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 7 ounces. a. the process standard deviation is 0.2, and the process control is set at plus or minus 0.75 standard deviation units with weights less than 6.85 or greater than 7.15 ounces will be classified as defects. what is the probability of a defect?

Answer 1

see mtorolaa is american company

Answer 2

Answer:

Probability of defects = 0.4532

Step-by-step explanation:

Mean = 7 ounces

Standard deviation = 0.2 ounces

The process control is set at +/- 0.75

Defects are classified if the weights are less than 6.85 or greater than 7.15 ounces. We will solve using normal distribution.  Z = $\frac{X- \mu}{\sigma}$

P(X <6.85) = P( Z < $\frac{6.85-7}{0.2}$) = P(Z < -0.75)

Using standard normal distribution table, P(Z< -0.75) = 0.2266

P(Z > 7.15) = P(Z > $\frac{7.15-7}{0.2}$) = P(Z > 0.75)

                                                                         = 0.2266

Probability of defects = P(Z <-0.75) +P(Z>0.75)

                                  = 0.2266+0.2266 = 0.4532