A fiber-spinning process currently produces a fiber whose strength is normally distributed with a mean of 75 N/
. The minimum acceptable strength is 65 N/.
a. Ten percent of the fiber produced by the current method fails to meet the minimum specification. What is the standard deviation of fiber strengths in the current process?
b. If the mean remains at 75 N/
, what must the standard deviation be so that only 1% of the fiber will fail to meet the specification?
c. If the standard deviation is 5 N/, to what value must the mean be set so that only 1% of the fiber will fail to meet the specification?
(a) The proportion of strengths that are less than 65 is 0.10. Therefore 65 is the 10th percentile of strength. The z-score of the 10th percentile is approximately z= ?1.28. Let ? be the required standard deviation. The value of ? must be chosen so that the z-score for 65 is ?1.28. Therefore ?1.28 = (65 ? 75)/?. Solving for ? yield s?= 7.8125 N/
.
(b) Let ? be the required standard deviation. The value of ? must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 = (65?75)/?. Solving for ?yields ?= 4.292 N/.
(c) Let ? be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 = (65 ? ?)/5. Solving for ? yields ?= 76.65 N/.
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