Question

A quality characteristic X follows a Normal distribution with mean 100 and standard deviation 2 when the production process is in control.  

Part (a): Suppose that the lower and upper specification limits are 94 and 106, respectively.  What is the percentage of produced product items that are non-conforming, i.e., do not conform to the specification limits, when the process is indeed in control?

Part (b): The management has decided to monitor the process with and R control charts with a sample size of 4.  What should be the lower limit, centerline and the upper limit for each of the two control charts?

Answer 1

a)As we know that the formula for finding prob from normal distribution is first to find z value and find the respective probability from Z table

Z= (Xbar-mu)/(sd/sqrt(n))

The lower and upper specification limits are 94 and 106, respectively

Hence using Xbar in the above formula with n=1

Z= (94-100)/(2) =-3

Z=(106-100)/2 = 3

Looking for the area under the curve from standard Normal distribution with Z=(-3,3) we get Area under the curve as 0.997

Hence we can say that there is 99.7% Probability that of produced product items that are conforming

Hence we can say that there is very rare probability of 0.3% of produced product items that are non-conforming

Ans=> 0.3%

b) To control the same

Z= (Xbar-mu)/(sd/sqrt(n))

Where we have Z=-3 and +3 for lower and upper limit respetively

We have n=4

Hence for lower limit

-3= (Xbar-100)/(2/sqrt(4))

Xbar=100-3=97 (Lower limit)

Same way

3= (Xbar-100)/(2/sqrt(4))

Xbar=100+3=103 (Upper limit)

Central Line = (Lower+Upper)/2 = (97+103)/2 =100

Hence Upper limit=103

Central line = 100

Lower Limit=97

For R charts

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