Question

1) In a chemical processing plant, it is important that the yield of a certain type of batch product stay above 85%. If it stays below 85% for an extended period of time, the company loses money. Occasional defective batches are of little concern. But if several batches per day are defective, the plant shuts down and adjustments are made. It is known that the yield is normally distributed with standard deviation 3%. (a) What is the probability of a "false alarm" (yield below 85%) when the mean yield is 90%? (b) What is the probability that a batch will have a yield that exceeds 85% when in fact the mean yield is 82%? (c) Assume that the process is producing batches with an average yield of 84%. What is the probability that none of the next 10 batches has a yield of less than 85%? (d) Assume that the process is producing batches with an average yield of 84%. What is the probability that you have to inspect exactly 8 batches in order to find two batches with a yield of less than 85%? (e) Assume that the process is producing batches with an average yield of 84%. What is the probability that you have to inspect at most 8 batches in order to find two batches with a yield of less than 85%?

Answer 1

Answer:

a)

To determine the probability of a false alarm(yield below 85%) when the mean yield is 90%

Given,

Mean = 90

Standard deviation = 3

we know,

z = (x - $\mu$) / $\sigma$

substitute values

= (85 - 90) / 3

= - 5 / 3

z = - 1.67

Required probability = P(X < 85 | $\mu$ = 90) = P(Z < - 1.67)

= 0.0474597

Required probability = 0.0475

b)

To determine the probability that a batch will have a yield that exceeds 85% when in fact the mean yield is 82%

$\mu$ = 82

$\sigma$ = 3

z = (x - $\mu$) / $\sigma$

substitute values

= (85 - 82) / 3

= 3 / 3

z = 1

Now to give the required probability

P(X > 85 | $\mu$ = 82) = 1 - P(X <= 85)

= 1 - P(Z <= 1)

= 1 - 0.8413447

= 0.1586553

= 0.1587

Required probability = 0.1587