Question

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a particular production process. Assume that the production process manufactures items with a mean weight of 10 ounces. Calculate the probability of a defect and the suspected number of defects for a 1,000-unit production run in the following situations. (a) The process standard deviation is 0.30, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places. (b) Through process design improvements, the process standard deviation can be reduced to 0.10. Assume that the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. If required, round your answer to four decimal places. (c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean? Reducing the proces standard deviation causes a in the number of defects.

Answer 1

Sol:

(a)

Consider a random variable X which represents the amount of defectives and follows standard normal distribution whose mean \left( \mu \right)(μ) is 10 ounces and standard deviation \left( \sigma \right)(σ) is 0.30.

The values of the random variable differ from mean by plus minus 1\sigma1σ that is the values are either greater than 10.15 or less than 9.85.

So, the probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be computed as:

P(X<9.85orX>10.15)=1−P(((9.85−10)/0.30)​<((X−10)/0.30)​<((10.15−10)/0.30)​)

=1−(Φ(1)−Φ(−1))​

Excel is used to calculate the values of the probability z = 1z=1 and z = - 1z=−1 . The screenshots of the formula used is shown below:

And

Therefore, the required probability can be calculated as:

P(X<9.85orX>10.15)=1−(Φ(1)−Φ(−1))=1−(0.841345−0.158655)=0.31731≈0.32​

(b)

The standard deviation is reduced and it is 0.10.

So, the probability that the amount of defectives which are either greater than 10.15 or less than 9.85 can be computed as:

P(X<9.85orX>10.15)=1−P(((9.85−10​)/0.10)<((X−10​)/0.10)<((10.15−10)/0.10)​)

=1−(Φ(3)−Φ(−3))​

Excel is used to calculate the values of the probability z = 3z=3 and z = - 3z=−3 . The screenshots of the formula used is shown below:

And

Therefore, the required probability can be calculated as:

P(X<9.85orX>10.15)=1−(Φ(3)−Φ(−3))=1−(0.99865−0.00135)=0.0027≈0.003​

(c)

The probability that the amount of defectives which are either greater than 10.15 or less than 9.85 is approximately 0.32 and 0.003 when the standard deviation is 0.30 and 0.10 respectively. That is, the probability of getting defective items will be reduced if the process standard deviation is reduced.