Independent solution

How to solve this Survival Functions question

Setup

Setup

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

S(x)=∫x∞8t3 dt=4x2,x>2S(x)=\int_x^{\infty}\frac8{t^3}\,dt=\frac4{x^2},\qquad x>2

Model

Model

The favorable interval is the part of the conditioning tail that ends at three.

Pr⁡(2.5≤X≤3∣X≥2.5)=S(2.5)−S(3)S(2.5)\Pr(2.5\le X\le3\mid X\ge2.5)=\frac{S(2.5)-S(3)}{S(2.5)}

Compute

Compute

Substitute the two survival values and reduce the ratio.

S(2.5)=46.25=1625,S(3)=49S(2.5)=\frac4{6.25}=\frac{16}{25},\qquad S(3)=\frac49
16/25−4/916/25=1−2536=1136=0.3055555556\frac{16/25-4/9}{16/25}=1-\frac{25}{36}=\frac{11}{36}=0.3055555556

Answer

Answer

The conditional probability rounds to 0.31.

0.31(B)\boxed{0.31\quad\text{(B)}}