Sums of Independent Random Variables

  • Thread starter WHB3
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  • #1
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Homework Statement



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?


Homework 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)


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!
 

Answers and Replies

  • #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...
 
  • #3
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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?
 
  • #4
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After several more attempts (and re-reading your analysis), I think I finally solved the problem. Thanks, Lanedance!
 
  • #5
lanedance
Homework Helper
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good one, glad it worked
 

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