Homework SourseThe Precision Die Company quality control manager shuts down
The Precision Die Company quality control manager shuts down an automatic lathe for corrective maintenance whenever a sample of the parts it produces has an average diameter greater than 2.01 inches or smaller than 1.99 inches. The lathe is designed to produce parts with a mean distance of 2.00 inches, and the sample averages have a standard deviation of .005 inches. Assume the normal distribution applies.
(a) What is the probability that the quality control manager will stop the process when the lathe is operating as designed, with m=2.00 inches?
(b) If the lathe begins to produce parts that on the average are too wide, with m=2.01 inches and
s = .01 inches, what is the probability that the lathe will continue to operate?
Solution
A)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 1.99
x2 = upper bound = 2.01
u = mean = 2
s = standard deviation = 0.005
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2
z2 = upper z score = (x2 - u) / s = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.022750132
P(z < z2) = 0.977249868
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.954499736
Thus, those outside this interval is the complement = 0.045500264 [ANSWER]
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B)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 1.99
x2 = upper bound = 2.01
u = mean = 2.01
s = standard deviation = 0.005
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -4
z2 = upper z score = (x2 - u) / s = 0
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 3.16712E-05
P(z < z2) = 0.5
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5000 [ANSWER]
