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Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 4F / Solution

Normalization of the probability density function gives

1 = c \int_{0}^{3} \frac{d x}{1 + x ^{2}} = c arctan (3) = c \frac{π}{3},

so

c = \frac{3}{π} .

The expected value is

E X = \frac{3}{π} \int_{0}^{3} \frac{x}{1 + x ^{2}} d x = \frac{3 lo g 2}{π} .

The second moment is

E X^{2} = \frac{3}{π} \int_{0}^{3} \frac{x ^{2}}{1 + x ^{2}} d x = \frac{3 3}{π} - 1.

Therefore the variance is

var X = \frac{3 3}{π} - 1 - \frac{9 ( lo g 2 ) ^{2}}{π ^{2}} .

For the median

m

,

\frac{3}{π} arctan m = \frac{1}{2},

so

m = tan \frac{π}{6} = \frac{1}{3} .

Solved by gpt-5.6-sol high.

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