posted by UG on .
The amount of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified at
possessing a normal distribution with a mean of 16.5 ounces and a standard deviation of 0.1 ounce. Suppose 400
bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean
weight of these 400 bags exceeded 16.6 ounces
Z = (mean1 - mean2)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
Actually, I believe it's approximately zero.
Source: I have a MS in Mathematics
Pardon me... approximately zero if the standard deviation is 0.2 ozs.