Random Variables & Probability Distributions (Mean, Variance)

MATRandom Variables & Distributions

A (discrete) random variable $X$ assigns a numerical value to each outcome, and its probability distribution lists every value $x_i$ with its probability $p_i$ . From this table you compute the mean (expectation) $E(X)$ and variance $\operatorname{Var}(X)$ , the standard summary measures examined via coin-tossing, card-drawing and spinner distributions.

Valid probability distribution

p_i = P(X = x_i) \ge 0, \qquad \sum_{i} p_i = 1

x_i

are the distinct values of

X

; each probability is non-negative and they sum to

1

. Use this to find an unknown constant

k

Mean (expectation)

\mu = E(X) = \sum_{i} x_i\,p_i

Sum of each value times its probability;

\mu

is the long-run average value of

X

Variance

\operatorname{Var}(X) = E(X^{2}) - \big(E(X)\big)^{2} = \sum_{i} x_i^{2}\,p_i - \mu^{2}

E(X^{2}) = \sum x_i^{2} p_i

; the form

E(X^2) - \mu^2

is the quick computational version, equal to

\sum (x_i-\mu)^2 p_i

Standard deviation

\sigma = \sqrt{\operatorname{Var}(X)}

\sigma \ge 0

; the positive square root of the variance, in the same units as

X

First build a clean table of all values $x_i$ and probabilities $p_i$ from the experiment before computing anything.
To find an unknown constant $k$ , set $\sum p_i = 1$ and solve the resulting equation.
Compute $E(X) = \sum x_i p_i$ and $E(X^2) = \sum x_i^2 p_i$ in the same table (add columns $x_i p_i$ and $x_i^2 p_i$ ).
Use $\operatorname{Var}(X) = E(X^2) - \big(E(X)\big)^2$ — it is faster and less error-prone than $\sum (x_i-\mu)^2 p_i$ , though both are equal.
Variance is always $\ge 0$ ; a negative result signals an arithmetic error (usually subtracting $\mu^2$ wrongly).
For "number of heads in 3 tosses" or "number of kings drawn", the distribution can be found by counting or, where trials are identical and independent, as a special binomial case.
The mean need not be one of the attainable values of $X$ (e.g. $1.5$ heads), which is expected for an average.

Building a probability distribution from an experiment

1Say in words what $X$ counts, then list every value it can take — this is the row the examiner marks first.
2Compute $P(X = x)$ for each value separately, keeping 'with replacement' or 'without replacement' fixed throughout.
3Check $\sum p_i = 1$ before going further. If it does not, a case is missing or double-counted, and every later mark depends on this.
4Tabulate $x_i$ against $p_i$ , then extend the table with an $x_i p_i$ row for the mean and an $x_i^2 p_i$ row for the variance.
5Mean $\mu = \sum x_i p_i$ ; variance $\sigma^2 = \sum x_i^2 p_i - \mu^2$ . Never square the mean before summing.

Worked example · 1 mark

A discrete random variable

X

takes the values

30

10

and

-10

with probabilities

\dfrac{1}{5}

\dfrac{3}{10}

and

\dfrac{1}{2}

. Find

E(X)

Check the distribution is valid: $\dfrac{1}{5} + \dfrac{3}{10} + \dfrac{1}{2} = \dfrac{2 + 3 + 5}{10} = 1$ . ✓
$E(X) = \sum x_i p_i = 30\times\dfrac{1}{5} + 10\times\dfrac{3}{10} + (-10)\times\dfrac{1}{2}$ .
$= 6 + 3 - 5$ .

Answer

E(X) = 4

. The negative value carries its sign into the sum — it is not a magnitude.

Where the marks go

Computing variance as $E(X^2) - E(X)$ instead of $E(X^2) - \big(E(X)\big)^2$ — you must subtract the square of the mean.
Forgetting to square the mean, or squaring $\sum x_i p_i$ wrongly, when applying $\operatorname{Var}(X) = E(X^2) - \mu^2$ .
Probabilities not summing to $1$ (or a negative or greater-than- $1$ value of $k$ ) — re-check the table before proceeding.
Treating without-replacement card draws as binomial; the count of Kings is then hypergeometric, so build the table by direct counting.

How the board asks it

Numericalmean and variance from a distribution table1 mkAsked 2025
A random variable $X$ has the following probability distribution: $X = 0, 1, 2, 3$ with $P(X) = 0.1, 0.3, 0.4, 0.2$ respectively. Find the mean $E(X)$ and the variance $\operatorname{Var}(X)$ of $X$ .
Applicationbuilding the distribution from an experiment6 mkAsked 2024 · 2026
Three coins are tossed simultaneously. If $X$ denotes the number of heads obtained, find the probability distribution of $X$ and hence calculate its mean and standard deviation.
Numericalfinding an unknown constant using $\sum p_i = 1$
The probability distribution of a random variable $X$ is given by $P(X = x) = kx$ for $x = 1, 2, 3, 4, 5$ , and $P(X = x) = 0$ otherwise. Find the value of $k$ and hence evaluate $E(X)$ .
Applicationwithout-replacement drawing (counting probabilities)6 mkAsked 2023 · 2026
Two cards are drawn successively without replacement from a well-shuffled pack of $52$ cards. If $X$ is the number of kings drawn, find the probability distribution of $X$ and its mean.
Give reasonsvalidity of a probability distribution
State, with reasons, whether the following can be the probability distribution of a random variable: $X = 0, 1, 2$ with $P(X) = 0.4, 0.4, 0.3$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.