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 0.10 ounces, and the process control is set at plus or minus 1.75 standard deviations. Units with weights less than 11.825 or greater than 12.175 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 (round to the nearest whole number)?

b. Through process design improvements, the process standard deviation can be reduced to 0.06 ounces. Assume the process control remains the same, with weights less than 11.825 or greater than 12.175 ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

c. What is the advantage of reducing process variation, thereby causing a problem limits to be at a greater number of standard deviations from the mean?

a. it can substantially reduce the number of defects

b. it may slightly reduce the number of defects

c. it has no effect on the number of defects

Answer 1

a.

Let the random variable X = weight of an item in ounce in a production process,

X follows normal distribution with,

An item is said to defective if it lies outside the control limit such that, outside 1.75 standard deviation from mean, . Hence the probability for the defective item is obtained using the z score as shown below,

In terms of z-score

The probability for z score is obtained from standard normal distribution table,

In a production run of 1000 parts, the expected number of defects are,

b.

The standard deviation reduced to 0.06 ounces. Now, the probability for the defective item is obtained by calculating the z score as shown below,

Converting to Z score,

The probability for z score is obtained from standard normal distribution table,

In a production run of 1000 parts, the expected number of defects are,

c.

Answer: a. it can substantially reduce the number of defects

Explanation:

From the first two part we can observe that as the variability of the production process decrease, the number of defects also decreases. Hence we say that reducing the standard deviations from the mean will reduce the number of defects in production process