problem 1

The PDF is

f (x) = x e^{- x^{2} ∕ 2}

\begin{matrix} P (X \leq a) & = \int_{0}^{a} f (x) dx = \int_{0}^{a} x e^{- x^{2} ∕ 2} dx = 1 - e^{- a^{2} ∕ 2} \end{matrix}

The quantile function $Q$ , equal to the inverse of the CDF, is given by

Q (a) = \sqrt{- 2 log (1 - a)}

(a)

\begin{matrix} P (1 < X < 3) & = P (X \leq 3) - P (X \leq 1) = e^{- 1 ∕ 2} - e^{- 9 ∕ 2} \end{matrix}

(b) The general formula for calculating the quartiles is as follows

P (X \leq q_{j}) = \frac{j}{4}

q_{j} = Q (\frac{j}{4}) = \sqrt{- 2 log (\frac{3 j}{4})}, for j = 1, 2, 3

First quartile $q_{1}$

q_{1} = 0.759

Second quartile $q_{2}$ , also known as median

q_{2} = 1.18

Third quartile $q_{3}$

q_{3} = 1.67