A loss X on a car has a mixed distribution with p = 0.95 on zero and p = 0.05 on an exponential distribution with a mean of $5000. If the loss X on a car is greater than the deductible of $500, the difference X−500 is paid to the owner of the car. Considering zero (if X ≤ 500) as a possible payment, determine the mean and the standard deviation of the payment.

Solution 13E

Step1 of 3:

We have a random variable X which has mixed distribution with p = 0.95 on zero and p = 0.05

on an exponential distribution with a mean of $5000.

If the loss X on a car is greater than the deductible of $500, the difference X−500 is paid to the owner of the car.

We need to determine the mean and the standard deviation of the payment by Considering zero (if X ≤ 500) as a possible payment.

Step2 of 3:

random variable X which follows exponential distribution with parameter

The probability density function of exponential distribution is given by

Mean of the Exponential distribution is given by

=

=

=

=

=

=

=

= 226.2093.

Hence,

Step3 of 3:

Variance of the Exponential distribution is given by

Var(X) =

We have

We need to find

Consider,

=

=

=

= 10000

= 10000226.2093

= 2262093

Hence,

Now,

Var(X) =

= 2262093 - (226.2093)2

= 2262093 - 51170.64741

= 2210922.353

Hence,Var(X) = 2210922.353.

Standard deviation of exponential distribution is given by

=

= 1486.9170

Hence, = 1486.9170.

Conclusion:

Mean of the Exponential distribution is

Standard deviation of exponential distribution is = 1486.9170.