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. (a) The process standard deviation is 0.15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. (Round your answer to the nearest integer.) Calculate the probability of a defect. (Round your answer to four decimal places.) Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.) defects (b) Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. Calculate the probability of a defect. (Round your answer to four decimal places.) Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.) defects (c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean

Answer:

a) 0.3174 = 31.74% probability of a defect. The number of defects for a 1,000-unit production run is 317.

b) 0.0026 = 0.26% probability of a defect. The expected number of defects for a 1,000-unit production run is 26.

c) Less variation means that the values are closer to the mean, and farther from the limits, which means that more pieces will be within specifications.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Assume a production process produces items with a mean weight of 10 ounces.

This means that [tex]\mu = 10[/tex].

Question a:

Process standard deviation of 0.15 means that [tex]\sigma = 0.15[/tex]

Calculate the probability of a defect.

Less than 9.85 or more than 10.15. Since they are the same distance from the mean, these probabilities is the same, which means that we find 1 and multiply the result by 2.

Probability of less than 9.85.

pvalue of Z when X = 9.85. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{9.85 - 10}{0.15}[/tex]

[tex]Z = -1[/tex]

[tex]Z = -1[/tex] has a pvalue of 0.1587

2*0.1587 = 0.3174

0.3174 = 31.74% probability of a defect.

Calculate the expected number of defects for a 1,000-unit production run.

Multiplication of 1000 by the probability of a defect.

1000*0.3174 = 317.4

Rounding to the nearest integer,

The number of defects for a 1,000-unit production run is 317.

Question b:

Now we have that [tex]\sigma = 0.05[/tex]

Probability of a defect:

Same logic as question a.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{9.85 - 10}{0.05}[/tex]

[tex]Z = -3[/tex]

[tex]Z = -3[/tex] has a pvalue of 0.0013

2*0.0013 = 0.0026

0.0026 = 0.26% probability of a defect.

Expected number of defects:

1000*0.0026 = 26

The expected number of defects for a 1,000-unit production run is 26.

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

Less variation means that the values are closer to the mean, and farther from the limits, which means that more pieces will be within specifications.

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