A cola-dispensing machine is set to dispense 9 ounces of cola per cup, with a st

Question

A cola-dispensing machine is set to dispense 9 ounces of cola per cup, with a standard deviation of 0.6 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 31, 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. 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 If the population mean shifts to 8.7, what is the probability that the change will be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.) Probability If the population mean shifts to 9.4, what is the probability that the change will be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.) Probability

Explanation / Answer

a) The sample means for sample size of 31 has mean of 9 ounces and a standard deviation of 0.6/sqrt(31)=0.108 ounces

The z score corresponding to the lower control limit is the 0.05 quantile of the standard gaussian(normal) distribution. That is the z score is -1.65. So the lower control limit is 0.108 x -1.65 +9 =8.8218

The z score corresponding to the upper control limit is the 0.95 quantile of the standard gaussian(normal) distribution. That is the z score is 1.65. So the lower control limit is 0.108 x 1.65 +9 =9.1782

Control limits (8.82,9.18)

b) Probability that change is detected is probability that a sample of size 31 has sample mean below the lower limit, for mean 8.7 is 0.8667. This is the probability of detection.

c) Probability that change is detected is probability that a sample of size 31 has sample mean above the upper limit, for mean 9.4 is 0.9792. This is the probability of detection.

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