 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/9/b/v/solution

The standard normal distribution has mean zero and variance one, so

E [Z^{2}] = var (Z) + (E [Z])^{2} = 1

. Part (iv) therefore gives

E [X] = \frac{π}{2} .

Using the second moment from part (iii),

var (X) = E [X^{2}] - (E [X])^{2} = 2 - \frac{π}{2} = \frac{4 - π}{2} .

 New to topics? Read the docs here!