SQA AH style: E and Var-style practice: A discrete random variable X has P(X=0)=0.2, P(X=1)=0.5, P(X=2)=0.3. Find E(X) and Var(X).

Expectation: E(X)=sum x\,P(X=x)=0(0.2)+1(0.5)+2(0.3)=0.5+0.6=1.1 (1 mark). Compute E(X^2)=sum x^2 P(X=x)=0(0.2)+1(0.5)+4(0.3)=0.5+1.2=1.7 (1 mark). Variance: Var(X)=E(X^2)-[E(X)]^2=1.7-1.1^2=1.7-1.21=0.49 (2 marks). Markers reward E(X), E(X^2) and the variance formula applied correctly.

ScotlandStatisticsSyllabus dot point

How do you model a count with a discrete random variable and find its expected value and variance?

Work with discrete probability distributions, calculate the expectation and variance of a discrete random variable and apply the laws of expectation and variance, and use the binomial, Poisson and geometric distributions as models.

A focused answer to the SQA Advanced Higher Statistics discrete random variables content: probability distributions, the expectation and variance of a discrete random variable, the laws of expectation and variance, and the binomial, Poisson and geometric distributions with their means and variances.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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Quick answer

For a discrete random variable, $E(X)=\sum x\,P(X=x)$ and $\text{Var}(X)=E(X^2)-[E(X)]^2$ . The laws give $E(aX+b)=aE(X)+b$ , $\text{Var}(aX+b)=a^2\text{Var}(X)$ , and for independent variables $E(X+Y)=E(X)+E(Y)$ , $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ . The three named models are binomial $B(n,p)$ with $E=np$ , $\text{Var}=np(1-p)$ , used for the number of successes in $n$ independent trials; Poisson $\text{Po}(\lambda)$ with $E=\text{Var}=\lambda$ , used for counts in a fixed interval; and geometric with $E=\dfrac{1}{p}$ , the number of trials to the first success.

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What this dot point is asking
Expectation and variance of a discrete random variable
The laws of expectation and variance
The binomial distribution
The Poisson distribution
The geometric distribution
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What this dot point is asking

A random variable turns the outcomes of a random experiment into numbers, and a discrete one takes separate values, typically counts. The SQA wants you to handle a general discrete distribution, find its expectation and variance, apply the laws of expectation and variance to linear transformations and sums, and recognise when the binomial, Poisson or geometric model applies.

Expectation and variance of a discrete random variable

Expectation is the long-run mean; variance measures spread about that mean.

Build a distribution and find its mean and variance

We are given a probability distribution and need to calculate two key summaries. The expectation is a weighted average of the values, and the variance uses the shortcut formula $\text{Var}(X)=E(X^2)-[E(X)]^2$ , which requires computing $E(X^2)$ as an intermediate step.

Step 1: Confirm the distribution is valid

Every probability must be non-negative and the probabilities must sum to $1$ . For $P(X=1)=0.4$ , $P(X=2)=0.4$ , $P(X=3)=0.2$ :

0.4+0.4+0.2=1 \checkmark

Step 2: Compute the expectation $E(X)$

Multiply each value by its probability and sum. This is the long-run average value of $X$ .

E(X)=1(0.4)+2(0.4)+3(0.2)=0.4+0.8+0.6=1.8

Step 3: Compute the variance using $\text{Var}(X)=E(X^2)-[E(X)]^2$

First find $E(X^2)$ by multiplying each squared value by its probability and summing, then subtract $[E(X)]^2$ .

E(X^2)=1^2(0.4)+2^2(0.4)+3^2(0.2)=0.4+1.6+1.8=3.8

\text{Var}(X)=3.8-1.8^2=3.8-3.24=0.56

Final answer: $E(X)=1.8$ and $\text{Var}(X)=0.56$ .

The laws of expectation and variance

These laws let you transform and combine random variables without recomputing the whole distribution.

Apply the laws to a linear transformation

When a random variable $X$ is scaled and shifted to form $Y=aX+b$ , we do not need to rebuild the whole distribution. The laws of expectation and variance tell us directly how the mean and spread change: multiplying by $a$ scales both the mean and the standard deviation, while adding a constant $b$ shifts the mean but leaves the spread unchanged.

Step 1: Identify the transformation

Starting from $X$ with $E(X)=1.8$ and $\text{Var}(X)=0.56$ , let $Y=3X-2$ . Here $a=3$ and $b=-2$ .

Step 2: Transform the mean using $E(aX+b)=aE(X)+b$

The mean is scaled by $a$ and shifted by $b$ .

E(Y)=3E(X)-2=3(1.8)-2=5.4-2=3.4

Step 3: Transform the variance using $\text{Var}(aX+b)=a^2\text{Var}(X)$

The variance is scaled by $a^2$ . The constant $b=-2$ contributes nothing to the variance because shifting a distribution does not change its spread.

\text{Var}(Y)=3^2\,\text{Var}(X)=9\times 0.56=5.04

Final answer: $E(Y)=3.4$ and $\text{Var}(Y)=5.04$ .

The binomial distribution

The binomial counts successes in a fixed number of independent yes/no trials.

The conditions are: a fixed number $n$ of trials, two outcomes per trial, a constant success probability $p$ , and independent trials. If any condition fails, the binomial does not apply.

The Poisson distribution

The Poisson models the number of events in a fixed interval of time or space when events occur independently at a constant average rate $\lambda$ .

A signature feature is that the mean equals the variance. Independent Poisson counts add: if $X\sim\text{Po}(\lambda_1)$ and $Y\sim\text{Po}(\lambda_2)$ are independent then $X+Y\sim\text{Po}(\lambda_1+\lambda_2)$ .

The geometric distribution

The geometric models the number of independent trials up to and including the first success.

Try this

Q1. $X\sim\text{Po}(3)$ . Find $P(X=2)$ . [2 marks]

Cue. $P(X=2)=\dfrac{e^{-3}3^2}{2!}=\dfrac{9e^{-3}}{2}\approx 0.224$ .

Q2. A fair die is rolled until the first six appears. State the expected number of rolls. [1 mark]

Cue. Geometric with $p=\tfrac{1}{6}$ , so $E(X)=\dfrac{1}{p}=6$ rolls.

Exam-style practice questions

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

AH style: E and Var4 marksA discrete random variable

X

has

P(X=0)=0.2

P(X=1)=0.5

P(X=2)=0.3

. Find

E(X)

and

\text{Var}(X)

Show worked answer →

Expectation: $E(X)=\sum x\,P(X=x)=0(0.2)+1(0.5)+2(0.3)=0.5+0.6=1.1$ (1 mark).

Compute $E(X^2)=\sum x^2 P(X=x)=0(0.2)+1(0.5)+4(0.3)=0.5+1.2=1.7$ (1 mark).

Variance: $\text{Var}(X)=E(X^2)-[E(X)]^2=1.7-1.1^2=1.7-1.21=0.49$ (2 marks). Markers reward $E(X)$ , $E(X^2)$ and the variance formula applied correctly.

AH style: binomial3 marksA biased coin lands heads with probability

0.3

. It is tossed

10

times. Find the probability of exactly

4

heads, and state the mean number of heads.

Show worked answer →

Model the number of heads as $X\sim B(10, 0.3)$ (1 mark).

$P(X=4)=\binom{10}{4}(0.3)^4(0.7)^6=210\times 0.0081\times 0.117649\approx 0.200$ (1 mark).

Mean of a binomial: $E(X)=np=10\times 0.3=3$ heads (1 mark). Markers reward the binomial model, the probability calculation and the mean.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)