Question 1026245: A store receives a consignment of bolts whose diameters have a normal distribution with a mean diameter of 12 inches and a standard deviation of 0.2 inches. The consignment will be considered substandard and returned if the mean diameter of 64 bolts is less than 11.97 inches or greater than 12.04 inches. Find the probability that the consignment will not be returned.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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