A cola-dispensing machine is set to dispense 10 ounces of cola per cup, with a s

Question

A cola-dispensing machine is set to dispense 10 ounces of cola per cup, with a standard deviation of 0.8 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. 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 9.76 and 10.24 b. If the population mean shifts to 9.7, 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 c. If the population mean shifts to 10.4, 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

solution=

a) The SD of the sample mean is 0.8/31 =0.144 The z value corresponding to 0.95 probability is 1.645

So the answer to question a is 10±1.645*0.144 9.763 to 10.237.

b) , we must compute P( 9.763<x<10.237.) taking into account that x is normally distributed with mean 9.7 and SD 0.8. The cumulative probabilities corresponding to z values of

0.079 and 0.671 are 0.74889 and 0.531484 The difference is 0.217406   0.217

c), we must compute  P( 9.763<x<10.237.) taking into account that x is normally distributed with mean 10.4 and SD 0.8 The cumulative probabilities corresponding to z values of

0.796 and 0.204 are 0.419176 and 0.213016 The difference is 0.20616   0.206

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