A dam is proposed for a river that is currently used for salmon breeding. You have modeled:

• For each hour the dam is opened the number of salmon that will pass through and reach the breeding grounds has a distribution with mean 100 and variance 900.
• The number of eggs released by each salmon has a distribution with mean 5 and variance 5.
• The number of salmon going through the dam each hour it is open and the numbers of eggs released by the salmon are independent.

Calculate the least number of whole hours the dam should be left open so the probability that 10,000 eggs will be released is greater than 95% using the normal approximation for the aggregate number of eggs released.

• 25
• 23
• 26
• 29
• 32

Key: B

N = number of salmon in t hours

X = eggs from one salmon

S = total eggs.

[math]\operatorname{E}(N) = 100t [/math]

[math]\operatorname{Var}(N) = 900t [/math]

[math]\operatorname{E}( S ) = \operatorname{E}( N ) \operatorname{E}(X) = (100 t)(5) = 500 t [/math]

[math]0.95 \lt \operatorname{Pr}( S \gt 10, 000) = \operatorname{Pr}( Z \gt \frac{10,000 - 500t}{\sqrt{23,000t}} ) \Rightarrow \frac{10,000 - 500t}{\sqrt{23,000t}}[/math]

[math] 10, 000 − 500t −1.645(151.66) \sqrt{t} = −249.48 \sqrt{t}[/math]

[math]500t - 249.48 \sqrt{t} - 10,000 = 0[/math]

[math]\sqrt{t} = \frac{249.48 \pm \sqrt{(-249.48)^2 - 4(500)(-10,000)}}{2(500)} = 4.73[/math]

[math] t = 22.6 [/math]

Round up to 23

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