A cola-dispensing machine is set to dispense 9.00 ounces of cola per cup, with a

Question

A cola-dispensing machine is set to dispense 9.00 ounces of cola per cup, with a standard deviation of 1.00 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 36, 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 3 decimal places.) Value should be in between

b. If the population mean shifts to 8.6, what is the probability that the change will not be detected? (Round z values to two decimal places. Round your answer to 4 decimal places.) Probability =

. If the population mean shifts to 9.6, what is the probability that the change will not be detected? (Round z values to two decimal places. Round your answer to 3 decimal places.) Probability=

Explanation / Answer

Solution:-

a). The SD of the sample mean is 1/36 = 1/6.

The z value corresponding to 0.95 probability is 1.645.

So the answer to question is 9 ± 1.645/6 8.725 9.275

LCL = 8.725

UCL = 9.275

b). Here, we must compute P(8.725 < x < 9.275)

taking into account that x is normally distributed with mean 8.6 and SD 1.

The cumulative probabilities corresponding to z values

are 0.54974 and 0.75016.

The difference is 0.20042.

c). Here,we must compute P(8.725 < x < 9.275) taking into account that x is normally distributed with mean 9.6 and SD 1.

The cumulative probabilities corresponding to z values are 0.19079 and 0.37259.

The difference is 0.1818.

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