For a certain insurance company, 10% of its policies are Type A, 50% are Type B, and 40% are Type C. The annual number of claims for an individual Type A, Type B, and Type C policy follow Poisson distributions with respective means 1, 2, and 10.

Let [math]X[/math] represent the annual number of claims of a randomly selected policy. Calculate the variance of [math]X[/math].

• 5.10
• 16.09
• 21.19
• 42.10
• 47.20

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

Solution: C

For the Poisson distribution the variance is equal to the mean and hence the second moment is the mean plus the square of the mean. Then,

[[math]] \operatorname{E}[ X ] =0.1(1) + 0.5(2) + 0.4(10) =5.1 [[/math]]

[[math]] \operatorname{E}[ X^2 ] = 0.1(1 + 12 ) + 0.5(2 + 22 ) + 0.4(10 + 102 )= 47.2 [[/math]]

[[math]] \operatorname{Var}( X ) = 47.2 − 5.12 = 21.19. [[/math]]

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