Independent solution

How to solve this Uniform Distribution question

Setup

Setup

For a uniform variable on [a,2a], its rth percentile, with r expressed as a number from 0 to 100, advances r/100 of the width a from the lower endpoint.

Qr=a+r100(2a−a)=a(1+r100)Q_r=a+\frac{r}{100}(2a-a)=a\left(1+\frac{r}{100}\right)

Model

Model

Translate the stated equality of ratios into a squared equation for the unknown middle quantile.

Q40=1.4a,Q80=1.8a,Qp=a(1+p100)Q_{40}=1.4a,\qquad Q_{80}=1.8a,\qquad Q_p=a\left(1+\frac{p}{100}\right)
Q40Qp=QpQ80⟹Qp2=Q40Q80\frac{Q_{40}}{Q_p}=\frac{Q_p}{Q_{80}}\quad\Longrightarrow\quad Q_p^2=Q_{40}Q_{80}

Compute

Compute

Cancel the positive scale a and solve for the percentile rank.

(1+p100)2=(1.4)(1.8)=2.52\left(1+\frac{p}{100}\right)^2=(1.4)(1.8)=2.52
p=100(2.52−1)=58.74507866…p=100\left(\sqrt{2.52}-1\right)=58.74507866\ldots

Answer

Answer

The requested percentile rank rounds to 58.7.

58.7(B)\boxed{58.7\quad\text{(B)}}