Developing Theoretical Probability Distributions, Expected Value - Statistics
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What is $P(X=0)$ when guessing on $5$ questions with $p=\frac{1}{4}$?
What is $P(X=0)$ when guessing on $5$ questions with $p=\frac{1}{4}$?
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$\left(\frac{3}{4}\right)^5$. All 5 wrong: $(1-p)^5$ where $p=\frac{1}{4}$.
$\left(\frac{3}{4}\right)^5$. All 5 wrong: $(1-p)^5$ where $p=\frac{1}{4}$.
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State the formula for the expected value of a discrete random variable $X$.
State the formula for the expected value of a discrete random variable $X$.
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$E(X)=\sum x,P(X=x)$. Sum each outcome value times its probability.
$E(X)=\sum x,P(X=x)$. Sum each outcome value times its probability.
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What two conditions must a probability distribution for discrete $X$ satisfy?
What two conditions must a probability distribution for discrete $X$ satisfy?
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$0 \le P(X=x) \le 1$ and $\sum P(X=x)=1$. Probabilities must be between 0 and 1, and sum to 1 for all outcomes.
$0 \le P(X=x) \le 1$ and $\sum P(X=x)=1$. Probabilities must be between 0 and 1, and sum to 1 for all outcomes.
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What is $P(X=5)$ when guessing on $5$ questions with $p=\frac{1}{4}$?
What is $P(X=5)$ when guessing on $5$ questions with $p=\frac{1}{4}$?
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$\left(\frac{1}{4}\right)^5$. All 5 correct: $p^5$ where $p=\frac{1}{4}$.
$\left(\frac{1}{4}\right)^5$. All 5 correct: $p^5$ where $p=\frac{1}{4}$.
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What is $P(X=1)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
What is $P(X=1)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
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$\binom{5}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4$. Uses binomial formula with $n=5$, $k=1$, $p=\frac{1}{4}$.
$\binom{5}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4$. Uses binomial formula with $n=5$, $k=1$, $p=\frac{1}{4}$.
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What is $P(X=2)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
What is $P(X=2)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
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$\binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3$. Uses binomial formula with $n=5$, $k=2$, $p=\frac{1}{4}$.
$\binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3$. Uses binomial formula with $n=5$, $k=2$, $p=\frac{1}{4}$.
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What is $P(X=3)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
What is $P(X=3)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
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$\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$. Uses binomial formula with $n=5$, $k=3$, $p=\frac{1}{4}$.
$\binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2$. Uses binomial formula with $n=5$, $k=3$, $p=\frac{1}{4}$.
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What is $P(X=4)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
What is $P(X=4)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
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$\binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)$. Uses binomial formula with $n=5$, $k=4$, $p=\frac{1}{4}$.
$\binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)$. Uses binomial formula with $n=5$, $k=4$, $p=\frac{1}{4}$.
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What is the expected number correct when guessing on $5$ questions with $p=\frac{1}{4}$?
What is the expected number correct when guessing on $5$ questions with $p=\frac{1}{4}$?
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$E(X)=5\cdot\frac{1}{4}=\frac{5}{4}$. For binomial, expected value equals $np$.
$E(X)=5\cdot\frac{1}{4}=\frac{5}{4}$. For binomial, expected value equals $np$.
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State the expected value formula for $X\sim\text{Binomial}(n,p)$.
State the expected value formula for $X\sim\text{Binomial}(n,p)$.
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$E(X)=np$. Mean of binomial distribution is number of trials times success probability.
$E(X)=np$. Mean of binomial distribution is number of trials times success probability.
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If score is $S=2X$, what is $E(S)$ in terms of $E(X)$?
If score is $S=2X$, what is $E(S)$ in terms of $E(X)$?
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$E(S)=2E(X)$. Expected value scales linearly with constant multiplier.
$E(S)=2E(X)$. Expected value scales linearly with constant multiplier.
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State the linearity rule for expected value: what is $E(aX+b)$?
State the linearity rule for expected value: what is $E(aX+b)$?
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$E(aX+b)=aE(X)+b$. Constants factor out; adding $b$ shifts expected value by $b$.
$E(aX+b)=aE(X)+b$. Constants factor out; adding $b$ shifts expected value by $b$.
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If each correct is $+1$ point and no penalty, what is expected score for $5$ questions with $p=\frac{1}{4}$?
If each correct is $+1$ point and no penalty, what is expected score for $5$ questions with $p=\frac{1}{4}$?
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$E(S)=E(X)=\frac{5}{4}$. Score equals number correct, so $E(S)=E(X)=np$.
$E(S)=E(X)=\frac{5}{4}$. Score equals number correct, so $E(S)=E(X)=np$.
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If each correct is $+4$ points and wrong is $0$, what is expected score for $5$ questions with $p=\frac{1}{4}$?
If each correct is $+4$ points and wrong is $0$, what is expected score for $5$ questions with $p=\frac{1}{4}$?
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$E(S)=4\cdot\frac{5}{4}=5$. Score is $4X$, so $E(S)=4E(X)=4\cdot\frac{5}{4}$.
$E(S)=4\cdot\frac{5}{4}=5$. Score is $4X$, so $E(S)=4E(X)=4\cdot\frac{5}{4}$.
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If each correct is $+1$ and each wrong is $-\frac{1}{3}$, what is expected score for $5$ questions with $p=\frac{1}{4}$?
If each correct is $+1$ and each wrong is $-\frac{1}{3}$, what is expected score for $5$ questions with $p=\frac{1}{4}$?
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$5\left(\frac{1}{4}-\frac{1}{3}\cdot\frac{3}{4}\right)=0$. Expected gain per question is $1\cdot p-\frac{1}{3}(1-p)=0$ when $p=\frac{1}{4}$.
$5\left(\frac{1}{4}-\frac{1}{3}\cdot\frac{3}{4}\right)=0$. Expected gain per question is $1\cdot p-\frac{1}{3}(1-p)=0$ when $p=\frac{1}{4}$.
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Which expression gives $P(X\ge 3)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
Which expression gives $P(X\ge 3)$ for $X\sim\text{Binomial}(5,\frac{1}{4})$?
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$1-[P(X=0)+P(X=1)+P(X=2)]$. Complement rule: $P(X\ge 3)=1-P(X\le 2)$.
$1-[P(X=0)+P(X=1)+P(X=2)]$. Complement rule: $P(X\ge 3)=1-P(X\le 2)$.
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What is the theoretical probability of a correct guess on a $4$-choice multiple-choice question?
What is the theoretical probability of a correct guess on a $4$-choice multiple-choice question?
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$\frac{1}{4}$. One correct choice out of four equally likely options.
$\frac{1}{4}$. One correct choice out of four equally likely options.
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Identify the distribution of $X$ = number correct when guessing on $n$ independent questions with success prob $p$.
Identify the distribution of $X$ = number correct when guessing on $n$ independent questions with success prob $p$.
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$X\sim\text{Binomial}(n,p)$. Counting successes in $n$ independent trials follows binomial distribution.
$X\sim\text{Binomial}(n,p)$. Counting successes in $n$ independent trials follows binomial distribution.
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State the binomial probability formula for $P(X=k)$ when $X\sim\text{Binomial}(n,p)$.
State the binomial probability formula for $P(X=k)$ when $X\sim\text{Binomial}(n,p)$.
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$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Gives probability of exactly $k$ successes in $n$ trials.
$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Gives probability of exactly $k$ successes in $n$ trials.
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What is $P(\text{correct})$ when guessing on one 4-choice multiple-choice question?
What is $P(\text{correct})$ when guessing on one 4-choice multiple-choice question?
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$\frac{1}{4}$. One correct choice out of four equally likely options.
$\frac{1}{4}$. One correct choice out of four equally likely options.
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What is $P(X=5)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
What is $P(X=5)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
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$\left(\frac{1}{4}\right)^5$. All 5 correct means $p^5$ where $p=\frac{1}{4}$.
$\left(\frac{1}{4}\right)^5$. All 5 correct means $p^5$ where $p=\frac{1}{4}$.
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What is $P(X=0)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
What is $P(X=0)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
Tap to reveal answer
$\left(\frac{3}{4}\right)^5$. All 5 wrong means $(1-p)^5$ where $p=\frac{1}{4}$.
$\left(\frac{3}{4}\right)^5$. All 5 wrong means $(1-p)^5$ where $p=\frac{1}{4}$.
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What is $P(X=2)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
What is $P(X=2)$ for guessing on $5$ questions with $p=\frac{1}{4}$?
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$\binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3$. Choose 2 positions for successes, multiply by $p^2$ and $(1-p)^3$.
$\binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3$. Choose 2 positions for successes, multiply by $p^2$ and $(1-p)^3$.
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What is the expected number correct when guessing on $5$ questions with $p=\frac{1}{4}$?
What is the expected number correct when guessing on $5$ questions with $p=\frac{1}{4}$?
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$E(X)=5\cdot\frac{1}{4}=\frac{5}{4}$. For binomial, $E(X)=np=5\cdot\frac{1}{4}$.
$E(X)=5\cdot\frac{1}{4}=\frac{5}{4}$. For binomial, $E(X)=np=5\cdot\frac{1}{4}$.
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State the expected value for $X\sim\text{Binomial}(n,p)$.
State the expected value for $X\sim\text{Binomial}(n,p)$.
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$E(X)=np$. Average number of successes equals trials times success probability.
$E(X)=np$. Average number of successes equals trials times success probability.
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What is the expected score if each correct answer is worth $2$ points and $X$ is # correct?
What is the expected score if each correct answer is worth $2$ points and $X$ is # correct?
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$E(2X)=2E(X)$. Linearity of expectation: constant multiplier scales the expected value.
$E(2X)=2E(X)$. Linearity of expectation: constant multiplier scales the expected value.
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What is the expected score for $5$ questions, $2$ points each correct, guessing with $p=\frac{1}{4}$?
What is the expected score for $5$ questions, $2$ points each correct, guessing with $p=\frac{1}{4}$?
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$2\cdot\frac{5}{4}=\frac{5}{2}$. Apply $E(2X)=2E(X)$ with $E(X)=\frac{5}{4}$.
$2\cdot\frac{5}{4}=\frac{5}{2}$. Apply $E(2X)=2E(X)$ with $E(X)=\frac{5}{4}$.
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What is the expected score for scheme $S=1X+0(5-X)$ when guessing on $5$ with $p=\frac{1}{4}$?
What is the expected score for scheme $S=1X+0(5-X)$ when guessing on $5$ with $p=\frac{1}{4}$?
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$E(S)=E(X)=\frac{5}{4}$. Only correct answers score, so $S=X$ and $E(S)=E(X)$.
$E(S)=E(X)=\frac{5}{4}$. Only correct answers score, so $S=X$ and $E(S)=E(X)$.
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What is the expected percent score if percent $=20X$ for $5$ questions and $p=\frac{1}{4}$?
What is the expected percent score if percent $=20X$ for $5$ questions and $p=\frac{1}{4}$?
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$E(20X)=20\cdot\frac{5}{4}=25$. Each correct is worth 20%, so multiply expected correct by 20.
$E(20X)=20\cdot\frac{5}{4}=25$. Each correct is worth 20%, so multiply expected correct by 20.
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Find $E(X)$ if $X\in{0,1,2}$ with $P(0)=0.2$, $P(1)=0.5$, $P(2)=0.3$.
Find $E(X)$ if $X\in{0,1,2}$ with $P(0)=0.2$, $P(1)=0.5$, $P(2)=0.3$.
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$E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$. Multiply each value by its probability and sum: $0.5+0.6=1.1$.
$E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$. Multiply each value by its probability and sum: $0.5+0.6=1.1$.
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