Question 1026245
the mean being compared is 12 inches.
the standard deviation is .2 inches.
the limits are between 11.97 and 12.04 inches.


you want to calculate the z-score of the limits.


the formula for z-score is:


z = (x-m)/s


x is the limit
m is the mean
s is the standard deviation.


since the consignment of 64 is equivalent to a sample of size 64, you need to calculate the standard deviation of the distribution of sample means.


this is called the standard error.


the formula for standard error is sd / sqrt(n).


sd is the standard deviation of the population, if known, or the stnadard deviation of the sample, if the standard deviation of the population is not known.


n is the sample size.


in this case, the formula becomes standard error = .2 / sqrt(64) = .2 / 8 = .025


now you need to calculate the z-score of the limits.


the limits are 11.97 to 12.04


the formula for z-score is z = (x-m) / s


since you are dealing with a sample mean, then s represents the standard error.


x represents the mean limit.
m represents the mean
s represents the standard error of sample means.


you get:


z1 = (11.97-12)/.025 and z2 =(12.04-12)/.025.


this will get you z1 = -1.2 and z2 = 1.6


your acceptable range is between a z-score of -1.2 and a z-score of 1.6.


if you look in the z-score table, you will find that there is a .1151 probability that a z-score will be less than -1.2, assuming the distribution is normal.


you will also find that there is a .9452 probability that a z-score will be less than 1.6, assuming the distribution is normal.


the probability that the z-score will be between these limits is therefore equal to .9452 - .1151 = .8301.


the probability that the z-score will not be between these limits is therefore equal to 1 - .8301 = .1699.


the probability that the consignment will not be returned means the z-score is between these limits.


that probability is .8301.