# Thread: Probability Random Variables Double Check Please

1. ## Probability Random Variables Double Check Please

A small business just leased a new computer and color laser printer for three years. The service contract for the computer offers unlimited repairs for a fee of $100 a year plus a$25 service charge for each repair needed. The company’s research suggested that during a given year 86% of these computers needed no repairs, 9% needed to be repaired once, 4% twice, 1% three times, and none required more than three repairs.
a)Find the expected number of repairs this kind of computer is expected to need each year. Show your work.

X | P(x) | xP(x) |
0 | .86 | 0 |
1 | .09 | .09 |
2 | .04 | .08 |
3 | .01 | .03 |

0 + .09 + .08 + .03 = .200

b)Find the standard deviation of the number of repairs each year.
.0344+.0576+.1296+.0784 = .54772

c)What are the mean and standard deviation of the company’s annual expense for the service contract? HINT: Annual expense Y = 100 + 25 ´ X, where X = number of repairs.

Mean= 105
Std Dev = 13.675

d)The service contract for the printer estimates a mean annual cost of $120 with standard deviation of$30. What is the expected value and standard deviation of the total cost for the service contracts on computer and printer? On what assumption does your calculation rest?
My Thoughts:

Would I just add 120 to the annual expenses for a computer in the previous problem and then calculate the expected value and standard deviation? If so, what is the point of the std deviation of \$30 given to me in the problem?

2. Originally Posted by rice4lifelegit
b)Find the standard deviation of the number of repairs each year.