Claim amounts at an insurance company are independent of one another. In year one, claim amounts are modeled by a normal random variable X with mean 100 and standard deviation 25. In year two, claim amounts are modeled by the random variable [math]Y = 1.04X + 5[/math].

Calculate the probability that a random sample of 25 claim amounts in year two average between 100 and 110.

• 0.48
• 0.53
• 0.54
• 0.67
• 0.68

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

Solution: B

Y is a normal random variable with mean 1.04(100) + 5 = 109 and standard deviation 1.04(25) = 26. The average of 25 observations has mean 109 and standard deviation 26/5 = 5.2. The requested probability is

[[math]] \begin{align*} \operatorname{P}(100 \lt \textrm{sample mean} \lt 110 ) &= \operatorname{P} \left( \frac{100-109}{5.2} = -1.73 \lt Z \lt \frac{110-109}{5.2} = 0.19 \right) \\ &= 0.5753 − (1 − 0.9582) = 0.5335. \end{align*} [[/math]]

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