If a cubical die is thrown,then the mean and | vedclass

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(D) The random variable $X$ takes values $\{1, 2, 3, 4, 5, 6\}$ with probability $P(X=x_i) = \frac{1}{6}$ for each $i$. Mean $E(X) = \sum x_i P(x_i) =

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$\frac{7}{2}, \frac{35}{12}$

Vedclass · Answer

(D) The random variable $X$ takes values $\{1, 2, 3, 4, 5, 6\}$ with probability $P(X=x_i) = \frac{1}{6}$ for each $i$. Mean $E(X) = \sum x_i P(x_i) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = \frac{7}{2}$. Variance $Var(X) = E(X^2) - [E(X)]^2$. $E(X^2) = \sum x_i^2 P(x_i) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$. $Var(X) = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$. Thus,the mean is $\frac{7}{2}$ and the variance is $\frac{35}{12}$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

If a cubical die is thrown,then the mean and variance of the random variable $X$,representing the number on the face that shows up,are respectively:

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