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A loss X on a car has a mixed distribution with p = 0.95

Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman ISBN: 9780321923271 41

Solution for problem 13E Chapter 3.4

Probability and Statistical Inference | 9th Edition

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Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman

Probability and Statistical Inference | 9th Edition

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Problem 13E

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.

Step-by-Step Solution:
Step 1 of 3

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.

Step 2 of 3

Chapter 3.4, Problem 13E is Solved
Step 3 of 3

Textbook: Probability and Statistical Inference
Edition: 9
Author: Robert V. Hogg, Elliot Tanis, Dale Zimmerman
ISBN: 9780321923271

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