Question

For a life insurance company, it is important to construct life tables (consisting of the probability that a person will survive in the next year, conditional on a person that has been survived up to current). A life insurance company uses life tables to assist calculation of life insurance premium. Assume that the lifetime of randomly selected person is approximately normally distributed with a mean 68 years and a standard deviation 4 year.

a) A whole life insurance implies that insurance company needs to pay out the death benefit if the insured die, no matter when the death event is happened.  For a group of 10 independent newborn insureds who are currently all under whole life insurances, what is the probability that at least 9 death benefits will be paid after 75 years.

b) What is the probability that a randomly selected person will survive beyond 75 years?

c)A whole life insurance implies that insurance company needs to pay out the death benefit if the insured die, no matter when the death event is happened.  For a group of 10 independent newborn insureds who are currently all under whole life insurances, What is the probability that a randomly selected person will die before 60 years?

Answer 1

let X denotes the lifetime (in years) of a randomly selected person.

then X~N(68,42)

a) at first we need to find the probability that a newborn baby will live more than 75 years.

that is P[X>75]=1-P[X<75]=1-P[(X-68)/4<(75-68)/4]=1-P[Z<1.75] where Z~N(0,1)

=1-0.9599408=0.0400592~0.04

let Y be the number of newboarn babies who die after 75 years of age out the 10 independent babies.

then Y~Bin(10,P[X>75]) or, Y~Bin(10,0.04)

hence P[ at least 9 death benefits will be paid after 75 years ] =P[Y>=9]

=P[Y=9]+P[Y=10]=10C90.049(1-0.04)+10C100.0410=2.527068*10-12 [answer]

b) probability that a randomly selected person will survive beyond 75 years is

P[X>75]=1-P[X<75]=1-P[(X-68)/4<(75-68)/4]=1-P[Z<1.75] where Z~N(0,1)

=1-0.9599408=0.0400592~0.04

c) probability that a randomly selected person will die before 60 years

is P[X<60]=P(X-68)/4<(60-68)/4]=P[Z<-2]=0.02275013 [answer] Z~N(0,1)