An insurance policy covers losses incurred by a policyholder, subject to a deductible of 10,000. Incurred losses follow a normal distribution with mean 12,000 and standard deviation [math]c.[/math] The probability that a loss is less than [math]k[/math] is 0.9582, where [math]k[/math] is a constant. Given that the loss exceeds the deductible, there is a probability of 0.9500 that it is less than [math]k[/math].

Calculate [math]c[/math].

• 2045
• 2267
• 2393
• 2505
• 2840

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

Solution: A

We have

[[math]] \begin{align*} 0.95 = \operatorname{P}(X \lt k | X \gt 10000) = \frac{\operatorname{P}(X \lt k) - \operatorname{P}(X \leq 10000)}{1-\operatorname{P}(X \leq 10000)} \\ 0.95[1 − \operatorname{P}( X ≤ 10, 000)]= 0.9582 − \operatorname{P}( X ≤ 10, 000) \\ \operatorname{P}(X \leq 10000 ) = \frac{0.9582-0.95}{1-0.95} = 0.164 \\ 0.164 = \Phi(\frac{10000-12000}{c}). \end{align*} [[/math]]

The z-value that corresponds to 0.164 is between -0.98 and -0.97. Interpolating leads to z = –0.978. Then,

[[math]] 0.164 = \Phi \left( \frac{10000 - 12000}{c}\right) \Rightarrow -0.978 = \frac{-2000}{c} \Rightarrow c = 2045. [[/math]]

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