 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/9/c/i/solution

Write the card values as

a_{1}, \dots, a_{1000}

. Uniform selection gives

E [X] = \frac{1}{1000} \sum_{i} a_{i} = 0, E [X^{2}] = \frac{1}{1000} \sum_{i} a_{i}^{2} = 1.

Thus $E [X] = 0$ and $Var (X) = 1$ , by the variance identity

Var (X) = E [X^{2}] - E [X]^{2}

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