Well, observe that average box weight (W) is approximately normal with mean 8.25 and sd=3/3.92=0.765. So, P[W<8]=P[Z<(8-8.25)/0.765]=P[Z=-3.267]=0.0006.
This is a problem from the book Statistics for Experimenters by Box, Hunter, Box book, page 151.
For each sample of 16 boxes of cereal an average box weight is recorded. Past records indicate that on the average 1 out of 40 of these average box weights exceeds 8.4 ounces and 1 out of 40 is less than 8.1 ounces. What is the prob that an individual box of cereal selected at random will weight less than 8 ounces?
Does my solution make sense?
Pr(x>8.4ozs) = 1/40=0.025
Pr(x<8.1ozs) = 1/40=0.025
z value associated with pr 0.025 is 1.96
therefore, Pr(x<8.0) for 1 of 16 boxes at random is df = 15, ...and now i am getting confused. Please help. Thanks.
Well, observe that average box weight (W) is approximately normal with mean 8.25 and sd=3/3.92=0.765. So, P[W<8]=P[Z<(8-8.25)/0.765]=P[Z=-3.267]=0.0006.
Last edited by Devanil; 04-09-2013 at 05:30 PM.
Uooops. observe that average box weight (W) is approximately normal with mean 8.25 and sd=0.3/3.92=0.0765. So, P[W<8]=P[Z<(8-8.25)/0.0765]=P[Z<-3.267]=0.0006
thanks so much!
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