Exercise


ABy Admin
Jun 25'24

Answer

Solution: B

We have

[[math]]P(X=i) = \frac{i}{\sum_{i=1}^6 i } = \frac{i}{21}[[/math]]

,

[[math]] E[X] = \frac{1}{21}\sum_{i=1}^6 i^2 = \frac{91}{21} [[/math]]

and

[[math]] \operatorname{E}[X^2] = \frac{1}{21}\sum_{i=1}^6 i^3 = 21. [[/math]]

Hence the variance of [math]X[/math] equals 21 - (91/21)2 = 2.22.

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