Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Random variables

  1. Feb 20, 2005 #1
    Hi I need some help. I don't think I did any of this right.

    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.

    1. Find the expected number of repairs this kind of computer is expected to need each year.

    100(.86) + 125(.09) + 150(.04) + 175(.01) = 105

    2. Find the standard deviation of the number of repairs each year.

    .86(100-105)^2 + .09(125-105)^2 + .04(150-105)^2 + .01(175-105)^2 = 187.5
    sqrt(187.5) = 13.69

    3. What are the mean and standard deviation of the company's annual expense for the service contract?
    I have no clue how to do this one.

    4. How many times should the company expect to have to get this computer repaired over the three-year term of lease?
    None?

    5. What is the standard deviation of the number of repairs that may be required during the three-year lease period?
    105*3 = 315

    Thanks
     
  2. jcsd
  3. Feb 21, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    An answer like 105 should tell you something is wrong.
    Use the right information: A computer needs no repairs with a probability of 86%. 1 repair with a prob. of 9%, 2 repairs with 4% and three with 1%.
    Find the expectation [itex]P(X)[/itex] from this.

    1. Find the expectation of the number of repairs squared: [itex]P(X^2)[/itex].
    Then use: [itex]\sigma_X^2=P(X^2)-P(X)^2[/itex]
     
  4. Feb 21, 2005 #3

    xanthym

    User Avatar
    Science Advisor

    HINTS GIVEN ABOVE IN RED.
    ~~
     
    Last edited: Feb 21, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook