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 6 ounces.A) The process standard deviation is 0.1, and the process control is set at plus or minus 2.5 standard deviations. Units with weights less than 5.75 or greater than 6.25 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?B) In a production run of 1000 parts, how many defects would be found (to 0 decimals)?C) Through process design improvements, the process standard deviation can be reduced to 0.09. Assume the process control remains the same, with weights less than 5.75 or greater than 6.25 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)?D) In a production run of 1000 parts, how many defects would be found (to 0 decimals)?E) What is the advantage of reducing process variation?

Answer 1

a. 0.0124

b. 12

c. 0.0054

d. 5

e. advantage of reducing process variation is that probability of defect item is decreased

Step-by-step explanation:

mean = 6

SD = 0.1

a)probability of a defect = 1 - P(5.75 <X <6.25)

= 1 - P(-2.5<Z <2.5)= = 1 - 2(0.9938-0.5) = 2(1-0.9938) = 2*0.0062 = 0.0124

b)In a production run of 1000 parts,number of defects would be found = 1000*0.124 = 12.4 = 12

c)1 - P(5.75 <X <6.25) = 1- P(-0.25/0.09 <Z < 0.25/0.09) = 1- P(-2.777 <Z<2.777)= 1 - 2 (0.9973 -0.5)

= 2 ( 1 - 0.9973) = 2*(0.0027) =0.0054

d)In a production run of 1000 parts,number of defects would be found = 1000*0.0054 = 5.4 = 5

e) advantage of reducing process variation is that probability of defect item is decreased.