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 0.20 ounces, and the process control is set at plus or minus 0.75 standard deviations. Units with weights less than 5.85 or greater than 6.15 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 0.07 ounces. Assume the process control remains the same, with weights less than 5.85 or greater than 6.15 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 (round to the nearest whole number)?
Solution:
a. The process standard deviation is 0.20 ounces, and the process control is set at plus or minus 0.75 standard deviations. Units with weights less than 5.85 or greater than 6.15 ounces will be classified as defects.
What is the probability of a defect (to 4 decimals)?
Answer: The probability of a defect is:
Now using the z-score formula, we have:
Now using the standard normal table, we have:
Therefore, the probability of a defect is 0.4532
In a production run of 1000 parts, how many defects would be found
b. Through process design improvements, the process standard deviation can be reduced to 0.07 ounces. Assume the process control remains the same, with weights less than 5.85 or greater than 6.15 ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?
Answer: The probability of a defect is:
Now using the z-score formula, we have:
Now using the standard normal table, we have:
Therefore, the probability of a defect is 0.0324
In a production run of 1000 parts, how many defects would be found
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