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 7 ounces.

a. The process standard deviation is 0.10, and the process control is set at plus or minus 2 standard deviation s . Units with weights less than 6.8 or greater than 7.2 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 (round to the nearest whole number)?

b. 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 6.8 or greater than 7.2 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)?

Answer 1

a. Let X be the random variable denoting the mean weight.

Thus, X ~ N(7, 0.1) i.e. (X - 7)/0.1 ~ N(0,1)

Hence, the probability of defect = P(X < 6.8) + P(X > 7.2)

= 1 - P(6.8 < X < 7.2)

= 1 - P[(6.8 - 7)/0.1 < (X - 7)/0.1 < (7.2-7)/0.1]

= 1 - P[-2 < (X - 7)/0.1 < 2] = 1 - (2) + (-2)

[(.) is the cdf of N (0,1)]

= 1 - 0.9772 + 0.0228 = 0.0456. (Ans).

Let Y be the random variable denoting the number of defects

in 1000 parts. Thus, Y ~ Binomial (1000, 0.0456).

Thus, E(Y) = 1000 * 0.0456 = 456.

Thus, In a production run of 1000 parts, 456 defects will be

found. (Ans).

b. Now, X ~ N(7, 0.08).

Hence, the probability of a defect = 1 - P(6.8 < X < 7.2)

= 1 - P(6.8 - 7)/0.08 < (X - 7)/0.08 < (7.2-7)/0.08]

= 1 - P[-2.5 < (X - 7)/0.08 < 2.5] = 1 - (2.5) + (-2.5)

= 1 - 0.9938 + 0.0062 = 0.0124. (Ans).

Now, Y ~ Binomial (1000, 0.0124)

Hence, E(Y) = 1000 * 0.0124 =124.

Thus, in a production of 1000 parts, 124 defects will be found.

(Ans).