Question

NEED ANSWER TO B & C

A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 45, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit.

a.	At what value should the control limit be set? (Round z values to two decimal places. Round your answers to 2 decimal places.)

Value should be in between

and

b.	If the population mean shifts to 7.7, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

Probability

c.	If the population mean shifts to 8.6, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

Probability

Answer 1

Here we have

First we need to find the limits. Now we need z-score that has 0.05 area to its left. z-score -1.645 has 0.05 area to its left. So lower limit of sample mean is

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Now we need z-score that has 0.05 area to its right. z-score 1.645 has 0.05 area to its right. So upper limit of sample mean is

(b)

Now new populaiton mean is 7.7 so z-score with and is

Now new populaiton mean is 7.7 so z-score with and is

So the probability that the change will be detected is

P(0.34< z < 3.69)= 0.3668

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(c)

Now new populaiton mean is 8.6 so z-score with and is

Now new populaiton mean is 7.7 so z-score with and is

So the probability that the change will be detected is

P(-5.70< z <-2.35)= 0.0094