Question

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violations for which the individual was cited during the last 3 years. The pmf of Y is y0123 p(y) .60 .25 .10 .05 a. Compute E(Y). b. Suppose an individual with Y violations incurs a sur- charge of $100Y2. Calculate the expected amount of the surcharge

Answer 1

Answer:

a) $E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

b) $E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

Step-by-step explanation:

We assume the following distribution given:

Y       0       1        2        3

P(Y) 0.60 0.25  0.10  0.05

Part a

We can find the expected value with this formula:

$E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

Part b

If we want to find the expected value of $100 Y^2$ we need to find the expected value of Y^2 and we have:

$ E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)$

And replacing we got:

$E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$