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Sums of Independent Random Variables

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Vicki owns two separtment stores. Delinquent charge accounts at store #1 show a normal distribution, with mean $90 and std. deviation $30, whereas at store #2, they show a normal distribution with mean $100 and std. deviation $50. If 10 delinquent accounts are selected randomly at store #1 and 15 at store #2, what is the probability that the average of the accounts selected at store #1 exceeds the average of those accounts at store #2?


    2. Relevant equations
    Let X1=N(90,30); where n=10, XBar1=N(90,30/10)=N(90/3). Let X2=N(100,50); where n=15, XBar2=N(100,50/15)=N(100,3.33)


    3. The attempt at a solution

    We need P(XBar1-XBar2>=1)=P((XBar1-XBar2+10)/Sqrt(-.33))>=(1+10)/Sqrt(-.33)
    =11/(Sqrt(-.333))???? This doesn't make any sense to me!!

    Any assistance would be very much appreciated!
     
  2. jcsd
  3. Sep 3, 2009 #2

    lanedance

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    hmmm... haven't done much probability for a while

    but based on your sampling I would try & come up with a probabilty distribution for the average of the measurement...

    so to start, say we pick X n times, then let Y be the average of the n samples, so the random variable Y is given by:
    Y = (X +...+ X)/n (sum of n times)

    What is the expected value & variance of Y, assuming each picked X is independent? Shouldn't be too hard to calculate... (though independence is an important assumtion....)

    If you have done this for both cases, then consider independent the random variables Y1 & Y2, how do you find the probabilty of Z = Y1-Y2 >0.

    Maybe something similar to first process, to find expectation & variance, then sums of guassian distributions are also gaussian...
     
  4. Sep 4, 2009 #3
    I see what you're saying, that the sample means should reflect the multiple of the sample size. However, don't I still end up with a denominator equal to the square root of minus 1/3?
     
  5. Sep 4, 2009 #4
    After several more attempts (and re-reading your analysis), I think I finally solved the problem. Thanks, Lanedance!
     
  6. Sep 5, 2009 #5

    lanedance

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    good one, glad it worked
     
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