Motorola used the normal distribution to determine the probability of defects an

Question

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 12 ounces.

A. The process standard deviation is 0.2, and the process control is set at plus or minus 0.5 standard deviation. Units with weights less than 11.9 or greater than 12.1 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

B. Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 11.9 or greater than 12.1 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)?

In a production run of 1000 parts, how many defects would be found (to 0 decimals)?

Explanation / Answer

A) = 12
= 0.2
standardize x to z = (x - ) /
P( 11.9 < x < 12.1) = P[( 11.9 - 12) / 0.2 < Z < (12.1-12) / 0.2]
P( -0.5 < Z < 0.5) = 0.3829
(From Normal probability table)

The area we want is Rejection region ( >12.1 or <11.9)

Probability of a defect = 1-0.3829= 0.6171
In a production run of 1000 parts, 1000(0.6171) =617.1= 617 defects would be found.

B)

= 12
= 0.05
standardize x to z = (x - ) /
P( 11.9 < x < 12.1) = P[( 11.9 - 12) / 0.05 < Z < (12.1-12) / 0.05]
P( -2 < Z < 2) = 0.9545
(From Normal probability table)

The area we want is Rejection region ( >12.1 or <11.9)

Probability of a defect = 1-0.9545= 0.0455
In a production run of 1000 parts, 1000(0.0455) =45.5= 46 defects would be found.

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