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rev | Admin | (Created page with "'''Solution: B''' We have <math display = "block">P(X=i) = \frac{i}{\sum_{i=1}^6 i } = \frac{i}{21}</math>, <math display = "block"> E[X] = \frac{1}{21}\sum_{i=1}^6 i^2 = \frac{91}{21} </math> and <math display = "block"> \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)<sup>2</sup> = 2.22.") | Jun 25'24 at 1:32 | +372 |