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.
