Statistics Flashcards: Developing Empirical Probability Distributions Expected Value

What this deck covers

This deck focuses on Developing Empirical Probability Distributions Expected Value, 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: What is the expected value formula for a discrete random variable $X$?

Answer: $E(X)=\sum x\,P(x)$. Multiply each value by its probability and sum all products.

Flashcard 2: What is the sample space for $X=\text{number of TV sets per household}$ if observed values are $0$ to $5$?

Answer: $\{0,1,2,3,4,5\}$. Sample space contains all possible values the random variable can take.

Flashcard 3: What is $E(X)$ for $X\in\{1,2,3\}$ with $P=\{0.2,0.5,0.3\}$?

Answer: $E(X)=2.1$. $1(0.2)+2(0.5)+3(0.3)=0.2+1.0+0.9=2.1$

Flashcard 4: Find $E(X)$ if $P(X=1)=0.25$, $P(X=2)=0.50$, $P(X=3)=0.25$.

Answer: $1(0.25)+2(0.50)+3(0.25)=2$. Weighted average: $1(0.25)+2(0.50)+3(0.25)=0.25+1+0.75$.

Flashcard 5: What is $E(X)$ for $X\in\{0,1,2,3\}$ with $P=\{0.1,0.2,0.3,0.4\}$?

Answer: $E(X)=2.0$. $0(0.1)+1(0.2)+2(0.3)+3(0.4)=0+0.2+0.6+1.2=2.0$

Flashcard 6: What two conditions must a discrete probability distribution $P(X=x)$ satisfy?

Answer: $0\le P(X=x)\le 1$ and $\sum P(X=x)=1$. Probabilities must be between 0 and 1, and sum to 1 for all outcomes.

Flashcard 7: Identify the random variable in: "number of TV sets per household" for a probability distribution.

Answer: $X=\text{number of TV sets in a randomly selected household}$. Random variable represents the quantity being measured in each trial.

Flashcard 8: Find and correct the error: "$E(X)=\sum P(X=x)$" for a discrete random variable.

Answer: Correct: $E(X)=\sum x\,P(X=x)$. Error: formula missing the $x$ values; must multiply $x$ by $P(X=x)$.

Flashcard 9: What is the expected value if $X$ takes values $0,1,2$ with probabilities $0.2,0.5,0.3$?

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

Flashcard 10: What is the expected value if a distribution is $P(0)=0.25$, $P(2)=0.75$?

Answer: $E(X)=1.5$. $0(0.25)+2(0.75)=0+1.5=1.5$

Flashcard 11: What does $E(X)$ represent in context for TVs per household?

Answer: Long-run average TVs per household. The average number of TVs we expect to find per household.

Flashcard 12: What is the expected number in $100$ households if $P(0)=0.2$, $P(1)=0.5$, $P(2)=0.3$?

Answer: $110$. $E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$, so $100(1.1)=110$

Flashcard 13: What is $E(X)$ if $X\in\{0,2\}$ with $P(X=0)=0.6$ and $P(X=2)=0.4$?

Answer: $0(0.6)+2(0.4)=0.8$. Only two outcomes: $0$ contributes nothing, $2$ contributes $2(0.4)$.

Flashcard 14: What is the expected number of TV sets in $100$ households if $E(X)=1.7$ sets per household?

Answer: $100\cdot 1.7=170$ sets. Multiply expected value per household by number of households.

Flashcard 15: Identify the random variable in: "number of TV sets in a randomly selected household."

Answer: $X=\text{number of TV sets in one household}$. The variable counts TV sets in a single household.

Flashcard 16: What is $E(X)$ for $X\in\{0,1,2\}$ with empirical counts $10,30,60$ out of $100$?

Answer: $0(0.1)+1(0.3)+2(0.6)=1.5$. Convert counts to probabilities: $\frac{10}{100}=0.1$, $\frac{30}{100}=0.3$, $\frac{60}{100}=0.6$.

Flashcard 17: What is an empirical probability for an outcome with frequency $f$ in $n$ trials?

Answer: $P(\text{outcome})=\frac{f}{n}$. Divides frequency of occurrence by total number of trials.

Flashcard 18: Find the expected number in $200$ households if $E(X)=2.3$ TV sets per household.

Answer: $200\cdot 2.3=460$ sets. Scale up single household expectation to 200 households.

Flashcard 19: Identify the missing probability if $P(X=0)=0.1$, $P(X=1)=0.4$, and $P(X=2)=?$ for a distribution.

Answer: $P(X=2)=1-0.1-0.4=0.5$. Probabilities must sum to 1, so missing probability is $1-0.1-0.4$.

Flashcard 20: What is $E(X)$ if $X\in\{0,1,2,3\}$ with probabilities $0.1,0.2,0.3,0.4$?

Answer: $0(0.1)+1(0.2)+2(0.3)+3(0.4)=2.0$. Sum products: $0+0.2+0.6+1.2=2.0$.

Flashcard 21: What is the expected count in $n$ trials if the expected value per trial is $E(X)$?

Answer: $\text{Expected total}=n\,E(X)$. Multiply expected value per trial by number of trials.

Flashcard 22: State the formula for the expected total over $n$ independent observations of $X$ with mean $E(X)$.

Answer: $E\left(\sum_{i=1}^{n} X_i\right)=nE(X)$. Expected sum of $n$ independent observations is $n$ times single expectation.

Flashcard 23: What must be true about the probabilities in a discrete probability distribution?

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

Flashcard 24: Identify the error: a "distribution" has $P(0)=0.4$, $P(1)=0.5$, $P(2)=0.3$.

Answer: $\sum P(x)=1.2\ne 1$. Probabilities sum to more than 1, violating distribution rules.

Flashcard 25: What is the expected total in $100$ trials if $E(X)=1.2$ per trial?

Answer: $120$. $100 \times 1.2 = 120$

Flashcard 26: What is the expected count in $n$ trials if the probability of success is $p$?

Answer: $E(\text{count})=np$. Expected count in $n$ trials equals trials times success probability.

Flashcard 27: What is the expected number in $100$ households if $E(X)=2.3$ TVs per household?

Answer: $230$. $100 \times 2.3 = 230$

Flashcard 28: Find the missing probability if $P(0)=0.12$, $P(1)=0.38$, $P(2)=0.27$, $P(3)=0.15$.

Answer: $P(4)=0.08$. $1-0.12-0.38-0.27-0.15=0.08$

Flashcard 29: What is $E(X)$ if $P(X=0)=0.1$, $P(X=1)=0.6$, and $P(X=2)=0.3$?

Answer: $E(X)=1.2$. $0(0.1)+1(0.6)+2(0.3)=0+0.6+0.6=1.2$

Flashcard 30: State the formula for the expected value of a discrete random variable $X$.

Answer: $E(X)=\sum x\,P(X=x)$. Expected value is the weighted average using probabilities as weights.

Flashcard 31: What is $E(X)$ if $X$ is always $4$ (that is, $P(X=4)=1$)?

Answer: $E(X)=4$. When a variable is constant, its expected value equals that constant.

Flashcard 32: What is the empirical probability if $18$ of $200$ households have $3$ TVs?

Answer: $P(X=3)=0.09$. $\frac{18}{200}=0.09$

Flashcard 33: What is the empirical probability if $7$ of $50$ households have $0$ TVs?

Answer: $P(X=0)=0.14$. $\frac{7}{50}=0.14$

Flashcard 34: What is the empirical distribution if counts for $X=0,1,2$ are $5,15,30$ out of $50$?

Answer: $P(0)=0.1,\;P(1)=0.3,\;P(2)=0.6$. Divide each count by total 50: $\frac{5}{50}=0.1$, $\frac{15}{50}=0.3$, $\frac{30}{50}=0.6$.

Flashcard 35: What is the sample space for $X$ = number of TV sets per household (discrete, nonnegative)?

Answer: $\{0,1,2,3,\dots\}$. All possible non-negative integer counts.