Question

T-Moble 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

1.

Weights are normally distributed with mean 10 ounce and standard deviation 0.15. Weights less than 9.85 oz or greater than 10.15 oz are classified as defects.

Let W be a random variable denoting the Weight of an item.

Find P(defect in an item) = P(W < 9.85 or W>10.15) = P(W < 9.85) + P(W>10.15)

= P(W < 9.85) + [1 - P(W <= 10.15) ] = CDF(9.85) + [1 - CDF(10.15)] = 0.1587 + [1 - 0.8413] (refer to CDF tables for Normal distribution to get these values)

= 0.3174

Hence, P(defect in an item) = 0.3174

Expected number of defects in 1000 units = 1000*P(defect in an item) = 1000*0.3174 = 317.4

2.

Process standard deviation is now 0.05. Mean is still 10 ounce and process controls are still less than 9.85 oz or greater than 10.15 oz.

P(defect in an item) = P(W < 9.85 or W>10.15) = P(W < 9.85) + P(W>10.15)

= P(W < 9.85) + [1 - P(W <= 10.15) ] = CDF(9.85) + [1 - CDF(10.15)] = 0.0013 + [1 - 0.9987]

= 0.0026

Expected number of defects in 1000 units = 1000*P(defect in an item) = 1000*0.0026 = 2.6

3.

The points 9.85 and 10.15 represent 1 standard deviation distance form the mean of 10 ounce and standard deviation of 0.15. The area contained within 1 standard deviation of a normal distribution is 0.6827. We are accepting all items within the values 9.85 and 10.15. Thus we will accept 68.27% of the items.

When the standard deviation is reduced to 0.05 from 0.15, the points representing 1 standard deviation away from the mean (10 ounce) are 9.95 and 10.05. Thus the area contained within the limits 9.95 and 10.05 (area within 1 standard deviation of mean) is 0.6827. But our process limits are still 9.85 and 10.15. 9.85 is further away from the mean than 9.95. Similarly, 10.15 is further away from mean than 10.05. Hence, the area of normal distribution curve contained within these limits (9.85 and 10.15) is larger. Hence the percentage of accepted items is also larger than 68.27%. Therefore, the percentage of rejected items is smaller than before and hence better production outcomes.