An insurance company’s annual profit is normally distributed with mean 100 and variance 400. Let Z be normally distributed with mean 0 and variance 1 and let F be the cumulative distribution function of Z.

Determine the probability that the company’s profit in a year is at most 60, given that the profit in the year is positive.

• [math]1 – F(2)[/math]
• [math]F(2)/F(5)[/math]
• [math][1 – F(2)]/F(5)[/math]
• [math][F(0.25) – F(0.1)]/F(0.25)[/math]
• [math][F(5) – F(2)]/F(5)[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

The profit variable [math]X[/math] is normal with mean 100 and standard deviation 20. Then,

[[math]] \operatorname{P}( X ≤ 60 | X \gt 0 ) = \frac{\operatorname{P}(0 \lt X ≤ 60)}{\operatorname{P}( X \gt 0)} = \frac{\operatorname{P}(\frac{0-100}{20} \lt Z \leq \frac{60-100}{20})}{\operatorname{P}(Z \gt \frac{0-100}{20})} = \frac{F(-2)-F(-5)}{1-F(-5)} [[/math]]

For the normal distribution, [math]F(–x) = 1 – F(x)[/math] and so the answer can be rewritten as

[[math]] [1 – F(2) – 1 + F(5)]/[1 – 1 + F(5)] = [F(5) – F (2)]/F(5). [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.