Question

The probability function for the number of insurance policies John will sell to a customer is given by f(X)=.5−(X/6) for X=0,1, or 2

a. Is this a valid probability function? Explain your answer
b. What is the probability that John will sell exactly 2 policies to a customer?
c. What is the probability that John will sell at least 2 policies to a customer?
d. What is the expected number of policies John will sell?
e. What is the variance of the number of policies John will sell?

Answer 1

a. X represents the number of insurance policies John will sell to a customer. b. The probability of selling exactly 2 policies is 0.167. c. The probability of selling at least 2 policies is 0.667. d. The expected number of policies John will sell is 0.667. e. The variance of the number of policies John will sell is 0.437.

Step-by-step explanation:

a. The random variable X represents the number of insurance policies that John will sell to a customer.

b. To find the probability that John will sell exactly 2 policies to a customer, we substitute X=2 into the probability function: f(2) = 0.5 - (2/6) = 0.5 - 0.333 = 0.167.

c. To find the probability that John will sell at least 2 policies to a customer, we sum the probabilities of selling 2 policies and selling 0 policies: f(2) + f(0) = 0.167 + 0.5 = 0.667.

d. To find the expected number of policies John will sell, we multiply each possible outcome by its probability and sum the results: Expected value = 0 * f(0) + 1 * f(1) + 2 * f(2) = 0 * 0.5 + 1 * 0.333 + 2 * 0.167 = 0.667.

e. To find the variance of the number of policies John will sell, we subtract the expected value from each possible outcome, square the result, multiply by the probability, and sum the results: Variance = (0-0.667)^2 * f(0) + (1-0.667)^2 * f(1) + (2-0.667)^2 * f(2) = 0.437.