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
Wartość oczekiwana i wariancja 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
Uczciwa moneta wypłaca \(1\) dolara za orła i \(0\) dolarów za reszkę. Jaka jest oczekiwana wypłata?
Correct answer: C. \(\tfrac12\)
Explanation: Wartość oczekiwana to średnia: \(E = \tfrac{1+0}{2} = \tfrac12\).
Question 2Not answered
Wirówka zatrzymuje się na \(2\) z prawdopodobieństwem \(0.2\), na \(4\) z prawdopodobieństwem \(0.3\) i na \(8\) z prawdopodobieństwem \(0.5\). Jaka jest wartość oczekiwana?
Jaka jest wartość oczekiwana uczciwej sześciennej kostki do gry z oczkami od \(1\) do \(6\)?
Correct answer: B. \(3.5\)
Explanation: Średnia z oczek: \((1+2+3+4+5+6)/6 = 21/6 = 3.5\).
Question 4Not answered
Jaka jest wariancja uczciwej monety, która wypłaca \(1\) za orła i \(0\) za reszkę?
Correct answer: C. \(0.25\)
Explanation: Wariancja wynosi \(p(1-p) = 0.5\times0.5 = 0.25\).
Question 5Not answered
Moneta obciążona wypada orłem (wartym \(2\)) z prawdopodobieństwem \(0.7\) oraz reszką (wartą \(0\)) z prawdopodobieństwem \(0.3\). Jaka jest jej wartość oczekiwana?
Correct answer: A. \(1.4\)
Explanation: Wartość oczekiwana: \(2\times0.7 + 0\times0.3 = 1.4\).
Question 6Not answered
Jaka jest oczekiwana suma przy rzucie dwiema uczciwymi sześciennymi kostkami do gry?
Correct answer: D. \(7\)
Explanation: Każda kostka ma wartość oczekiwaną \(3.5\), więc razem \(3.5 + 3.5 = 7\).
Question 7Not answered
Jeśli rzucisz dwiema uczciwymi monetami i policzysz liczbę orłów, jaka jest oczekiwana liczba orłów?
Correct answer: A. \(1\)
Explanation: Rozkład dwumianowy z \(n=2, p=0.5\): wartość oczekiwana \(= np = 2\times0.5 = 1\).
Question 8Not answered
Jaka jest wariancja liczby orłów przy dwóch rzutach uczciwą monetą?
Correct answer: D. \(0.5\)
Explanation: Var = \(np(1-p) = 2\times0.5\times0.5 = 0.5\).
Question 9Not answered
Gra wypłaca \(+3\) z prawdopodobieństwem \(0.5\) oraz \(-1\) z prawdopodobieństwem \(0.5\). Jaka jest oczekiwana wypłata?
Correct answer: C. \(1\)
Explanation: Wartość oczekiwana: \(0.5\times3 + 0.5\times(-1) = 1\).
Question 10Not answered
Loteria wypłaca \(5\) z prawdopodobieństwem \(0.2\) oraz \(0\) z prawdopodobieństwem \(0.8\). Jaka jest jej wartość oczekiwana?
Correct answer: A. \(1\)
Explanation: Wartość oczekiwana: \(5\times0.2 + 0\times0.8 = 1\).
