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 $\frac{5}{4}$.

Weather	Clear	Fair	Dubious	Rainy
Probability	$0.50$	$0.30$	$0.15$	$0.05$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

$x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$0.2$	$0.5$	$0.15$	$0.25$	$0.1$

$A$ 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

If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$
$P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$

Then the variance of $X$ is:

The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then, $P(0 < X < 4)$ is equal to:

If the probability distribution of a random variable $X$ is given by the following table,then $F(0) =$ . . . . . .
$x_i$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $0.2$ $0.5$ $0.15$ $0.25$ $0.1$

Two coins are tossed simultaneously. Then,the value of $E(X)$,where $X$ denotes the number of heads is

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