Question

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 .002 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 =2.00 inches? (b) If the lathe begins to produce parts that on the average are too wide, with =2.01 inches and  = .01 inches, what is the probability that the lathe will continue to operate?

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 2
standard Deviation ( sd )= 0.002
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To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.99) = (1.99-2)/0.002
= -0.01/0.002 = -5
= P ( Z <-5) From Standard Normal Table
= 0
P(X < 2.01) = (2.01-2)/0.002
= 0.01/0.002 = 5
= P ( Z <5) From Standard Normal Table
= 1
P(1.99 < X < 2.01) = 1-0 = 1
So the probability that the quality control manager will stop the process is
P(X< 1.98) + P(X > 2.02) = 1 - P(1.99 < X < 2.01) = 1 - 1 = 0
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(b)
The z-score for X = 2.02
The z-score for X = 1.98
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1.99) = (1.99-2.01)/0.01
= -0.02/0.01 = -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(X < 2.01) = (2.01-2.01)/0.01
= 0/0.01 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(1.99 < X < 2.01) = 0.5-0.0228 = 0.4772
So the probability that the quality control manager will stop the process is
P(X< 1.98) + P(X > 2.01) = 1 - P(1.99 < X < 2.01) = 1 - ( 0.4772 + 0.228) = 1 - 0.5 = 0.5