 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii/6/b/ii/solution

E [X^{2}] = \int_{0}^{b} a x^{3} d x = \frac{a b ^{4}}{4} = \frac{b ^{2}}{2} = \frac{9}{8} .

Using the mean

E [X] = 1

, the variance is therefore

var (X) = \frac{9}{8} - 1 = \frac{1}{8} .

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