A random variable X takes values -1, 0, 1, 2 | vedclass

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(D) The probabilities must satisfy $0 \leq P(X=x_i) \leq 1$ for all $i$.Thus,$0 \leq \frac{1+3p}{4} \leq 1$,$0 \leq \frac{1-p}{4} \leq 1$,$0 \leq \fra

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$-\frac{1}{16}$ and $\frac{5}{4}$

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(D) The probabilities must satisfy $0 \leq P(X=x_i) \leq 1$ for all $i$.Thus,$0 \leq \frac{1+3p}{4} \leq 1$,$0 \leq \frac{1-p}{4} \leq 1$,$0 \leq \frac{1+2p}{4} \leq 1$,and $0 \leq \frac{1-4p}{4} \leq 1$.Solving these inequalities:$1$) $1+3p \geq 0 \Rightarrow p \geq -1/3$$2$) $1-p \geq 0 \Rightarrow p \leq 1$$3$) $1+2p \geq 0 \Rightarrow p \geq -1/2$$4$) $1-4p \geq 0 \Rightarrow p \leq 1/4$Combining these,we get $-\frac{1}{3} \leq p \leq \frac{1}{4}$.The mean $E(X) = \sum x_i P(X=x_i) = (-1)\left(\frac{1+3p}{4}\right) + 0\left(\frac{1-p}{4}\right) + 1\left(\frac{1+2p}{4}\right) + 2\left(\frac{1-4p}{4}\right)$.$E(X) = \frac{-1-3p+1+2p+2-8p}{4} = \frac{2-9p}{4}$.Given $-\frac{1}{3} \leq p \leq \frac{1}{4}$,we find the range of $E(X)$:For $p = 1/4$,$E(X) = \frac{2-9(1/4)}{4} = \frac{2-2.25}{4} = -\frac{0.25}{4} = -\frac{1}{16}$.For $p = -1/3$,$E(X) = \frac{2-9(-1/3)}{4} = \frac{2+3}{4} = \frac{5}{4}$.Thus,the minimum value is $-\frac{1}{16}$ and the maximum value is $\frac{5}{4}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

Similar Questions

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|---|---|---|---|---|---|
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