What is a discrete random variable in the context of Edexcel International A Levels Statistics 1?

In the Edexcel International A Levels Statistics 1 curriculum, a discrete random variable is a type of random variable that can take on a finite or countably infinite set of values. These values are distinct and separate, meaning there are gaps between them. Examples include the number of students in a class or the result of rolling a die.

How do you calculate the expected value of a discrete random variable?

To calculate the expected value (or mean) of a discrete random variable, you multiply each possible value of the variable by its probability and then sum all these products. Mathematically, it is expressed as E(X) = Σ [x * P(x)], where x represents the values of the random variable and P(x) is the probability of each value.

What is the difference between discrete and continuous random variables in Statistics 1?

The main difference between discrete and continuous random variables lies in the type of values they can take. Discrete random variables have specific, countable values, such as the number of heads in coin tosses. In contrast, continuous random variables can take any value within a given range, such as the height of students in a class. This distinction is crucial in the Edexcel International A Levels Statistics 1 syllabus.

How do you find the probability distribution of a discrete random variable?

To find the probability distribution of a discrete random variable, list all possible values the variable can take and assign the probability of each value occurring. The sum of all probabilities should equal 1. This distribution helps in understanding the likelihood of different outcomes and is a key concept in the Edexcel International A Levels Statistics 1 course.

Why is understanding discrete random variables important in Statistics 1?

Understanding discrete random variables is crucial in Statistics 1 because they form the basis for analyzing situations where outcomes are countable and distinct. This knowledge helps in solving real-world problems involving probability and statistics, such as predicting outcomes and making informed decisions, which are essential skills in the Edexcel International A Levels Mathematics curriculum.

Why is "Forgetting to enforce $\sum P(X = x) = 1$ when finding a constant" a common mistake in Discrete random variables?

Why it happens: Skipping the normalisation step. How to avoid it: First check: P = 1. Solve for the unknown ONLY using this condition. Then verify probabilities are non-negative. Source: WST01/01 examiner reports — annual

Why is "Claiming $\text{Var}(aX + b) = a\,\text{Var}(X) + b$" a common mistake in Discrete random variables?

Why it happens: Treating variance like expectation. How to avoid it: E is linear; Var is not. Var(aX + b) = a^2 Var(X) — the additive constant disappears, and the multiplicative constant is SQUARED.

Why is "Writing $\text{Var}(X) = E(X)^{2} - E(X^{2})$" a common mistake in Discrete random variables?

Why it happens: Sign error. How to avoid it: Correct form: Var(X) = E(X^2) - [E(X)]^2 — E(X^2) first, then subtract [E(X)]^2. Variance is non-negative, so this gives a check.

Why is "Using $n/2$ instead of $(n+1)/2$ for the discrete uniform mean" a common mistake in Discrete random variables?

Why it happens: Confusion with the continuous uniform on [0, n]. How to avoid it: For the DISCRETE uniform on \1, 2, , n\: mean = (n+1)/2. A six-sided die has mean 3.5, not 3.

Why is "Interpolating between values for a discrete median" a common mistake in Discrete random variables?

Why it happens: Carry-over from continuous data. How to avoid it: For a discrete random variable, the median is the smallest x with F(x) 0.5. It IS one of the values X can take — no interpolation.

Statistics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Discrete random variables

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Discrete Random Variables Study Notes — Edexcel IAL Mathematics S1 (WST01/01, 2018 spec — 2026 onwards)

PMFs, $E(X)$ , $\text{Var}(X)$ , linear transformations $E(aX + b)$ and $\text{Var}(aX + b)$ , the discrete uniform distribution, and the cumulative distribution function. The chapter introduces the language of random variables that the normal-distribution chapter and S2 build on.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

S1 6.1 — Understand discrete random variables and probability mass functions.
S1 6.2 — Calculate the expected value $E(X)$ and variance $\text{Var}(X)$ .
S1 6.3 — Use the linear transformation rules $E(aX + b) = aE(X) + b$ and $\text{Var}(aX + b) = a^{2}\text{Var}(X)$ .
S1 6.4 — Recognise and use the discrete uniform distribution, including the formulae $E(X) = (n+1)/2$ and $\text{Var}(X) = (n^{2} - 1)/12$ .
S1 6.5 — Calculate and use the cumulative distribution function $F(x) = P(X \le x)$ .

Probability mass functions

$P(X = x) \ge 0$ and probabilities sum to $1$ . That's the whole rule.

Definition. A discrete random variable $X$ takes values in a countable set. The probability mass function is

$P(X = x) \ge 0 \text{ for all } x, \qquad \sum_{x} P(X = x) = 1.$

Specifying a PMF. Three common forms:

Table: list each value of $x$ with $P(X = x)$ .
Formula: e.g. $P(X = x) = kx$ for $x = 1, 2, 3, 4$ .
Verbal/contextual: 'a fair coin is tossed three times; $X$ = number of heads'.

Finding an unknown constant. Use the normalisation $\sum P = 1$ .

Example. $P(X = x) = kx$ for $x = 1, 2, 3, 4$ . Sum $= k(1+2+3+4) = 10k = 1$ , so $k = 1/10$ .

Checks before answering.

All $P(X = x) \ge 0$ .
$\sum P(X = x) = 1$ exactly.
The support is the right set (e.g. $\{0, 1, 2, 3\}$ for $3$ coin tosses, not $\{1, 2, 3\}$ ).

$P(X = x) \ge 0$ for all $x$ .
$\sum P(X = x) = 1$ .
Use normalisation to solve for unknown constants.
Verify probabilities are non-negative.

Expectation and variance

$E(X)$ = probability-weighted mean. $\text{Var}(X) = E(X^{2}) - [E(X)]^{2}$ .

Expectation.

$E(X) = \sum_{x} x \cdot P(X = x).$

Interpretation: the long-run average of $X$ over many trials. The 'centre of mass' of the PMF.

For any function $g(X)$ :

$E[g(X)] = \sum_{x} g(x) P(X = x).$

In particular $E(X^{2}) = \sum x^{2} P(X = x)$ , used for variance.

Variance. Two equivalent forms:

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

The shortcut (right-hand side) is in the booklet — use it. Standard deviation $\sigma = \sqrt{\text{Var}(X)}$ .

Worked example. PMF $P(X = x) = x/10$ for $x = 1, 2, 3, 4$ .

$E(X) = (1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4)/10 = (1 + 4 + 9 + 16)/10 = 30/10 = 3$ . Equivalently: $1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3$ .
$E(X^{2}) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$ .
$\text{Var}(X) = 10 - 9 = 1$ . SD $= 1$ .

$E(X) = \sum x P(X = x)$ .
$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}$ — booklet shortcut.
$\sigma = \sqrt{\text{Var}(X)}$ .
Variance is non-negative — quick sanity check.

Linear transformations: $aX + b$

$E$ is linear. $\text{Var}$ squares the multiplier and ignores the shift.

Rules.

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

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

Why the variance rule. Adding a constant $b$ shifts every value of $X$ by $b$ , which moves the mean by $b$ but does not change the spread. Multiplying by $a$ scales each deviation from the mean by $a$ , so squared deviations scale by $a^{2}$ .

Worked example. $E(X) = 3$ , $\text{Var}(X) = 1$ . Let $W = 3X - 2$ .

$E(W) = 3 \cdot 3 - 2 = 7$ .
$\text{Var}(W) = 9 \cdot 1 = 9$ .
SD $(W) = 3$ .

Common application — converting units. If $X$ is in hours and $W = 60X$ converts to minutes, then $E(W) = 60 E(X)$ , $\text{Var}(W) = 3600 \text{Var}(X)$ .

Common application — payouts. A game pays $5X - 2$ dollars where $X$ is the result of a die roll. Use the rules to find expected payout and variance directly without re-computing the PMF.

$E(aX + b) = aE(X) + b$ — linear.
$\text{Var}(aX + b) = a^{2}\text{Var}(X)$ — shift drops out, scale squares.
SD $(aX + b) = |a| \cdot \text{SD}(X)$ .
Useful for unit conversion and payout questions.

Discrete uniform distribution

$X$ on $\{1, \ldots, n\}$ with $P = 1/n$ . $E(X) = (n+1)/2$ , $\text{Var}(X) = (n^{2}-1)/12$ .

Setup. $X$ takes values $1, 2, \ldots, n$ each with probability $1/n$ .

Mean. Using $\sum_{k=1}^{n} k = n(n+1)/2$ :

$E(X) = \dfrac{1}{n} \cdot \dfrac{n(n+1)}{2} = \dfrac{n+1}{2}.$

Variance. Using $\sum_{k=1}^{n} k^{2} = n(n+1)(2n+1)/6$ :

$E(X^{2}) = \dfrac{(n+1)(2n+1)}{6}.$

Then $\text{Var}(X) = E(X^{2}) - [(n+1)/2]^{2}$ . Algebra gives

$\text{Var}(X) = \dfrac{n^{2} - 1}{12}.$

Six-sided die. $n = 6$ : $E(X) = 3.5$ , $\text{Var}(X) = 35/12 \approx 2.917$ , SD $\approx 1.708$ .

Memorise or derive. These formulae are NOT in the IAL formula booklet. You can either memorise them or derive on the spot using the sum formulae (which ARE in the booklet under Pure Mathematics).

General version $\{a, a+1, \ldots, a+n-1\}$ . Shift the formula: $E(X) = a + (n-1)/2$ . Variance unchanged at $(n^{2} - 1)/12$ .

Each value has probability $1/n$ .
$E(X) = (n+1)/2$ .
$\text{Var}(X) = (n^{2} - 1)/12$ .
NOT in the formula booklet — derive or memorise.

Cumulative distribution function $F$

$F(x) = P(X \le x)$ — cumulate from below. Use to find tails and medians.

Definition.

$F(x) = P(X \le x) = \sum_{u \le x} P(X = u).$

For a discrete random variable, $F$ is a step function: constant between values of $X$ , with a jump of size $P(X = x_i)$ at each value $x_i$ in the support.

Worked example. PMF $0.1, 0.3, 0.4, 0.2$ on $\{0, 1, 2, 3\}$ .

$x$	$P(X = x)$	$F(x) = P(X \le x)$
$0$	$0.1$	$0.1$
$1$	$0.3$	$0.4$
$2$	$0.4$	$0.8$
$3$	$0.2$	$1.0$

Properties.

$F$ is non-decreasing.
$F(-\infty) = 0$ , $F(+\infty) = 1$ .
$P(a < X \le b) = F(b) - F(a)$ .

Median. Smallest $x$ with $F(x) \ge 0.5$ . From the example: $F(1) = 0.4 < 0.5$ , $F(2) = 0.8 \ge 0.5$ , so median $= 2$ .

Going the other way. From a given CDF, recover the PMF by differencing: $P(X = x_i) = F(x_i) - F(x_{i-1})$ .

$F(x) = P(X \le x)$ .
Step function for discrete $X$ , jumps at each support value.
$P(a < X \le b) = F(b) - F(a)$ .
Median $=$ smallest $x$ with $F(x) \ge 0.5$ .

How it’s examined

Discrete random variables is a Q5-Q6 fixture worth $8$ - $12$ marks per WST01/01 paper. Typical asks: complete a PMF table to satisfy $\sum P = 1$ ( $3$ - $4$ marks); compute $E(X)$ and $\text{Var}(X)$ from a PMF ( $4$ - $5$ marks); apply linear-transformation rules to find $E(aX + b)$ and $\text{Var}(aX + b)$ ( $3$ - $4$ marks); recognise a discrete uniform and quote its mean/variance ( $3$ - $4$ marks); compute a CDF, find a median or use the CDF to compute a tail probability ( $4$ - $5$ marks). Sign errors and the variance-shift rule are the most-flagged mark losses.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WST01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Discrete random variables

Step-by-step solutions to past-paper-style questions on discrete random variables, written exactly the way a tutor would explain them at the board.

1Find the constant in a PMF and compute $E(X)$ (5 marks, S1)
Core• Adapted from WST01/01 January 2024 Q5• PMF, expectation
Question
Discrete random variable $X$ has $P(X = x) = kx$ for $x = 1, 2, 3, 4$ and $P(X = x) = 0$ otherwise. Find (a) $k$ , (b) $E(X)$ , (c) $P(X \ge 3)$ .
Step-by-step solution
1. Step 1
  Probabilities sum to $1$ :
  $k(1) + k(2) + k(3) + k(4) = 10k = 1 \Rightarrow k = \dfrac{1}{10}$
2. Step 2
  So $P(X = 1) = 0.1$ , $P(X = 2) = 0.2$ , $P(X = 3) = 0.3$ , $P(X = 4) = 0.4$ .
3. Step 3
  Expectation:
  $E(X) = \sum xP(X = x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3$
4. Step 4
  Probability tail:
  $P(X \ge 3) = P(X = 3) + P(X = 4) = 0.3 + 0.4 = 0.7$
Answer
(a) $k = 1/10$ . (b) $E(X) = 3$ . (c) $P(X \ge 3) = 0.7$ .
Examiner tip
M1 for $\sum P = 1$ . A1 for $k$ . M1 for $\sum x P(X = x)$ . A1 for $E(X) = 3$ . B1 for $P(X \ge 3) = 0.7$ . Many candidates forget to check that all probabilities are non-negative — implicit when $k > 0$ here but worth confirming on the page.
2Variance of $X$ and of a linear transform (6 marks, S1)
Extended• Adapted from WST01/01 June 2024 Q4• variance, linear transformation
Question
Using the distribution from the previous example ( $P(X = x) = x/10$ for $x = 1, 2, 3, 4$ ), find (a) $\text{Var}(X)$ , (b) $E(3X - 2)$ , (c) $\text{Var}(3X - 2)$ .
Step-by-step solution
1. Step 1
  Compute $E(X^{2})$ .
  $E(X^{2}) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$
2. Step 2
  Variance:
  $\text{Var}(X) = E(X^{2}) - [E(X)]^{2} = 10 - 3^{2} = 1$
3. Step 3
  Linear-transform expectation $E(aX + b) = aE(X) + b$ :
  $E(3X - 2) = 3E(X) - 2 = 3 \cdot 3 - 2 = 7$
4. Step 4
  Linear-transform variance $\text{Var}(aX + b) = a^{2} \text{Var}(X)$ (the shift $b$ has no effect):
  $\text{Var}(3X - 2) = 3^{2} \cdot 1 = 9$
Answer
(a) $\text{Var}(X) = 1$ . (b) $E(3X - 2) = 7$ . (c) $\text{Var}(3X - 2) = 9$ .
Examiner tip
M1 for $E(X^{2})$ . A1 for $10$ . M1A1 for $\text{Var}(X) = 1$ . B1 for $E(3X - 2)$ . B1 for $\text{Var}(3X - 2)$ . Watch the sign — students sometimes write $\text{Var}(3X - 2) = 9 \cdot 1 - 2$ , which is wrong: the additive constant disappears under variance.
3Discrete uniform distribution — derive $E$ and $\text{Var}$ (6 marks, S1)
Extended• Adapted from WST01/01 January 2023 Q6• discrete uniform, derivation
Question
$X$ has a discrete uniform distribution on $\{1, 2, \ldots, n\}$ . Show, by direct calculation, that $E(X) = (n + 1)/2$ and $\text{Var}(X) = (n^{2} - 1)/12$ . Then evaluate $E(X)$ and $\text{Var}(X)$ for a fair six-sided die.
Step-by-step solution
1. Step 1
  Each outcome has probability $1/n$ . Expectation:
  $E(X) = \sum_{x=1}^{n} x \cdot \dfrac{1}{n} = \dfrac{1}{n} \cdot \dfrac{n(n+1)}{2} = \dfrac{n+1}{2}$
2. Step 2
  Compute $E(X^{2})$ using $\sum_{x=1}^{n} x^{2} = n(n+1)(2n+1)/6$ :
  $E(X^{2}) = \dfrac{1}{n} \cdot \dfrac{n(n+1)(2n+1)}{6} = \dfrac{(n+1)(2n+1)}{6}$
3. Step 3
  Variance:
  $\text{Var}(X) = E(X^{2}) - [E(X)]^{2} = \dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^{2}}{4}$
4. Step 4
  Common denominator $12$ :
  $\text{Var}(X) = \dfrac{2(n+1)(2n+1) - 3(n+1)^{2}}{12} = \dfrac{(n+1)[2(2n+1) - 3(n+1)]}{12}$
5. Step 5
  Simplify bracket: $2(2n+1) - 3(n+1) = 4n + 2 - 3n - 3 = n - 1$ . So
  $\text{Var}(X) = \dfrac{(n+1)(n-1)}{12} = \dfrac{n^{2} - 1}{12}$
6. Step 6
  Six-sided die: $n = 6$ . $E(X) = 7/2 = 3.5$ . $\text{Var}(X) = 35/12 \approx 2.917$ .
Answer
$E(X) = (n+1)/2$ and $\text{Var}(X) = (n^{2}-1)/12$ . For $n = 6$ : $E(X) = 3.5$ , $\text{Var}(X) = 35/12 \approx 2.92$ .
Examiner tip
M1 for $\sum_{x=1}^{n} x$ formula. A1 for $E(X) = (n+1)/2$ . M1 for $\sum x^{2}$ formula. A1 for $E(X^{2})$ . M1 for combining and factorising. A1 for $\text{Var}(X) = (n^{2}-1)/12$ . B1 for the die values. The formulae for $E$ and $\text{Var}$ are NOT in the IAL formula booklet — derive or memorise.
4Cumulative distribution function $F$ (5 marks, S1)
Extended• Adapted from WST01/01 June 2023 Q6• CDF, F(x)
Question
$X$ has PMF $P(X = 0) = 0.1$ , $P(X = 1) = 0.3$ , $P(X = 2) = 0.4$ , $P(X = 3) = 0.2$ . (a) Write down the cumulative distribution function $F(x) = P(X \le x)$ for $x \in \{0, 1, 2, 3\}$ . (b) Find the median of $X$ (smallest value such that $F \ge 0.5$ ).
Step-by-step solution
1. Step 1
  Cumulate the probabilities.
  $F(0) = 0.1, \;\; F(1) = 0.4, \;\; F(2) = 0.8, \;\; F(3) = 1.0$
2. Step 2
  The median is the smallest $x$ with $F(x) \ge 0.5$ .
3. Step 3
  $F(1) = 0.4 < 0.5$ and $F(2) = 0.8 \ge 0.5$ , so median $= 2$ .
Answer
$F(0) = 0.1$ , $F(1) = 0.4$ , $F(2) = 0.8$ , $F(3) = 1.0$ . Median $= 2$ .
Examiner tip
B1 for each cumulative value (max $2$ ). M1 for using the smallest- $x$ definition. A1 for median $= 2$ . The median of a discrete random variable can be ambiguous if $F(x) = 0.5$ exactly — Edexcel takes the convention that the median is the smallest $x$ with $F \ge 0.5$ .
5Construct a PMF from a context — coin tosses (7 marks, S1)
Extended• Adapted from WST01/01 June 2024 Q5• PMF construction, expectation, context
Question
A fair coin is tossed $3$ times. $X$ = number of heads. (a) Write down the PMF of $X$ . (b) Find $E(X)$ and $\text{Var}(X)$ . (c) If a 'payout' is defined as $W = 5X - 2$ , find $E(W)$ and $\text{Var}(W)$ .
Step-by-step solution
1. Step 1
  $X \in \{0, 1, 2, 3\}$ . By counting: $P(X = 0) = (1/2)^{3} = 1/8$ , $P(X = 1) = 3/8$ , $P(X = 2) = 3/8$ , $P(X = 3) = 1/8$ .
2. Step 2
  $E(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = (0 + 3 + 6 + 3)/8 = 12/8 = 1.5$ .
3. Step 3
  $E(X^{2}) = 0 + 1(3/8) + 4(3/8) + 9(1/8) = (3 + 12 + 9)/8 = 24/8 = 3$ .
4. Step 4
  $\text{Var}(X) = 3 - 1.5^{2} = 3 - 2.25 = 0.75$ .
5. Step 5
  $E(W) = 5 E(X) - 2 = 5 \times 1.5 - 2 = 7.5 - 2 = 5.5$ .
6. Step 6
  $\text{Var}(W) = 5^{2} \text{Var}(X) = 25 \times 0.75 = 18.75$ .
Answer
(a) $P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8$ . (b) $E(X) = 1.5$ , $\text{Var}(X) = 0.75$ . (c) $E(W) = 5.5$ , $\text{Var}(W) = 18.75$ .
Examiner tip
M1A1 for PMF. M1A1 for $E(X)$ . M1A1 for $\text{Var}(X)$ . B1 for $E(W)$ . B1 for $\text{Var}(W)$ . ECF is generous: if PMF is wrong but subsequent calculations follow consistently, A-marks may still be awarded.

Key Formulae — Discrete random variables

The formulae you need to memorise for discrete random variables on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Expectation $E(X)$
$E(X) = \sum_{x} x \cdot P(X = x)$
$E(X)$
expected value of $X$
$P(X = x)$
probability mass function
When to use
Average value of $X$ over many trials. IS in the IAL formula booklet (under expected value of a function $g(X)$ ).
Example
PMF $P(X = x) = x/10$ for $x = 1, 2, 3, 4$ : $E(X) = (1 + 4 + 9 + 16)/10 = 3$ .
Variance $\text{Var}(X)$
$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}, \quad E(X^{2}) = \sum x^{2} P(X = x)$
$\text{Var}(X)$
variance of $X$
When to use
Spread of $X$ about its mean. Shortcut formula avoids deviation-from-mean form. IS in the IAL formula booklet.
Example
$E(X) = 3$ , $E(X^{2}) = 10$ : $\text{Var}(X) = 10 - 9 = 1$ .
Linear transformation of $E$ and $\text{Var}$
$E(aX + b) = aE(X) + b, \quad \text{Var}(aX + b) = a^{2} \text{Var}(X)$
$a, b$
constants
When to use
Whenever a question rescales or shifts $X$ (e.g. a payout, a converted unit). NOT in the formula booklet — memorise. The shift $b$ has NO effect on variance.
Example
$E(X) = 3$ , $\text{Var}(X) = 1$ : $E(3X - 2) = 7$ , $\text{Var}(3X - 2) = 9$ .
Discrete uniform on $\{1, 2, \ldots, n\}$
$E(X) = \dfrac{n + 1}{2}, \quad \text{Var}(X) = \dfrac{n^{2} - 1}{12}$
$n$
number of equally-likely outcomes
When to use
Fair dice, lottery balls $1$ through $n$ , equally-likely choices. NOT in the IAL formula booklet — derive (using $\sum x$ and $\sum x^{2}$ ) or memorise.
Example
Six-sided die ( $n = 6$ ): $E(X) = 3.5$ , $\text{Var}(X) = 35/12 \approx 2.92$ .
Cumulative distribution function (discrete)
$F(x) = P(X \le x) = \sum_{u \le x} P(X = u)$
$F(x)$
CDF evaluated at $x$
When to use
Probability of $X$ not exceeding $x$ . Cumulate the PMF from below. Used for medians and tails. NOT in the booklet but standard.
Example
PMF $0.1, 0.3, 0.4, 0.2$ on $\{0, 1, 2, 3\}$ : $F = 0.1, 0.4, 0.8, 1.0$ .

Key Definitions and Keywords — Discrete random variables

Definitions to memorise and the exact keywords mark schemes credit for discrete random variables answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Discrete random variable
Examiner keyword
A variable $X$ whose possible values form a countable set (often finite) and whose probability mass function $P(X = x)$ satisfies $P(X = x) \ge 0$ and $\sum_{x} P(X = x) = 1$ .
Probability mass function (PMF)
Examiner keyword
Function $P(X = x)$ giving the probability that the discrete random variable $X$ takes the specific value $x$ . Must be non-negative and sum to $1$ .
Cumulative distribution function (CDF)
Examiner keyword
$F(x) = P(X \le x)$ . Step function for discrete $X$ : jumps at each value $x$ in the support by $P(X = x)$ .
Expectation
Examiner keyword
$E(X) = \sum x P(X = x)$ — the probability-weighted average value, also called the 'expected value' or 'mean' of the distribution. Linear: $E(aX + b) = aE(X) + b$ .
Variance and standard deviation of $X$
Examiner keyword
$\text{Var}(X) = E(X^{2}) - [E(X)]^{2}$ . Measures spread. SD = $\sqrt{\text{Var}}$ . Under linear transformation, $\text{Var}(aX + b) = a^{2} \text{Var}(X)$ — shifts have no effect.
Discrete uniform distribution
Examiner keyword
$X$ taking values $1, 2, \ldots, n$ each with probability $1/n$ . Models a fair die, a random integer drawn from a uniform list, etc. $E(X) = (n+1)/2$ , $\text{Var}(X) = (n^{2}-1)/12$ .

Common Mistakes and Misconceptions — Discrete random variables

The traps other students keep falling into on discrete random variables questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to enforce $\sum P(X = x) = 1$ when finding a constant
WST01/01 examiner reports — annual
Why it happens
Skipping the normalisation step.
How to avoid it
First check: $\sum P = 1$ . Solve for the unknown ONLY using this condition. Then verify probabilities are non-negative.
✕Claiming $\text{Var}(aX + b) = a\,\text{Var}(X) + b$
Why it happens
Treating variance like expectation.
How to avoid it
$E$ is linear; $\text{Var}$ is not. $\text{Var}(aX + b) = a^{2} \text{Var}(X)$ — the additive constant disappears, and the multiplicative constant is SQUARED.
✕Writing $\text{Var}(X) = E(X)^{2} - E(X^{2})$
Why it happens
Sign error.
How to avoid it
Correct form: $\text{Var}(X) = E(X^{2}) - [E(X)]^{2}$ — $E(X^{2})$ first, then subtract $[E(X)]^{2}$ . Variance is non-negative, so this gives a check.
✕Using $n/2$ instead of $(n+1)/2$ for the discrete uniform mean
Why it happens
Confusion with the continuous uniform on $[0, n]$ .
How to avoid it
For the DISCRETE uniform on $\{1, 2, \ldots, n\}$ : mean $= (n+1)/2$ . A six-sided die has mean $3.5$ , not $3$ .
✕Interpolating between values for a discrete median
Why it happens
Carry-over from continuous data.
How to avoid it
For a discrete random variable, the median is the smallest $x$ with $F(x) \ge 0.5$ . It IS one of the values $X$ can take — no interpolation.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Discrete random variables — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Discrete random variables

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Find the constant in a PMF and compute E(X)E(X)E(X) (5 marks, S1)

2Variance of XXX and of a linear transform (6 marks, S1)

3Discrete uniform distribution — derive EEE and Var\text{Var}Var (6 marks, S1)

4Cumulative distribution function FFF (5 marks, S1)

5Construct a PMF from a context — coin tosses (7 marks, S1)

Expectation E(X)E(X)E(X)

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

Linear transformation of EEE and Var\text{Var}Var

Discrete uniform on {1,2,…,n}\{1, 2, \ldots, n\}{1,2,…,n}

Cumulative distribution function (discrete)

Discrete random variable

Probability mass function (PMF)

Cumulative distribution function (CDF)

Expectation

Variance and standard deviation of XXX

Discrete uniform distribution

✕Forgetting to enforce ∑P(X=x)=1\sum P(X = x) = 1∑P(X=x)=1 when finding a constant

✕Claiming Var(aX+b)=a Var(X)+b\text{Var}(aX + b) = a\,\text{Var}(X) + bVar(aX+b)=aVar(X)+b

✕Writing Var(X)=E(X)2−E(X2)\text{Var}(X) = E(X)^{2} - E(X^{2})Var(X)=E(X)2−E(X2)

✕Using n/2n/2n/2 instead of (n+1)/2(n+1)/2(n+1)/2 for the discrete uniform mean

✕Interpolating between values for a discrete median

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Discrete random variables — frequently asked questions

1Find the constant in a PMF and compute $E(X)$ (5 marks, S1)

2Variance of $X$ and of a linear transform (6 marks, S1)

3Discrete uniform distribution — derive $E$ and $\text{Var}$ (6 marks, S1)

4Cumulative distribution function $F$ (5 marks, S1)

Expectation $E(X)$

Variance $\text{Var}(X)$

Linear transformation of $E$ and $\text{Var}$

Discrete uniform on $\{1, 2, \ldots, n\}$

Variance and standard deviation of $X$

✕Forgetting to enforce $\sum P(X = x) = 1$ when finding a constant

✕Claiming $\text{Var}(aX + b) = a\,\text{Var}(X) + b$

✕Writing $\text{Var}(X) = E(X)^{2} - E(X^{2})$

✕Using $n/2$ instead of $(n+1)/2$ for the discrete uniform mean