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 6 ounces. a. The process standard deviation is .20, and the process control is set at plus or minus .5 standard deviation . Units with weights less than 5.9 or greater than 6.1 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

Probability of defect=0.617

Explanation:

Given

Mean ∑= 6

Standard deviation σ = 0.20

The standardized scores can be assumed as value x which has to reduced by mean value and divided by standard deviation.

z= (x - ∑)/ σ

=(5.9-6)/0.20

=-0.5

z= (x - ∑)/ σ

=(6.1-6)/0.20

=0.5

Calculation of probability of defect

P(x≤5.9 or x ≥6.1)

=P(x≤-0.5 or x ≥0.5)

=2p(x≤-0.5)

=2(0.3085)

Probability of defect=0.617

To find the number of units of defect the manfucturing units have to be multiplied with the probability value.

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Explanation: