A life insurance salesman operates on the premise that the probability that a man reaching his sixtieth birthday will not live to his sixty-first birthday is 0.05. On visiting a holiday resort for seniors, he sells 12 policies to men approaching their sixtieth birthdays. Each policy comes into effect on the birthday of the insured, and pays a fixed sum on death. All 12 policies can be assumed to be mutually independent. Provide answers to the following to 3 decimal places.
a) What is the expected number of policies that will pay out before the insured parties have reached age 61?
b) What is the variance of the number of policies that will pay out before the insured parties have reached age 61?
c) What is the probability that at least two policies will pay out before the insured parties have reached age 61?
a)
To find the number of policies:
µX = np
Mean = 12
Probability is 0.05
Looking for N
So 12 multiplied by 0.05
equals 0.6
b) To find variance I did
= (0.6 - 12)^2 * 0.05
= 6.498
c)
P(X greater than or equal to 2) = 1 - (x < 2)
I use the following binomial distribution formula
P(X = x) = nCx * P^x * (1 - P)^(n - x)
= 1 -
12 C 0 * 0.05 ^ 0 * (1 - 0.05)^(12 - 0)+
12 C 1 * 0.05 ^ 1 * (1 - 0.05)^(12 - 1)
= 0.9804 to the 4th decimal
What are the correct answers and solutions? It seems like I am way off.
Answer
The variance formula is $np(1-p)=12*(0.05)*(.95)=.57$ by my calculation.
The second answer is $1-(.95)^{12}-12*(.05)*(.95)^{11}=.1183...$ Your addition sign is wrong.
Wrong first answer below
(On the second part, you forgot it's 1-(the sum) so you should get 1-0.9804 = 0.0196 which gives a very intuitive answer of about 2%.)
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