(8a). A process makes parts with a critical dimension which follows a Normal dis

Question

(8a). A process makes parts with a critical dimension which follows a Normal distribution with mean 20 meters and standard deviation 0.05 meters. If specification limits for the part is 20 plus or minus 0.12 meters, what percentage of the products would be non-defective? (Hint: if X is Normal with mean 20 and standard deviation 0.05, what is the probability that X is between 19.88 and 20.12?) (b). Consider a process that produces parts with a mean weight of 9 grams, and a standard deviation of 0.0134 grams. If samples of size ten are taken, what would be the mean of the sample means? IF

Explanation / Answer

Q1.
Specification: 20 m ± 0.12 m
Lower specification level (LSL) = 20 – 0.12 = 19.88 m
Upper specification level (USL) = 20 + 0.12 = 20.12 m

Process mean = µ = 20 m
Standard deviation of the process = ? = 0.05 m
Probability first time capable within tolerances = P(X <= USL) – P(X >= LSL)

Percent between 19.88 m to 20.12 m:
Probability of parts between 19.88 m to 20.12 m = P(X <= 20.12) – P(X >= 19.88)
P(X <= 20.12) = P(z <= [(20.12 – µ)/?]) = P(z <= [(20.12 – 20)/0.05]) = P(z <= 2.4) = 0.9918
(Using Excel function - =NORMSDIST(2.4) = 0.9918)
P(X >= 19.88) = P(z >= [(19.88 – µ)/?]) = P(z >= [(19.88 – 20)/0.05]) = P(z <= -2.4) = 1 – P(z >= 2.4) = 1 – 0.9918 = 0.0082

Probability of parts between 99 mm to 101 mm = 0.9918 – 0.0082 = 0.9836
Percent of parts to be nondefective (between 19.88 mm to 20.12 mm) = 98.36%

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