is the core skill of this section.

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

Continuous expectation and variance

The discrete sums become integrals of the density; the variance identity is unchanged.

Worked example

Mean and variance by integration

For the density $f(x) = \tfrac{1}{8}x$ on $[0, 4]$ , find $E(X)$ and $\text{Var}(X)$ .

Mean
$E(X) = \int_0^4 x\cdot\tfrac{x}{8}\,\mathrm{d}x = \tfrac{1}{8}\left[\tfrac{x^3}{3}\right]_0^4 = \tfrac{1}{8}\cdot\tfrac{64}{3} = \tfrac{8}{3}$ .
Second moment
$E(X^2) = \int_0^4 x^2\cdot\tfrac{x}{8}\,\mathrm{d}x = \tfrac{1}{8}\left[\tfrac{x^4}{4}\right]_0^4 = \tfrac{1}{8}\cdot 64 = 8$ .
Variance
$\text{Var}(X) = 8 - \left(\tfrac{8}{3}\right)^2 = 8 - \tfrac{64}{9} = \tfrac{8}{9}$ .
$\text{Var}(X) = 8 - \frac{64}{9} = \frac{8}{9}$

Result: $E(X) = \tfrac{8}{3} = 2.67$ and $\text{Var}(X) = \tfrac{8}{9} = 0.889$ (3 s.f.).

Typical mistakes

Integrating $f(x)$ instead of $x f(x)$ when finding the mean.
Using $\left(\int x f\right)^2$ for $E(X^2)$ rather than $\int x^2 f$ .

Active revision

For $f(x) = \tfrac{1}{8}x$ on $[0, 4]$ , find $E(X)$ and $\text{Var}(X)$ .

Active recall

Recall the key points — then reveal.

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

The continuous uniform distribution#

StandardPearson Edexcel 9ST0, Topic 9 (Paper 2)

The rectangular (uniform) density

Fig. 3Uniform density on [2, 8]: constant height 1/6; the shaded region P(3 < X < 5) = 1/3.

The continuous uniform (rectangular) distribution spreads probability evenly over an interval

[a, b]

, with constant density

f(x) = \dfrac{1}{b - a}

there and 0 elsewhere. The density is a rectangle of width

b - a

and height

\dfrac{1}{b - a}

, so its area is exactly 1, as the figure for

[2, 8]

(height

\tfrac{1}{6}

) shows. It models a quantity known only to lie somewhere in a range with no preference.

By symmetry the mean is the midpoint,

E(X) = \dfrac{a + b}{2}

, and integration gives the variance

\text{Var}(X) = \dfrac{(b - a)^2}{12}

. For

[2, 8]

these are

E(X) = 5

and

\text{Var}(X) = \dfrac{36}{12} = 3

. Probabilities are proportions of the interval:

P(c < X < d) = \dfrac{d - c}{b - a}

for any sub-interval

[c, d]

inside

[a, b]

The uniform is the continuous analogue of the discrete uniform met earlier, and it is a useful baseline model and a source of clean check-cases. It also appears as the distribution of rounding errors and of a randomly chosen point in an interval, and it is the distribution from which computer random numbers on

[0, 1]

are drawn.

Its simple, closed-form results make it ideal for testing the general machinery: computing

E(X)

and

\text{Var}(X)

from the integrals and checking them against the formulae reinforces the methods of the previous section, while the proportion rule for probabilities gives an intuitive feel for area-as-probability.

f(x) = \frac{1}{b - a} \;\text{on}\; [a, b]; \quad E(X) = \frac{a + b}{2}, \quad \text{Var}(X) = \frac{(b - a)^2}{12}

The continuous uniform distribution

Constant density; mean at the midpoint; probabilities are proportions of the interval.

Worked example

A uniform distribution

A bus is equally likely to arrive at any time in the 20-minute window $[0, 20]$ after the hour. Find the mean and variance of the waiting time and $P(X < 5)$ .

Mean
$E(X) = \dfrac{0 + 20}{2} = 10$ minutes.
Variance
$\text{Var}(X) = \dfrac{(20 - 0)^2}{12} = \dfrac{400}{12} = 33.3$ (3 s.f.).
$\text{Var}(X) = \frac{400}{12} = 33.3$
Probability
$P(X < 5) = \dfrac{5 - 0}{20 - 0} = 0.25$ .

Result: Mean 10 minutes, variance 33.3 (3 s.f.), and $P(X < 5) = 0.25$ .

Typical mistakes

Using $\frac{(b - a)^2}{12}$ but forgetting the square, or dividing by a different constant.
Applying the proportion rule to a sub-interval that extends outside $[a, b]$ .

Active revision

A quantity is uniform on $[10, 30]$ . Find its mean, variance and $P(X < 18)$ .

Active recall

Recall the key points — then reveal.

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

Combining independent random variables#

AdvancedPearson Edexcel 9ST0, Topic 9 (Paper 2)

Expectation is linear without any independence assumption:

E(aX + bY) = aE(X) + bE(Y)

, so means simply combine according to the coefficients. Variance, however, requires independence: for independent

X

and

Y

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

. The coefficients are squared, and — a point that catches many students — the variance of a DIFFERENCE is the SUM of the variances, because

(-1)^2 = 1

A crucial special case is the sum or mean of

n

independent copies of

X

(a random sample). The total

T = X_1 + \cdots + X_n

has mean

n\mu

and variance

n\sigma^2

, while the sample mean

\bar{X} = \tfrac{T}{n}

has mean

\mu

and variance

\dfrac{\sigma^2}{n}

. The standard deviation of the sample mean,

\dfrac{\sigma}{\sqrt{n}}

, is the standard error that governs all of estimation and testing.

When the variables being combined are normal, the combination is also normal, so its distribution is completely determined by the combined mean and variance. If

X \sim N(50, 4^2)

and

Y \sim N(30, 3^2)

independently, then

X + Y \sim N(80, 25)

and

X - Y \sim N(20, 25)

— the same variance 25 in each, since the variances add either way.

These rules are the algebraic engine of the whole inference course. That

\text{Var}(\bar{X}) = \dfrac{\sigma^2}{n}

decreases with

n

is why larger samples give more precise estimates; that a difference of independent means is normal with the summed variances is exactly what makes the two-sample tests of the later topics work.

\text{Var}(aX \pm bY) = a^2\,\text{Var}(X) + b^2\,\text{Var}(Y), \qquad \text{Var}(\bar{X}) = \frac{\sigma^2}{n}

Variance of a combination (independent)

Coefficients square; a difference sums the variances; the sample mean has variance $\sigma^2/n$ .

Worked example

A difference of independent normals

The masses of two independent components are $X \sim N(50, 4^2)$ and $Y \sim N(30, 3^2)$ grams. Find the distribution of the difference $D = X - Y$ and $P(D > 25)$ .

Mean of the difference
$E(D) = 50 - 30 = 20$ .
Variance of the difference
$\text{Var}(D) = 4^2 + 3^2 = 16 + 9 = 25$ , so $D \sim N(20, 25)$ with standard deviation 5.
Standardise
$P(D > 25) = P\!\left(Z > \dfrac{25 - 20}{5}\right) = P(Z > 1)$ .
$P(D > 25) = P(Z > 1) = 1 - 0.8413 = 0.1587$
Evaluate
$= 1 - \Phi(1) = 0.1587$ .

Result: $D \sim N(20, 25)$ and $P(D > 25) = 0.1587$ .

Typical mistakes

Subtracting variances for a difference; independent variances always add.
Forgetting to square the coefficients, or applying the variance rule without independence.

Active revision

$X \sim N(50, 4^2)$ and $Y \sim N(30, 3^2)$ are independent. Find the distribution of $X - Y$ and $P(X - Y > 25)$ .

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)

Probability distributions

Key points

Finding a density constant and a probability

Key points

Median from the cdf

Key points

Mean and variance by integration

Key points

A uniform distribution

Key points

A difference of independent normals

Sources