Independent solution

How to solve this Conditional Normal Probability question

Setup

Setup

Standardize both the upper event boundary and the lower conditioning boundary using mean 100 and standard deviation 20.

Z=X−10020,X=60↦−2,X=0↦−5Z=\frac{X-100}{20},\quad X=60\mapsto-2,\quad X=0\mapsto-5

Model

Model

The numerator is the standard-normal probability between the two standardized boundaries. The denominator is the probability above the lower boundary.

Pr⁡(X≤60∣X>0)=Pr⁡(−5<Z≤−2)Pr⁡(Z>−5)\Pr(X\le60\mid X>0)=\frac{\Pr(-5<Z\le-2)}{\Pr(Z>-5)}

Compute

Compute

Express the negative-tail probabilities by symmetry to obtain the ratio shown in choice E.

Pr⁡=Φ(−2)−Φ(−5)Φ(5)=Φ(5)−Φ(2)Φ(5)\Pr=\frac{\Phi(-2)-\Phi(-5)}{\Phi(5)}=\frac{\Phi(5)-\Phi(2)}{\Phi(5)}

Answer

Answer

The standardized conditional expression is the one in choice E.

Φ(5)−Φ(2)Φ(5)(E)\boxed{\frac{\Phi(5)-\Phi(2)}{\Phi(5)}\quad\text{(E)}}