SOLUTION: In a bottle-filling process, the amount of drink injected into 16 oz bottles is normally distributed with a mean of 16 oz and a standard deviation of .32 oz. Bottles containing les

Question 993926: In a bottle-filling process, the amount of drink injected into 16 oz bottles is normally distributed with a mean of 16 oz and a standard deviation of .32 oz. Bottles containing less than 15.86 oz do not meet the bottler�s quality standard. What percentage of filled bottles do not meet the standard? (Round the z value to 2 decimal places. Round your answer to 2 decimal places.)
P(x < 15.86) %
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the z factor is equal to (x - m) / s

x is the raw score.
m is the mean.
s is the standard deviation.

the critical z factor is the z-score of the cutoff value.

if the bottle contains less than 15.86 ounces of drink, then it is rejected.

that makes the cutoff z factor equal to (15.86 - 16) / .32 = -.44

look up a z-score of -.44 in the z-score table and you will get .33

that means that 33% of the z-scores in the normal distribution table are below a z-score of .44.

that means that 33% of the bottles would be rejected if the amount of drink in the bottles was normally distributed with a mean of 16 ounces and the cutoff was any bottle that had less than 15.86 ounces of drink in it.

you can see the results in the following z-score calculator outputs.

the first output is based on the raw score.

the second output is based on the z-score.

the different in percentage is due to rounding of the z-score.

$$$

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