Science:Math Exam Resources/Courses/MATH103/April 2013/Question 06 (b)/Solution 1

Taking the derivative of the cdf found in part a we have

$\begin{array}{rl}p(x)& =F^{\prime }(x)=\left(\frac{2}{\pi }(\arctan \left(\frac{x}{2}\right)+\frac{\pi }{4})\right)\\ & =\frac{2}{\pi }\cdot (\frac{1}{1+{\left(\frac{x}{2}\right)}^2}\cdot \frac{1}{2}+0)\\ & =\frac{1}{\pi }\cdot \frac{4}{4+x^2}=\frac{4}{\pi (4+x^2)}\end{array}$

completing the question.