Statistics Flashcards: Calculating Expected Value Of Random Variables

What this deck covers

This deck focuses on Calculating Expected Value Of Random Variables, giving you a quick way to review the definitions, rules, and examples that matter most for Statistics.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find and correct the formula error: $E(X)=\sum P(X=x)$.

Answer: Correct: $E(X)=\sum x\,P(X=x)$. Missing the multiplication by $x$ in each term.

Flashcard 2: Identify the condition that must hold for a valid discrete probability distribution.

Answer: $\sum P(X=x)=1$ and each $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.

Flashcard 3: Calculate $E(X)$ if $X=10$ with probability $0.1$ and $X=0$ otherwise.

Answer: $E(X)=1$. $E(X) = 10(0.1) + 0(0.9) = 1$.

Flashcard 4: Compute $E(X+Y)$ given $E(X)=1.2$ and $E(Y)=-0.7$.

Answer: $0.5$. $E(X+Y) = 1.2 + (-0.7) = 1.2 - 0.7$.

Flashcard 5: What does the expected value $E(X)$ represent in a probability distribution?

Answer: The mean (long-run average) of the distribution. It's the average value if repeated infinitely.

Flashcard 6: Which expression gives $E(X)$ when outcomes are $x_1,\dots,x_n$ with probabilities $p_1,\dots,p_n$?

Answer: $E(X)=\sum_{i=1}^{n} x_i p_i$. Alternative notation using indexed outcomes and probabilities.

Flashcard 7: Identify the expected value $E(X)$ if $P(X=2)=1$ (a certain outcome).

Answer: $E(X)=2$. When only one outcome is possible, $E(X)$ equals that outcome.

Flashcard 8: State the additivity rule for expected value of a sum $X+Y$ (no independence required).

Answer: $E(X+Y)=E(X)+E(Y)$. Expected values add even without independence.

Flashcard 9: Compute $E(X)$ for a fair coin where $X=1$ for heads and $X=0$ for tails.

Answer: $E(X)=0.5$. Fair coin: $E(X) = 0(0.5) + 1(0.5)$.

Flashcard 10: Calculate $E(X)$ if $P(X=0)=0.2$ and $P(X=5)=0.8$.

Answer: $E(X)=4$. $E(X) = 0(0.2) + 5(0.8) = 0 + 4 = 4$.

Flashcard 11: Compute the expected value of a fair six-sided die roll $X\in\{1,2,3,4,5,6\}$.

Answer: $E(X)=3.5$. $E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6}$.

Flashcard 12: Identify the expected value for a fair game with outcomes as net gains and losses.

Answer: $E(\text{net gain})=0$. Fair games have zero expected net gain/loss.

Flashcard 13: State the linearity rule for the expected value of $aX+b$.

Answer: $E(aX+b)=aE(X)+b$. Linear transformations preserve linearity in expectation.

Flashcard 14: Calculate $E(X)$ if $X\in\{-1,1\}$ with $P(X=-1)=0.7$ and $P(X=1)=0.3$.

Answer: $E(X)=-0.4$. $E(X) = -1(0.7) + 1(0.3) = -0.7 + 0.3 = -0.4$.

Flashcard 15: Compute $E(X)$ for $X\in\{-1,2\}$ with $P(X=-1)=0.6$ and $P(X=2)=0.4$.

Answer: $E(X)=0.2$. $E(X) = -1(0.6) + 2(0.4) = -0.6 + 0.8$.

Flashcard 16: Compute $E(X)$ for $X\in\{1,3\}$ with $P(X=1)=\frac{1}{4}$ and $P(X=3)=\frac{3}{4}$.

Answer: $E(X)=\frac{5}{2}$. $E(X) = 1(\frac{1}{4}) + 3(\frac{3}{4}) = \frac{1}{4} + \frac{9}{4} = \frac{10}{4}$.

Flashcard 17: Compute $E(X)$ for a uniform distribution on $\{2,4,6,8\}$.

Answer: $E(X)=5$. $E(X) = \frac{2+4+6+8}{4} = \frac{20}{4} = 5$.

Flashcard 18: Calculate $E(X)$ for $X\in\{1,2,3\}$ with probabilities $\{0.2,0.5,0.3\}$.

Answer: $E(X)=2.1$. $E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1$.

Flashcard 19: What does it mean if a game has expected value $E(G)=0$ for the player's net gain?

Answer: It is fair in the long run. Zero expected value means no advantage to either side.

Flashcard 20: State the shortcut for $E(X)$ when $X$ takes values $a$ and $b$ with probabilities $p$ and $1-p$.

Answer: $E(X)=ap+b(1-p)$. Direct formula for two-outcome distributions.

Flashcard 21: Calculate $E(X)$ for a fair coin where $X=1$ for heads and $X=0$ for tails.

Answer: $E(X)=0.5$. $E(X) = 0(0.5) + 1(0.5) = 0.5$.

Flashcard 22: Interpret $E(X)=2.4$ in a repeated-trials context.

Answer: Average outcome approaches $2.4$ per trial in the long run. Over many trials, the average result converges to $E(X)$.

Flashcard 23: What is $E(c)$ if $c$ is a constant random variable (always equals $c$)?

Answer: $E(c)=c$. A constant's expected value is itself.

Flashcard 24: What condition must a discrete probability distribution satisfy for $E(X)$ to be computed?

Answer: $\sum P(X=x)=1$ and $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.

Flashcard 25: What is the interpretation of $E(X)$ in a probability distribution for a discrete random variable?

Answer: The long-run mean (average) outcome. Expected value represents the average if repeated infinitely.

Flashcard 26: Compute $E(X)$ for $X\in\{0,1,2\}$ with $P(0)=0.25$, $P(1)=0.5$, $P(2)=0.25$.

Answer: $E(X)=1$. $E(X) = 0(0.25) + 1(0.5) + 2(0.25) = 0 + 0.5 + 0.5 = 1$.

Flashcard 27: Identify the expected value of net gain $G$ for a $2 ticket that pays $8 with probability $0.1$ and $0 otherwise.

Answer: $E(G)=-1.2$ dollars. $E(G) = (8-2)(0.1) + (0-2)(0.9) = 0.6 - 1.8$.

Flashcard 28: Find $E(X)$ if $X$ takes values $0$ and $1$ with $P(X=1)=p$ and $P(X=0)=1-p$.

Answer: $E(X)=p$. For Bernoulli: $E(X) = 0(1-p) + 1(p) = p$.

Flashcard 29: Compute expected winnings $E(W)$ if you win $10 with probability $0.3$ and lose $4 with probability $0.7$.

Answer: $E(W)=0.2$ dollars. $E(W) = 10(0.3) + (-4)(0.7) = 3 - 2.8$.

Flashcard 30: Identify the expected value of a constant random variable $X=c$.

Answer: $E(X)=c$. Constant values have expected value equal to the constant.

Flashcard 31: Compute $E(X)$ for $X\in\{0,1,2\}$ with probabilities $0.2,0.5,0.3$ respectively.

Answer: $E(X)=1.1$. $E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 0 + 0.5 + 0.6$.

Flashcard 32: If $E(X)=4$, what is $E(3X-2)$?

Answer: $E(3X-2)=10$. $E(3X-2) = 3(4) - 2 = 12 - 2 = 10$.

Flashcard 33: State the expected value rule for a linear transformation $Y=aX+b$.

Answer: $E(Y)=aE(X)+b$. Linear transformations scale and shift the expected value.

Flashcard 34: Which value must you compute to interpret the mean of a discrete probability distribution?

Answer: The expected value $E(X)$. The mean of a distribution is its expected value.

Flashcard 35: State the formula for the expected value $E(X)$ of a discrete random variable.

Answer: $E(X)=\sum x\,P(X=x)$. Sum each outcome times its probability.

Flashcard 36: Find $E(X)$ for a fair six-sided die roll with outcomes $1$ to $6$.

Answer: $E(X)=3.5$. $E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$.

Flashcard 37: State the formula for the expected value $E(X)$ of a discrete random variable $X$.

Answer: $E(X)=\sum x\,P(X=x)$. Sum each outcome times its probability.

Flashcard 38: Compute $E(3X-5)$ given that $E(X)=4$.

Answer: $7$. $E(3X-5) = 3E(X) - 5 = 3(4) - 5 = 12 - 5$.

Flashcard 39: Which value equals the mean of a discrete probability distribution: median, mode, or expected value?

Answer: Expected value. Expected value and mean are the same for probability distributions.