An insurance company issues 1250 vision care insurance policies. The number of claims filed by a policyholder under a vision care insurance policy during one year is a Poisson random variable with mean 2. Assume the numbers of claims filed by different policyholders are mutually independent.

Calculate the approximate probability that there is a total of between 2450 and 2600 claims during a one-year period.

• 0.68
• 0.82
• 0.87
• 0.95
• 1.00

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

Solution: B

A single policy has a mean and variance of 2 claims. For 1250 polices the mean and variance of the total are both 2500. The standard deviation is the square root, or 50. The approximate probability of being between 2450 and 2600 is the same as a standard normal random variable being between (2450 – 2500)/50 = –1 and (2600 – 2500)/50 = 2. From the tables, the probability is 0.9772 – (1 – 0.8413) = 0.8185.

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