Question

The probability distribution for damage claims paid by the Newton Automobile Insurance Company on collision insurance follows. Payment ($) Probability 0 0.85 500 0.04 1,000 0.04 3,000 0.03 5,000 0.02 8,000 0.01 10,000 0.01
(a) Use the expected collision payment to determine the collision insurance premium that would enable the company to break even. If required, round your answers to two decimal places. If your answer is zero, enter "0". x f(x) xf(x) 0 500 1000 3000 5000 8000 10000 Total
(b) The insurance company charges an annual rate of $520 for the collision coverage. What is the expected value of the collision policy for a policyholder? (Hint: It is the expected payments from the company minus the cost of coverage.) If required, enter negative value as negative number. $ Why does the policyholder purchase a collision policy with this expected value? The input in the box below will not be graded, but may be reviewed and considered by your instructor.

Answer 1

Answer:

a) $430

b) -$90

Step-by-step explanation:

Given a random variable X with possible values

$\large x_1,x_2,...x_n$

with respective probabilities of occurrence  

$\large P(x_1),P(x_2),...P(x_n)$

then the expected value E(X) of X is

$\large x_1*P(x_1)+x_2*P(x_2)+...+x_n*P(x_n)$

a)

The expected collision payment would then be

0*0.85 + 500*0.04 + 1000*0.04 + 3000*0.03 + 5000*0.02 + 8000*0.01 + 10000*0.01 = $430

So the insurance premium that would enable the company to break even is $430

b)

The expected value of the collision policy for a policyholder is the expected payments from the company minus the cost of coverage:

$430 - $520 = -$90

Why does the policyholder purchase a collision policy with this expected value?

Because the policyholder does not know what the probability of having an accident is in her particular case.

Besides, it is better to have the policy and not need it than to need it and not have it.