Question

Age group // Number in age group // Probability of claim per life // Mean of...

Age group // Number in age group // Probability of claim per life // Mean of the exponential distribution of claim amounts

18-35 // 400 // 0.03 // 5

36-50 // 300 // 0.07 // 3

51-65 // 200 // 0.10 // 2

A group life insurance contract covering independent lives is rated in three age groupings as given in the table above. The insurer prices the contract so that the probability that claims will exceed the premium is 0.05. Using the normal approximation, determine the premium that the insurer will charge.

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Answer #1

We will calculate mean and variance :

Means=E(X) =400*(0.03)*(5)+300*(0.07)*(3)+400*(0.1)*(2)

=163

Variance = Var(X)

=400[(0.03)(5)2 + (0.03)(0.97)52 ] +300[(0.07)(3)2 + 0.07(0.93)(3)2 ] + 200[0.10(2)2 + 0.10(0.9)(2)2 ] =1107.77.

We have

Pr(S > P) =0.05

Pr(S <= P) =0.95

Pr ( Z ≤ P − 163 / √ 1107.77 )=0.95

P − 163 / √ 1107.77 =1.645

P=1.645*√ 1107.77+163

P=217.75

Hope this will be helpful. Thanks and God Bless You:)

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