The fill volume of cans filled by a certain machine is normally distributed with mean 12.05 oz and standard deviation 0.03 oz.
a. What proportion of cans contain less than 12 oz?
b. The process mean can be adjusted through calibration. To what value should the mean be set so that 99% of the cans will contain 12 oz or more?
c. If the process mean remains at 12.05 oz, what must the standard deviation be so that 99% of the cans will contain 12 oz or more?
(a) The z-score of 12 is (12 ? 12.05)/0.03 = ?1.67. The area to the left of z= ?1.67 is 0.0475. The proportion is 0.0475.
(b) Let ? be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 12. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 = (12??)/0.03. Solving for ? yields ?= 12.07 ounces.
(c) Let ? be the required standard deviation. The value of ? must be chosen so that the 1st percentile of the distribution is 12. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 =(12 ? 12.05)??. Solving for ?yields ?= 0.0215 ounces.
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