Independent solution

How to solve this Continuous Random Variables question

Setup

Setup

The unknown proportionality constant cancels when the partial integral is divided by the full-support integral.

∫x1+x2 dx=12ln⁡(1+x2)\int \frac{x}{1+x^2}\,dx=\frac{1}{2}\ln(1+x^2)

Model

Model

Form the CDF on the support from the ratio of accumulated density shape to total density shape.

F(x)=12ln⁡(1+x2)12ln⁡(26)=ln⁡(1+x2)ln⁡(26),0<x<5F(x)=\frac{\frac12\ln(1+x^2)}{\frac12\ln(26)}=\frac{\ln(1+x^2)}{\ln(26)},\qquad 0<x<5

Compute

Compute

A median divides the probability mass into equal halves.

ln⁡(1+m2)ln⁡(26)=12\frac{\ln(1+m^2)}{\ln(26)}=\frac12
1+m2=261+m^2=\sqrt{26}
m=26−1=2.0246035…m=\sqrt{\sqrt{26}-1}=2.0246035\ldots

Answer

Answer

The closest listed value is 2.02.

m≈2.02(A)\boxed{m\approx 2.02\quad\text{(A)}}