What is probabilities not summing to one?

Always check sum P = 1 (or use it to find an unknown constant) before using the distribution.

X has E(X) = 4 and Var(X) = 3. Find Var(2X + 5). [2 marks]

Pearson Edexcel Edexcel 2021-style practice: The random variable X has E(X) = 5 and Var(X) = 4. Find E(3X - 2) and Var(3X - 2), and find E(X^2).

Apply the linear-coding rules, then rearrange the variance definition for E(X^2). E(3X - 2) = 3E(X) - 2 = 3(5) - 2 = 13 (M1 A1). Var(3X - 2) = 3^2Var(X) = 9(4) = 36 (the constant -2 has no effect on variance) (M1 A1). From Var(X) = E(X^2) - [E(X)]^2: 4 = E(X^2) - 25, so E(X^2) = 29 (A1).

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How do you describe a discrete random variable and compute its expectation and variance?

Discrete random variables and probability distributions, expectation and variance, the effect of linear coding, and expectation and variance of functions of a discrete variable.

A focused answer to the Edexcel A-Level Further Mathematics Further Statistics content on discrete probability distributions, covering discrete random variables and their distributions, expectation and variance, the effect of linear coding, and the expectation and variance of functions of a discrete random variable.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Distributions, expectation and variance
Functions of a discrete variable
Linear coding
Examples in context
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What this dot point is asking

Edexcel Further Statistics wants you to define a discrete random variable through its probability distribution, compute expectation $E(X)$ and variance $\operatorname{Var}(X)$ , find the expectation and variance of a function of $X$ , and use the effect of linear coding $E(aX + b)$ and $\operatorname{Var}(aX + b)$ . These are the foundational tools that the named distributions (Poisson, geometric, negative binomial) all build on.

Distributions, expectation and variance

A probability distribution lists each value the variable can take alongside its probability, and these probabilities must sum to $1$ . The expectation (mean) is the probability-weighted average of the values; the variance measures the spread about that mean, computed most efficiently as $E(X^2) - [E(X)]^2$ rather than from the deviations directly.

Expectation and variance of a discrete variable

$X$ takes values $1, 2, 3$ with probabilities $0.2, 0.5, 0.3$ . Find $E(X)$ and $\operatorname{Var}(X)$ .

Step 1: Check the probabilities are valid

Before computing anything, confirm that the probabilities sum to $1$ , which is the fundamental requirement for any probability distribution:

0.2 + 0.5 + 0.3 = 1. \checkmark

Step 2: Compute $E(X)$

The expectation is the probability-weighted average of the possible values. Multiply each value by its probability and add:

E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1.

Step 3: Compute $E(X^2)$

To find the variance we need $E(X^2) = \sum x^2 P(X = x)$ . Square each value of $x$ first, then weight by the same probabilities:

E(X^2) = 1^2(0.2) + 2^2(0.5) + 3^2(0.3) = 0.2 + 2.0 + 2.7 = 4.9.

Step 4: Apply the shortcut variance formula

The formula $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ avoids summing squared deviations. Substitute the values found above:

\operatorname{Var}(X) = 4.9 - (2.1)^2 = 4.9 - 4.41 = 0.49.

Final answer: $E(X) = 2.1$ and $\operatorname{Var}(X) = 0.49$ .

Functions of a discrete variable

For a function $g(X)$ , the expectation is $E(g(X)) = \sum g(x) P(X = x)$ , summing the function values weighted by the same probabilities. This is how $E(X^2)$ is computed, and it extends to any function, but note that in general $E(g(X)) \neq g(E(X))$ unless $g$ is linear.

Expectation of a function

For the variable above ( $X = 1, 2, 3$ with probabilities $0.2, 0.5, 0.3$ ), find $E(2X^2 + 1)$ .

Step 1: Evaluate the function $g(x) = 2x^2 + 1$ at each possible value

Compute $g(x)$ for each value $X$ can take:

g(1) = 2(1)^2 + 1 = 3, \quad g(2) = 2(2)^2 + 1 = 9, \quad g(3) = 2(3)^2 + 1 = 19.

Step 2: Weight by the corresponding probabilities and sum

The general rule is $E(g(X)) = \sum g(x) P(X = x)$ , weighting each function value by the probability of that outcome:

E(2X^2 + 1) = 3(0.2) + 9(0.5) + 19(0.3) = 0.6 + 4.5 + 5.7 = 10.8.

Step 3: Verify using linearity of expectation

Because $g$ is linear in $X^2$ , we can also write $E(2X^2 + 1) = 2E(X^2) + 1$ . Using $E(X^2) = 4.9$ from the previous example:

2(4.9) + 1 = 10.8. \checkmark

Final answer: $E(2X^2 + 1) = 10.8$ .

Linear coding

Scaling and shifting a variable changes its mean and variance in fixed, predictable ways, which is the basis of coding to simplify calculations. Adding a constant slides the whole distribution along the number line, moving the mean but leaving the spread unchanged. Multiplying by a constant stretches the distribution, scaling the standard deviation by the multiplier and hence the variance by its square.

Examples in context

These definitions are the engine room of Further Statistics. The named distributions all have their means and variances derived by these sums: the Poisson has $E(X) = \operatorname{Var}(X) = \lambda$ , the binomial $E(X) = np$ and $\operatorname{Var}(X) = np(1 - p)$ , and the geometric $E(X) = \frac{1}{p}$ . Linear coding is used to standardise data and to relate raw scores to coded ones in calculations. The summation techniques mirror the series work in further algebra, where $\sum r$ and $\sum r^2$ appear, and the expectation of a function underlies the moment calculations used in fitting distributions for chi-squared tests.

Try this

Q1. $X$ has $E(X) = 4$ and $\operatorname{Var}(X) = 3$ . Find $\operatorname{Var}(2X + 5)$ . [2 marks]

Cue. $\operatorname{Var}(2X + 5) = 2^2 \times 3 = 12$ .

Q2. For the same $X$ , find $E(2X + 5)$ . [1 mark]

Cue. $E(2X + 5) = 2(4) + 5 = 13$ .

Q3. A variable has $E(X) = 2$ and $E(X^2) = 7$ . Find its variance. [2 marks]

Cue. $\operatorname{Var}(X) = 7 - 2^2 = 3$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20187 marksThe discrete random variable

X

has probability distribution

P(X = x) = kx

for

x = 1, 2, 3, 4

. Find the value of

k

, then find

E(X)

and

\operatorname{Var}(X)

Show worked answer →

Use the total-probability condition to find $k$ , then the standard expectation and variance sums.

$\sum P(X = x) = k(1 + 2 + 3 + 4) = 10k = 1$ , so $k = \frac{1}{10}$ (M1 A1).

$E(X) = \sum x P(X = x) = \frac{1}{10}(1^2 + 2^2 + 3^2 + 4^2) = \frac{1 + 4 + 9 + 16}{10} = 3$ (M1 A1).

$E(X^2) = \sum x^2 P(X = x) = \frac{1}{10}(1 + 8 + 27 + 64) = \frac{100}{10} = 10$ (M1).

$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 10 - 9 = 1$ (A1, plus A1 for fully correct working).

Edexcel 20215 marksThe random variable

X

has

E(X) = 5

and

\operatorname{Var}(X) = 4

. Find

E(3X - 2)

and

\operatorname{Var}(3X - 2)

, and find

E(X^2)

Show worked answer →

Apply the linear-coding rules, then rearrange the variance definition for $E(X^2)$ .

$E(3X - 2) = 3E(X) - 2 = 3(5) - 2 = 13$ (M1 A1).

$\operatorname{Var}(3X - 2) = 3^2\operatorname{Var}(X) = 9(4) = 36$ (the constant $-2$ has no effect on variance) (M1 A1).

From $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ : $4 = E(X^2) - 25$ , so $E(X^2) = 29$ (A1).

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Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)