Motorola used the normal distribution to determine the probability of defects an

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 16 ounces. The process standard deviation is 0.1, and the process control is set at plus or minus 1.25 standard deviations. Units with weights less than 15.875 or greater than 16.125 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found (to 0 decimals)? Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 15.875 or greater than 16.125 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)? In a production run of 1000 parts, how many defects would be found (to 0 decimals)? What is the advantage of reducing process variation?

Explanation / Answer

the probability of a defect = 1 - P( -1.25 < Z< 1.25)

= 1 - 0.7887

= 0.2113

b)

In a production run of 1000 parts, how many defects would be found

= 1000*0.2113 = 211.3

c)

1 - P(15.875 < X < 16.125)

Z = (X - 16)/0.08

P(15.875 < X < 16.125)

= P ( 1.56<Z<1.56 )=0.8812

hence required probability = 1 - 0.8812 = 0.1188

d)

In a production run of 1000 parts, how many defects would be found

= 1000*0.1188 = 118.8 = 119

What is the advantage of reducing process variation

when we reduce process variation

the probability of getting a product away from mean (desired) decreases . (as their z-score gets away from 0 (in both direction))

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