Expected Value & Variance Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice expected value and variance in probability and statistics: computing the mean (expected value) of a discrete random variable with \(E[X]=\sum x\,p(x)\), using the fast variance identity \(\mathrm{Var}(X)=E[X^2]-(E[X])^2\), interpreting standard deviation as spread, and applying core rules like linearity of expectation \(E[aX+b]=aE[X]+b\) and the scaling rule \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\). If you want a refresher with worked examples (dice, coins, spinners, and small distributions), click Start lesson.
Answer the question set and review your mistakes at the end.
How this expected value & variance practice works
1. Take the practice set: answer the expected value and variance questions below.
2. Open the lesson (optional): review formulas, shortcuts, and common probability distributions with step-by-step examples.
3. Retry: return to the question set and apply \(E[X]\) and \(\mathrm{Var}(X)\) rules immediately.
What you will learn in the expected value and variance lesson
Expected value (mean) essentials
Discrete expected value: \(E[X]=\sum x\,p(x)\)
Interpretation: long-run average and “fair price” of a game
Linearity: \(E[X+Y]=E[X]+E[Y]\) (works even without independence)
Purpose: Build a clear, reliable understanding of expected value (mean) and variance (spread) so you can compute \(E[X]\), \(\mathrm{Var}(X)\), and standard deviation quickly and correctly for common probability distributions.
Success criteria
Compute expected value for a discrete random variable using \(E[X]=\sum x\,p(x)\).
Use linearity of expectation (including \(E[X+Y]=E[X]+E[Y]\)).
Compute variance using \(\mathrm{Var}(X)=E[(X-\mu)^2]\) and the shortcut \(\mathrm{Var}(X)=E[X^2]-\mu^2\).
Convert between variance and standard deviation: \(\sigma=\sqrt{\mathrm{Var}(X)}\).
Apply key rules: \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\) and (if independent) \(\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)\).
Handle classic contexts: dice, coins, spinners, and small payoff tables.
Key vocabulary
Random variable: a numerical outcome produced by chance (like number of heads, die roll, or payout).
Expected value (mean): \(E[X]\), the long-run average value of \(X\).
Variance: \(\mathrm{Var}(X)\), the average squared distance from the mean.
Standard deviation: \(\sigma\), the square root of variance, measured in the same units as \(X\).
Probability distribution: the list (or rule) of possible values and their probabilities.
Quick pre-check
Pre-check 1: What is the expected value of a fair six-sided die roll?
Hint: For a fair die, \(E[X]=\dfrac{1+2+3+4+5+6}{6}\).
Pre-check 2: Let \(X\) take values \(1\) and \(3\) each with probability \(1/2\). What is \(\mathrm{Var}(X)\)?
Hint: First find \(\mu=E[X]\), then compute \(E[(X-\mu)^2]\).
Expected Value Basics
Expected value: the weighted average
Learning goal: Compute expected value from a probability table and interpret it as a long-run mean.
Key idea
For a discrete random variable with possible values \(x_1,x_2,\dots\) and probabilities \(p(x_1),p(x_2),\dots\), the expected value is the weighted average: \[ E[X] = \sum_x x\,p(x). \] Think: "multiply each outcome by how often it happens, then add."
Worked example
Example: A bag contains \(\{0,5\}\) equally likely. What is the expected draw?
Each value has probability \(1/2\). So: \[ E[X] = 0\cdot \frac12 + 5\cdot \frac12 = 2.5. \] The mean outcome is \(2.5\) even though \(2.5\) is not an outcome you can draw.
Try it
Try it 1: A fair three-sided die has faces \(1,2,3\). What is the expected value?
Hint: \(E[X]=\dfrac{1+2+3}{3}\).
Try it 2: \(X\) takes values \(10, 20, 30\) with probabilities \(0.2, 0.3, 0.5\). What is \(E[X]\)?
Hint: Multiply each value by its probability and add: \(10(0.2)+20(0.3)+30(0.5)\).
Summary
Expected value is a weighted average: \(E[X]=\sum x\,p(x)\).
It represents the long-run mean, not necessarily a value you can observe in one trial.
Linearity & Sums
Linearity of expectation: your best shortcut
Learning goal: Use linearity to compute expected values of sums and scaled variables quickly.
Key idea
Linearity of expectation works always: \[ E[aX+b]=aE[X]+b,\quad E[X+Y]=E[X]+E[Y]. \] You do not need independence for expected values to add.
Worked example
Example: What is the expected number of heads in 2 fair coin flips?
Let \(X\) be the number of heads in 2 flips. Think of \(X\) as a sum: \[ X = I_1 + I_2, \] where \(I_k=1\) if flip \(k\) is heads and \(0\) otherwise. For a fair coin, \(E[I_k]=0.5\). So: \[ E[X]=E[I_1]+E[I_2]=0.5+0.5=1. \]
Try it
Try it 1: What is the expected sum when rolling two fair six-sided dice?
Hint: \(E[X+Y]=E[X]+E[Y]\) and each die has mean \(3.5\).
Try it 2: A lottery pays \(3\) with probability \(\tfrac13\) and \(0\) otherwise. What is the expected value?
Hint: \(E[X]=3\cdot \tfrac13 + 0\cdot \tfrac23\).
Summary
Linearity: \(E[aX+b]=aE[X]+b\).
Sums: \(E[X+Y]=E[X]+E[Y]\) (independence not required).
Variance Basics
Variance: measure spread around the mean
Learning goal: Compute variance using the definition and the fast identity \(\mathrm{Var}(X)=E[X^2]-(E[X])^2\).
Key idea
If \(\mu=E[X]\), then: \[ \mathrm{Var}(X) = E[(X-\mu)^2]. \] The most useful shortcut is: \[ \mathrm{Var}(X)=E[X^2]-\mu^2. \] This avoids expanding \((X-\mu)^2\) for every outcome.
Worked example
Example: A bag contains \(\{2,4,6\}\) equally likely. What is the variance?
First compute the mean: \[ \mu = E[X]=\frac{2+4+6}{3}=4. \] Now compute \(E[X^2]\): \[ E[X^2]=\frac{2^2+4^2+6^2}{3}=\frac{4+16+36}{3}=\frac{56}{3}. \] So: \[ \mathrm{Var}(X)=E[X^2]-\mu^2=\frac{56}{3}-16=\frac{8}{3}. \]
Try it
Try it 1: What is the variance of a fair three-outcome spinner with values \(0,1,2\)?
Hint: \(\mu=1\). Compute \(E[X^2]=\frac{0^2+1^2+2^2}{3}=\frac{5}{3}\), then subtract \(1^2\).
Try it 2: What is the variance of that coin payout (\(-1\) or \(+1\) equally likely)?
Hint: \(\mu=0\) and \(E[X^2]=1\).
Summary
Variance measures spread around the mean.
Fast identity: \(\mathrm{Var}(X)=E[X^2]-(E[X])^2\).
Variance Rules
Variance rules: shift, scale, and sums
Learning goal: Apply variance rules correctly and know when independence matters.
Key idea
Two essential rules: \[ \mathrm{Var}(aX+b)=a^2\mathrm{Var}(X), \] and for sums: \[ \mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)+2\mathrm{Cov}(X,Y). \] If \(X\) and \(Y\) are independent, then \(\mathrm{Cov}(X,Y)=0\) and variances add: \[ \mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y). \]
Worked example
Example: What is the variance of the number of heads in two fair coin flips?
If \(X\sim \text{Binomial}(n=2,p=0.5)\), then: \[ \mathrm{Var}(X)=np(1-p)=2(0.5)(0.5)=0.5. \] This matches the idea that counts of heads vary, but not wildly, across 2 flips.
Try it
Try it 1: If \(\mathrm{Var}(X)=9\), what is \(\mathrm{Var}(2X-5)\)?
Hint: Use \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\). The "\(-5\)" does not change variance.
Try it 2: Let \(X\) and \(Y\) be independent with \(\mathrm{Var}(X)=2\) and \(\mathrm{Var}(Y)=3\). What is \(\mathrm{Var}(X+Y)\)?
Hint: Independence makes covariance \(0\), so variances add.
Summary
Shift does not change variance; scaling by \(a\) multiplies variance by \(a^2\).
For independent sums, add variances.
Common Distributions
Common distributions and fastest formulas
Learning goal: Recognize standard models (Bernoulli, Binomial, Uniform) and use their mean/variance formulas.
Standard deviation is the square root of variance, measured in the same units as \(X\).
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the skill you need (expected value, variance, or the key rules).
Practice set
Verwachte waarde en variantie practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Een eerlijke munt levert \(1\) dollar op voor kop en \(0\) dollar voor munt. Wat is de verwachte uitbetaling?
Correct answer: C. \(\tfrac12\)
Explanation: De verwachte waarde is het gemiddelde: \(E = \tfrac{1+0}{2} = \tfrac12\).
Question 2Not answered
Een draaischijf landt op \(2\) met kans \(0.2\), op \(4\) met kans \(0.3\), en op \(8\) met kans \(0.5\). Wat is de verwachte waarde?
