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.

My answer: .200

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.

My answer: .54772

.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.*

My answer:

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?