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?
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!