In: Statistics and Probability
Calculate the Probability *Binomial Distribution*
An insurance company sells life insurance policies to 5 people with the same age (35) and the same health condition (Excellent). If the Insurance company was estimating at 66% the probability of a person of that age and health condition should live 30 more years. What is the probability that 30 years after buying the policy:
a. All 5 are alive
b. 4 out of 5 have died
c. No more than 3 survive
d. 2 or more survive
Calculate mean, variance and Standard Deviation
Let X be the random variable that denotes the number of people living 30 years after buying the policy.
An insurance company sells life insurance policies to 5 people with the same age and the same health condition. Therefore, n = 5
The probability of a person of that age and health condition living 30 more years is estimated at 66%. Therefore, p = 0.66
X Binomial (n = 5, p = 0.66)
The pmf of X is
P(X = x) = 5Cx * 0.66x * (1 - 0.66)5 - x ; x = 0, 1, 2, 3, 4, 5.
= 0 ; otherwise
a. Answer :
P(X = 5) = 5C5 * 0.665 * (1 - 0.66)5 - 5
= 0.665
= 0.1252
Therefore, the probability that 30 years after buying the policy, all 5 are alive is 0.1252
b. Answer :
P(X = 1) = 5C1 * 0.661 * (1 - 0.66)5 - 1
= 0.0441
Therefore, the probability that 30 years after the buying the policy, 4 out of 5 have died is 0.0441
c. Answer :
P(X 3) = 1 - P(X > 3)
= 1 - (P(X = 4) + P(X = 5))
= 1 - (5C4 * 0.664 * (1 - 0.66)5 - 4 + 5C5 * 0.665 * (1 - 0.66)5 - 5)
= 1 - (0.3226 + 0.1252)
= 0.5522
Therefore, the probability that 30 years after buying the policy, no more than 3 survive is 0.5522
d. Answer :
P(X 2) = 1 - P(X < 2)
= 1 - (P(X = 0) + P(X = 1))
= 1 - (5C0 * 0.660 * (1 - 0.66)5 - 0 + 5C1 * 0.661 * (1 - 0.66)5 - 1)
= 1 - (0.0045 + 0.0441)
= 0.9514
Therefore, the probability that 30 years after buying the policy, 2 or more survive is 0.9514
e. Answer :
E(X) = n * p
= 5 * 0.66
= 3.3
V(X) = n * p * (1 - p)
= 5 * 0.66 * (1 - 0.66)
= 1.122
SD(X) =
=
= 1.0593
Therefore, the mean is 3.3, variance is 1.122 and standard deviation is 1.0593