Independent solution

How to solve this Continuous Random Variables question

Answer in brief

Integrating after the substitution u=ln x gives F(x)=(ln x/ln 80)^2. Inverting F(x)=0.75 yields x=80^{sqrt(0.75)}=44.4759, corresponding to choice C.

Setup

Setup

Integrate the density from the lower endpoint, using the logarithm as the natural substitution.

u=ln⁡t,du=dttu=\ln t,\qquad du=\frac{dt}{t}

Model

Model

The accumulated probability is a squared ratio of logarithms.

F(x)=∫1x2ln⁡t(ln⁡80)2t dtF(x)=\int_1^x\frac{2\ln t}{(\ln 80)^2t}\,dt
F(x)=(ln⁡x)2(ln⁡80)2,1<x<80F(x)=\frac{(\ln x)^2}{(\ln 80)^2},\qquad 1<x<80

Compute

Compute

Set the CDF equal to the requested cumulative probability and take the positive logarithmic root.

(ln⁡xln⁡80)2=0.75\left(\frac{\ln x}{\ln 80}\right)^2=0.75
ln⁡x=0.75 ln⁡80\ln x=\sqrt{0.75}\,\ln 80
x=800.75=44.4758518…x=80^{\sqrt{0.75}}=44.4758518\ldots

Answer

Answer

The calculated percentile rounds to the value under choice C.

x0.75≈44.48(C)\boxed{x_{0.75}\approx 44.48\quad\text{(C)}}