Question

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a product process produces items with a mean weight of 10 ounces. a. The process standard deviation is 0.15 ounces, and the process control is set at plus or minus 1.6 standard deviations. Units with weights less than 9.76 greater than 10.24 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? .6827 In a production run of 1000 parts, how many defects would be found (round to the nearest whole number)? 317 ® b. Through process design improvements, the process standard deviation can be reduced to 0.09 ounces. Assume the process control remains the same, wi weights less than 9.76 or greater than 10.24 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? - Select your answer -

Answer 1

a)

Z = (X - mean)/sd

1- P(9.76 < X < 10.24)

= 1 - P(-1.6 <Z < 1.6)

= 0.1096

1000*p = 109.6 = 110

b)

1- P(9.76 < X < 10.24)

= 0.0077

1000*0.0077 = 7.7 = 8

c)

advantage is the probability of defect decreases

Please rate