Independent solution

How to solve this Continuous Random Variables question

Setup

Setup

Integrate the density above a threshold to obtain its survival function.

S(x)=∫x∞2t3 dt=1x2,x>1S(x)=\int_x^{\infty}\frac{2}{t^3}\,dt=\frac1{x^2},\qquad x>1

Model

Model

Express the conditional interval probability through survival values at three and four.

Pr⁡(X<4∣X≥3)=S(3)−S(4)S(3)\Pr(X<4\mid X\ge3)=\frac{S(3)-S(4)}{S(3)}

Compute

Compute

Substitute the two survival probabilities and simplify.

1/9−1/161/9=1−916=716=0.4375\frac{1/9-1/16}{1/9}=1-\frac9{16}=\frac7{16}=0.4375

Answer

Answer

The conditional probability rounds to 0.44.

0.44(E)\boxed{0.44\quad\text{(E)}}