 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/7/b/solution

E ∣ Z ∣ = \frac{2}{2 π} \int_{0}^{\infty} z e^{- z^{2} /2} d z = \frac{2}{2 π} [- e^{- z^{2} /2}]_{0}^{\infty} = \frac{2}{π} .

The integral is finite. Equivalently,

∣ Z ∣

has the half-normal distribution with unit scale.

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