Calculating Expected Value of Random Variables - Statistics
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Which value equals the mean of a discrete probability distribution: median, mode, or expected value?
Which value equals the mean of a discrete probability distribution: median, mode, or expected value?
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Expected value. Expected value and mean are the same for probability distributions.
Expected value. Expected value and mean are the same for probability distributions.
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What does the expected value $E(X)$ represent in a probability distribution?
What does the expected value $E(X)$ represent in a probability distribution?
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The mean (long-run average) of the distribution. It's the average value if repeated infinitely.
The mean (long-run average) of the distribution. It's the average value if repeated infinitely.
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Interpret $E(X)=2.4$ in a repeated-trials context.
Interpret $E(X)=2.4$ in a repeated-trials context.
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Average outcome approaches $2.4$ per trial in the long run. Over many trials, the average result converges to $E(X)$.
Average outcome approaches $2.4$ per trial in the long run. Over many trials, the average result converges to $E(X)$.
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Identify the expected value for a fair game with outcomes as net gains and losses.
Identify the expected value for a fair game with outcomes as net gains and losses.
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$E(\text{net gain})=0$. Fair games have zero expected net gain/loss.
$E(\text{net gain})=0$. Fair games have zero expected net gain/loss.
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Compute $E(X)$ for $X\in{0,1,2}$ with $P(0)=0.25$, $P(1)=0.5$, $P(2)=0.25$.
Compute $E(X)$ for $X\in{0,1,2}$ with $P(0)=0.25$, $P(1)=0.5$, $P(2)=0.25$.
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$E(X)=1$. $E(X) = 0(0.25) + 1(0.5) + 2(0.25) = 0 + 0.5 + 0.5 = 1$.
$E(X)=1$. $E(X) = 0(0.25) + 1(0.5) + 2(0.25) = 0 + 0.5 + 0.5 = 1$.
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Which value must you compute to interpret the mean of a discrete probability distribution?
Which value must you compute to interpret the mean of a discrete probability distribution?
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The expected value $E(X)$. The mean of a distribution is its expected value.
The expected value $E(X)$. The mean of a distribution is its expected value.
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Calculate $E(X)$ for $X\in{1,2,3}$ with probabilities ${0.2,0.5,0.3}$.
Calculate $E(X)$ for $X\in{1,2,3}$ with probabilities ${0.2,0.5,0.3}$.
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$E(X)=2.1$. $E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1$.
$E(X)=2.1$. $E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1$.
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Calculate $E(X)$ if $P(X=0)=0.2$ and $P(X=5)=0.8$.
Calculate $E(X)$ if $P(X=0)=0.2$ and $P(X=5)=0.8$.
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$E(X)=4$. $E(X) = 0(0.2) + 5(0.8) = 0 + 4 = 4$.
$E(X)=4$. $E(X) = 0(0.2) + 5(0.8) = 0 + 4 = 4$.
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Which expression gives $E(X)$ when outcomes are $x_1,\dots,x_n$ with probabilities $p_1,\dots,p_n$?
Which expression gives $E(X)$ when outcomes are $x_1,\dots,x_n$ with probabilities $p_1,\dots,p_n$?
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$E(X)=\sum_{i=1}^{n} x_i p_i$. Alternative notation using indexed outcomes and probabilities.
$E(X)=\sum_{i=1}^{n} x_i p_i$. Alternative notation using indexed outcomes and probabilities.
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Identify the condition that must hold for a valid discrete probability distribution.
Identify the condition that must hold for a valid discrete probability distribution.
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$\sum P(X=x)=1$ and each $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.
$\sum P(X=x)=1$ and each $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.
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State the formula for the expected value $E(X)$ of a discrete random variable.
State the formula for the expected value $E(X)$ of a discrete random variable.
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$E(X)=\sum x,P(X=x)$. Sum each outcome times its probability.
$E(X)=\sum x,P(X=x)$. Sum each outcome times its probability.
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Calculate $E(X)$ if $X=10$ with probability $0.1$ and $X=0$ otherwise.
Calculate $E(X)$ if $X=10$ with probability $0.1$ and $X=0$ otherwise.
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$E(X)=1$. $E(X) = 10(0.1) + 0(0.9) = 1$.
$E(X)=1$. $E(X) = 10(0.1) + 0(0.9) = 1$.
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State the shortcut for $E(X)$ when $X$ takes values $a$ and $b$ with probabilities $p$ and $1-p$.
State the shortcut for $E(X)$ when $X$ takes values $a$ and $b$ with probabilities $p$ and $1-p$.
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$E(X)=ap+b(1-p)$. Direct formula for two-outcome distributions.
$E(X)=ap+b(1-p)$. Direct formula for two-outcome distributions.
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Identify the expected value of a constant random variable $X=c$.
Identify the expected value of a constant random variable $X=c$.
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$E(X)=c$. Constant values have expected value equal to the constant.
$E(X)=c$. Constant values have expected value equal to the constant.
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Compute $E(X)$ for a uniform distribution on ${2,4,6,8}$.
Compute $E(X)$ for a uniform distribution on ${2,4,6,8}$.
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$E(X)=5$. $E(X) = \frac{2+4+6+8}{4} = \frac{20}{4} = 5$.
$E(X)=5$. $E(X) = \frac{2+4+6+8}{4} = \frac{20}{4} = 5$.
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Calculate $E(X)$ if $X\in{-1,1}$ with $P(X=-1)=0.7$ and $P(X=1)=0.3$.
Calculate $E(X)$ if $X\in{-1,1}$ with $P(X=-1)=0.7$ and $P(X=1)=0.3$.
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$E(X)=-0.4$. $E(X) = -1(0.7) + 1(0.3) = -0.7 + 0.3 = -0.4$.
$E(X)=-0.4$. $E(X) = -1(0.7) + 1(0.3) = -0.7 + 0.3 = -0.4$.
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State the linearity rule for the expected value of $aX+b$.
State the linearity rule for the expected value of $aX+b$.
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$E(aX+b)=aE(X)+b$. Linear transformations preserve linearity in expectation.
$E(aX+b)=aE(X)+b$. Linear transformations preserve linearity in expectation.
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Compute $E(X)$ for $X\in{1,3}$ with $P(X=1)=\frac{1}{4}$ and $P(X=3)=\frac{3}{4}$.
Compute $E(X)$ for $X\in{1,3}$ with $P(X=1)=\frac{1}{4}$ and $P(X=3)=\frac{3}{4}$.
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$E(X)=\frac{5}{2}$. $E(X) = 1(\frac{1}{4}) + 3(\frac{3}{4}) = \frac{1}{4} + \frac{9}{4} = \frac{10}{4}$.
$E(X)=\frac{5}{2}$. $E(X) = 1(\frac{1}{4}) + 3(\frac{3}{4}) = \frac{1}{4} + \frac{9}{4} = \frac{10}{4}$.
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Find $E(X)$ if $X$ takes values $0$ and $1$ with $P(X=1)=p$ and $P(X=0)=1-p$.
Find $E(X)$ if $X$ takes values $0$ and $1$ with $P(X=1)=p$ and $P(X=0)=1-p$.
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$E(X)=p$. For Bernoulli: $E(X) = 0(1-p) + 1(p) = p$.
$E(X)=p$. For Bernoulli: $E(X) = 0(1-p) + 1(p) = p$.
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Identify the expected value $E(X)$ if $P(X=2)=1$ (a certain outcome).
Identify the expected value $E(X)$ if $P(X=2)=1$ (a certain outcome).
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$E(X)=2$. When only one outcome is possible, $E(X)$ equals that outcome.
$E(X)=2$. When only one outcome is possible, $E(X)$ equals that outcome.
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What condition must a discrete probability distribution satisfy for $E(X)$ to be computed?
What condition must a discrete probability distribution satisfy for $E(X)$ to be computed?
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$\sum P(X=x)=1$ and $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.
$\sum P(X=x)=1$ and $P(X=x)\ge 0$. Probabilities must be non-negative and sum to 1.
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What is the interpretation of $E(X)$ in a probability distribution for a discrete random variable?
What is the interpretation of $E(X)$ in a probability distribution for a discrete random variable?
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The long-run mean (average) outcome. Expected value represents the average if repeated infinitely.
The long-run mean (average) outcome. Expected value represents the average if repeated infinitely.
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State the formula for the expected value $E(X)$ of a discrete random variable $X$.
State the formula for the expected value $E(X)$ of a discrete random variable $X$.
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$E(X)=\sum x,P(X=x)$. Sum each outcome times its probability.
$E(X)=\sum x,P(X=x)$. Sum each outcome times its probability.
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Compute $E(X)$ for $X\in{0,1,2}$ with probabilities $0.2,0.5,0.3$ respectively.
Compute $E(X)$ for $X\in{0,1,2}$ with probabilities $0.2,0.5,0.3$ respectively.
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$E(X)=1.1$. $E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 0 + 0.5 + 0.6$.
$E(X)=1.1$. $E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 0 + 0.5 + 0.6$.
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Find and correct the formula error: $E(X)=\sum P(X=x)$.
Find and correct the formula error: $E(X)=\sum P(X=x)$.
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Correct: $E(X)=\sum x,P(X=x)$. Missing the multiplication by $x$ in each term.
Correct: $E(X)=\sum x,P(X=x)$. Missing the multiplication by $x$ in each term.
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What does it mean if a game has expected value $E(G)=0$ for the player’s net gain?
What does it mean if a game has expected value $E(G)=0$ for the player’s net gain?
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It is fair in the long run. Zero expected value means no advantage to either side.
It is fair in the long run. Zero expected value means no advantage to either side.
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Identify the expected value of net gain $G$ for a $\$2$ ticket that pays $$8$ with probability $0.1$ and $\$0$ otherwise.
Identify the expected value of net gain $G$ for a $\$2$ ticket that pays $$8$ with probability $0.1$ and $\$0$ otherwise.
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$E(G)=-1.2$ dollars. $E(G) = (8-2)(0.1) + (0-2)(0.9) = 0.6 - 1.8$.
$E(G)=-1.2$ dollars. $E(G) = (8-2)(0.1) + (0-2)(0.9) = 0.6 - 1.8$.
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Compute expected winnings $E(W)$ if you win $\$10$ with probability $0.3$ and lose $$4$ with probability $0.7$.
Compute expected winnings $E(W)$ if you win $\$10$ with probability $0.3$ and lose $$4$ with probability $0.7$.
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$E(W)=0.2$ dollars. $E(W) = 10(0.3) + (-4)(0.7) = 3 - 2.8$.
$E(W)=0.2$ dollars. $E(W) = 10(0.3) + (-4)(0.7) = 3 - 2.8$.
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Compute $E(X)$ for a fair coin where $X=1$ for heads and $X=0$ for tails.
Compute $E(X)$ for a fair coin where $X=1$ for heads and $X=0$ for tails.
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$E(X)=0.5$. Fair coin: $E(X) = 0(0.5) + 1(0.5)$.
$E(X)=0.5$. Fair coin: $E(X) = 0(0.5) + 1(0.5)$.
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Compute the expected value of a fair six-sided die roll $X\in{1,2,3,4,5,6}$.
Compute the expected value of a fair six-sided die roll $X\in{1,2,3,4,5,6}$.
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$E(X)=3.5$. $E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6}$.
$E(X)=3.5$. $E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6}$.
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