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If a cubical die is thrown, then the mean and variance of the random variable $\mathrm{X}$, giving the number on the face that shows up, are respectively
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$\frac{7}{2}, \frac{35}{12}$
$\begin{aligned} & \text { } X=\{1,2,3,4,5,6\} \text { and } P(1)=P(2) \ldots P(6)=\frac{1}{6} \\ & \text { Mean }=\sum x_i \cdot P=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2} \\ & \text { Variance }=\sum x_i^2 \cdot P-(\text { Mean })^2 \\ & =\frac{1}{6}\left[1^2+2^2+3^2+4^2+5^2+6^2\right]-\frac{49}{4} \\ & =\frac{1}{6} \times 91-\frac{49}{4}=\frac{35}{12} \\ & \Rightarrow \text { Variance }=\frac{35}{12}\end{aligned}$
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