Exercises10 Problem 5

The distribution of values of the retirement package offered by a company to new employees is modeled by the probability density function

1

5

-

1

5

(x-5)

e

for

x>5

. Calculate the variance of the retirement package value for a new employee is greater than 15, given that it is at least 10.

Solution

In[]:=

retirementDist=ProbabilityDistribution[1/5*Exp[-1/5*(x-5)],{x,5,Infinity}]

Out[]=

ProbabilityDistribution

1

5

5-x.

5



,{x.,5,∞}

Then calculate the variance. Here, conditional variance is not an option of the Variance function. However, you can obtain variance from the Expectation, as

Var[X]=

μ

2

-

2

μ

1

=

E[

2

X

]-

2

E[X]

:

In[]:=

Expectation[x^2x>1.5,xretirementDist]-Expectation[xx>1.5,xretirementDist]^2

Out[]=

25.

I don’t see how the condition x>1.5 represents the problem statement of “greater that 15 given that it is at least 10.