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 0.12, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.88 or greater than 10.12 ounces will be classified as defects. (Round your answer to the nearest integer.)

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

2. Through process design improvements, the process standard deviation can be reduced to 0.04. Assume the process control remains the same, with weights less than 9.88 or greater than 10.12 ounces being classified as defects.

a. Calculate the probability of a defect. (Round your answer to four decimal places.)
b. Calculate the expected number of defects for a 1,000-unit production run.

1. a) 0.3174 = 31.74% probability of a defect

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

2. a) 0.0026 = 0.26% probability of a defect

2. b) The expected number of defects for a 1,000-unit production run is of 3.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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.

Question 1:

We have that:

a. Calculate the probability of a defect.

Less than 9.88 or greater than 10.12. These probabilities are equal, so we find one and multiply by 2.

Probability of less than 9.88:

This is the pvalue of Z when X = 9.88. So

has a pvalue of 0.1587

2*0.1587 = 0.3174

0.3174 = 31.74% probability of a defect

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

The expected number of defects is 31.74% of 1000. So

0.3174*1000 = 317.4

Rounding to the nearest integer

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

Question 2:

The mean remains the same, but the standard deviation is now

a. Calculate the probability of a defect.

Less than 9.88 or greater than 10.12. These probabilities are equal, so we find one and multiply by 2.

Probability of less than 9.88:

This is the pvalue of Z when X = 9.88. So

has a pvalue of 0.0013

2*0.0013 = 0.0026

0.0026 = 0.26% probability of a defect

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

The expected number of defects is 31.74% of 1000. So

0.0026*1000 = 2.6

Rounding to the nearest integer

The expected number of defects for a 1,000-unit production run is of 3.