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 13 ounces.

The process standard deviation is 0.2, and the process control is set at plus or minus 0.5 standard deviation. Units with weights less than 12.9 or greater than 13.1 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 0.05. Assume the process control remains the same, with weights less than 12.9 or greater than 13.1 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)?

Explanation / Answer

In first case, Mean weight = 13 ounce and standard deviation = 0.2 ounce

Probability of a defect = Pr (x < 12.9 and x> 13.1) = Pr ( x >13.1; 13; 0.2) - Pr( x< 12.9; 13 ; 0.2)

Z- values are +- 0.5

so Probability of defect = (1 - (0.5))+ (-0.5) = 1 - ((0.5) - (-0.5)) = 1 - [ 0.6915 - 0.3085)] = 0.6170

where is the cumulative standard normal probability distribution

Number of defects to be found out of 1000 parts = 1000 * 0.617 = 617

Now There is second case, where standard deviation reduced to 0.05

so Pr( x < 12.9 and x >13.1 ; 13; 0.05) = Pr ( x < 12.9 , 13, 0.05) + Pr( x > 13.1; 13; 0.05)

Z vlaues are +-2

Pr( x < 12.9 and x >13.1 ; 13; 0.05) =(1- (2)) + (-2) = 1 - 0.9772 + 0.0228 = 0.0456

So in production number of defects to be found out of 1000 = 1000 * 0.0456 = 46 defects.

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