 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/16/g/solution

E [X] = \int_{0}^{1} x 12 x^{2} (1 - x) d x = 12 (\frac{1}{4} - \frac{1}{5}) = \frac{3}{5} .

E [X^{2}] = 12 (1/5 - 1/6) = 2/5

Var (X) = \frac{2}{5} - (\frac{3}{5})^{2} = \frac{1}{25}, E [X] = \frac{3}{5}, σ = \frac{1}{5} .

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