A discrete random variable $X$ has an expected value $E(X) = 4$ and variance $\text{Var}(X) = 3$. Find the exact values of $E(5X - 2)$ and $\text{Var}(5X - 2)$.

1. Identify the rule for the expected value of a linear transformation: $E(aX + b) = aE(X) + b$. 2. Substitute the given expected value into the formula: $E(5X - 2) = 5(4) - 2$. 3. Evaluate the expected value: $20 - 2 = 18$. 4. Identify the rule for the variance of a linear transformation: $\text{Var}(aX + b) = a^2\text{Var}(X)$. 5. Substitute the given variance into the formula: $\text{Var}(5X - 2) = 5^2(3) = 25(3)$. 6. Evaluate to find the final exact variance: $\mathbf{E(5X - 2) = 18}$ and $\mathbf{\text{Var}(5X - 2) = 75}$.

A continuous random variable $X$ has a probability density function $f(x) = k(4x - x^2)$ for $0 \le x \le 4$, and $f(x)=0$ otherwise. Find the exact value of the constant $k$.

1. State the fundamental property of continuous probability density functions: the total area under the valid curve must equal $1$, so $\int_0^4 f(x) \,dx = 1$. 2. Set up the definite integral with the constant $k$ factored out: $k \int_0^4 (4x - x^2) \,dx = 1$. 3. Integrate the polynomial terms inside the bracket with respect to $x$: $k \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = 1$. 4. Substitute the upper limit (since the lower limit evaluates to $0$): $k \left( 2(16) - \frac{64}{3} \right) = 1$. 5. Simplify the algebraic expression inside the bracket: $k \left( 32 - \frac{64}{3} \right) = k \left( \frac{96 - 64}{3} \right) = k \left( \frac{32}{3} \right) = 1$. 6. Solve the resulting equation for the constant: $\mathbf{k = \frac{3}{32}}$.

A continuous random variable $X$ has the probability density function $f(x) = \frac{3}{8}x^2$ for $0 \le x \le 2$ . Calculate the expected value $\text{E}(X)$ and the variance $\text{Var}(X)$ .

Use $\text{E}(X)=\int x f(x)\,dx$ : $\text{E}(X)=\int_0^2\frac{3}{8}x^3\,dx$ M1 $\text{E}(X)=\frac{3}{8}\left[\frac{x^4}{4}\right]_0^2=\frac{3}{8}\times 4=1.5$ A1 Find $\text{E}(X^2)=\int_0^2\frac{3}{8}x^4\,dx$ M1 $\text{E}(X^2)=\frac{3}{8}\left[\frac{x^5}{5}\right]_0^2=\frac{3}{8}\times\frac{32}{5}=2.4$ A1 Use $\text{Var}(X)=\text{E}(X^2)-[\text{E}(X)]^2$ : $\text{Var}(X)=2.4-1.5^2$ M1 $\text{Var}(X)=0.15$ A1

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IB Maths AA HL · Unit 4: Statistics and Probability

IB Maths AA HL Advanced Random Variables Questions

Updated October 2026

Exam-style IB Maths AA HL advanced random variables questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step, showing the working that earns the marks — then try the practice questions and check your working against the mark scheme.

52 questions
Paper 1: 29
Paper 2: 23
13 easy
25 medium
14 hard
3 worked examples

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1 of 5

Question 1 of 5
Question 2 of 5
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Question 5 of 5

Start with one question on linear combinations of random variables. Stuck? Take a hint. No sign-in needed.

E (X) = 10

and

Var (X) = 4

. Find

Var (3 X + 2)

Choose your answer · keys 1–4

12

32

36

38

Find

E (X)

for the distribution in the table.

$x$	-1	0	3
$P (X = x)$	$\frac{2}{5}$	$\frac{2}{5}$	$\frac{1}{5}$

Choose your answer · keys 1–4

\frac{2}{3}

1

2.2

0.2

Find

E (X)

for the distribution in the table.

$x$	1	3	5	7
$P (X = x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Choose your answer · keys 1–4

19

1

4

16

What you need to know

E(X), Var(X), and the shortcut Var(X) = E(X²) - [E(X)]². HL AA Paper 1 tests these algebraically without a GDC. Discrete probability distributions — expectation and variance overview →

B(n, p) with mean np and variance np(1-p). HL AA combines the binomial with conditional probability, or asks you to find n or p from a given probability. Binomial distribution — mean, variance and unknown parameters overview →

What's examined in AA HL advanced random variables

The question bank covers these advanced random variables question types (number of questions in brackets):

Defining Probability Functions (18)
Sampling Distributions & CLT (15)
Moments of Random Variables (14)
Transformations & Normal Dist (5)

Key formulas

Variance (continuous): $Var (X) = E (X^{2}) - [E (X)]^{2}$

In the same notation as the IB formula booklet. All AA HL formulas →

Where students lose marks on advanced random variables

Examiner's noteContinuous random variables: mode, median and expectation from a pdf

Where marks are lost: In both HL sessions pdf questions lost marks: the mode found by differentiating instead of reading the graph (or not recognised as the x-value of the maximum), the median found algebraically when the GDC was quicker, wrong bounds for part of the domain, and very few able to set up an expectation. Comparing a probability with 0.5 to place the median was rarely used.

Examiners flagged this in 2 exam series (May 2024, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Examiner's noteDiscrete probability: independence assumed, E(X²) and Var misused

Where marks are lost: November 2025: many mishandled P(A ∩ B) and the conditional probability that followed, gave the sum-to-1 condition when a range of values of a parameter was asked, squared the probabilities to find E(X²), and did not use Var(a − X) = Var(X). May 2024: choosing one group of people fixed the other (a combinations point missed). The SL reports show the same probability errors.

Examiners flagged this in 3 exam series (May 2024, November 2024, November 2025). Summarised in our own words from IB subject reports, IB Programme Resource Centre. Read the full note →

Advanced Random Variables worked examples

Worked example 1: Linear transformations of expected value and variance · easy

A discrete random variable $X$ has an expected value $E (X) = 4$ and variance $Var (X) = 3$ . Find the exact values of $E (5 X - 2)$ and $Var (5 X - 2)$ .

Solution

6 steps · 1 to fill in

Examiner tip: A very common mistake is incorrectly subtracting the constant

b

from the variance or forgetting to square the coefficient

a

when applying the variance transformation rule.

Worked example 2: Finding constants for continuous probability density functions · medium

A continuous random variable $X$ has a probability density function $f (x) = k (4 x - x^{2})$ for $0 \leq x \leq 4$ , and $f (x) = 0$ otherwise. Find the exact value of the constant $k$ .

Solution

6 steps · 1 to fill in

Examiner tip: Students often forget that the total area under a continuous PDF must equal exactly 1, or they make arithmetic errors when subtracting fractions after substituting the integral limits.

Worked example 3: Calculating variance from a discrete probability distribution · hard

A discrete random variable $X$ has a probability distribution given by $P (X = x) = \frac{c x}{4}$ for $x \in {1, 2, 3}$ . Find the exact value of the constant $c$ , and hence calculate $Var (X)$ .

Solution

6 steps · 1 to fill in

Examiner tip: A frequent trap is calculating the sum of the squares

E (X^{2})

and stopping there, forgetting to subtract the square of the expected value

[E (X)]^{2}

to find the actual variance.

Try these IB Maths AA HL advanced random variables questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 2 marks · Paper 2

A continuous random variable $T$ follows a uniform distribution with probability density function $f (t) = 0.5$ for $2 \leq t \leq 4$ . Find $P (T \geq 3.5)$ .

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Question 2 · medium · 6 marks · Paper 2

A continuous random variable $X$ has the probability density function $f (x) = \frac{3}{8} x^{2}$ for $0 \leq x \leq 2$ . Calculate the expected value $E (X)$ and the variance $Var (X)$ .

Full worked solution Free, no sign-in

Use $E (X) = \int x f (x) d x$ : $E (X) = \int_{0}^{2} \frac{3}{8} x^{3} d x$ M1
$E (X) = \frac{3}{8} {[\frac{x^{4}}{4}]}_{0}^{2} = \frac{3}{8} \times 4 = 1.5$ A1
Find $E (X^{2}) = \int_{0}^{2} \frac{3}{8} x^{4} d x$ M1
$E (X^{2}) = \frac{3}{8} {[\frac{x^{5}}{5}]}_{0}^{2} = \frac{3}{8} \times \frac{32}{5} = 2.4$ A1
Use $Var (X) = E (X^{2}) - [E (X)]^{2}$ : $Var (X) = 2.4 - {1.5}^{2}$ M1
$Var (X) = 0.15$ A1

Every question here has a full worked solution. See them all →

Question 3 · hard · 6 marks · Paper 1

A continuous random variable $X$ has the probability density function $f (x) = \frac{1}{2} \sin x$ for $0 \leq x \leq π$ . Using integration by parts, prove that the expected value $E (X) = \frac{π}{2}$ .

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FAQ

How many IB Maths AA HL advanced random variables questions are there?

There are 52 exam-style advanced random variables questions in the AA HL question bank (Paper 1: 29 · Paper 2: 23), graded 13 easy, 25 medium, 14 hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is advanced random variables on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 29 · Paper 2: 23. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA HL Unit 4 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

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Free ready-to-teach lesson (slides and teacher notes): Continuous random variables (pdf, cdf).

IB Maths AA HL Advanced Random Variables Questions

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What you need to know

What's examined in AA HL advanced random variables

Key formulas

Where students lose marks on advanced random variables

Advanced Random Variables worked examples

Worked example 1: Linear transformations of expected value and variance · easy

Solution

Worked example 2: Finding constants for continuous probability density functions · medium

Solution

Worked example 3: Calculating variance from a discrete probability distribution · hard

Solution

Try these IB Maths AA HL advanced random variables questions

Question 1 · easy · 2 marks · Paper 2

Question 2 · medium · 6 marks · Paper 2

Full worked solution Free, no sign-in

Question 3 · hard · 6 marks · Paper 1

Revise advanced random variables for the exam

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