Question

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Chapter 6 Reflection:

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 10 ounces.

1. The process standard deviation is 0.15 and the process control is set at plus or minus on standard deviation, so units with weights less than 9.85 oz or greater than 10.15 oz will be classified as defects. Find the probability of a defect and the expected number of defects for a 1000-unit production run.
2. Through process design improvements, suppose the process standard deviation can be reduced to 0.05. Assume the process control remains the same, so products with weights less than 9.85 and more than 10.15 ounces are classified as defects. Find the probability of a defect and the expected number of defects for a 1000-unit production run.

3. What is the advantage of reducing process variation (question 2)? In other words, explain (mathematically) why reducing the standard deviation to 0.05 but keeping the process controls at 9.85 and 10.15 ounces led to better production outcomes.

Answer 1

Given Information:

Initially, the mean value is 10 ounces and the standard deviation is 0.15 ounces.

Process control = (9.85 , 10.15) = (10 - 0.15 , 10 +0.15) = (Mean - 1*S.D. , Mean + 1* S.D.)

After changing the standard deviation to 0.05

Process control = ( 9.85 , 10.15) = (10 - 3*0.05 , 10 + 3*0.05) = (Mean - 3*S.D. , Mean + 3* S.D.)

By referring to the empirical rule, the percentage of acceptable units is 99.73%.

Therefore, the probability of defects is 1 - 0.9973 = 0.0027

The expected number of defects in the unit production run is 0.0027*1000 = 2.7 defects

Question 2:

By reducing the process variation, the probability of defects gets reduced. When S.D. was 0.15, the process was a 1-sigma process which means the probability of defects is 0.3173 and when S.D changed to 0.05, the process became a 3-sgima process which means the probability of defects is 0.0027.