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? Value should be in between If the population mean shifts to 8.7, what is the probability that the change will be detected? Probability If the population mean shifts to 9 4. what is the probability that the change will be detected? Probability

Explanation / Answer

Solution :-

The SD of the sample mean is 1/31 = 1/5.57 = 0.1796.

The z value corresponding to 0.95 probability is 1.645.

a) 9 ± 1.645 / 5.57 8.70 9.275

Value should be in between 8.70 and 9.30

b) We must compute P(8.70 < x < 9.30) taking into account that x is normally distributed with mean 8.7 and SD of 0.6.

The cumulative probabilities corresponding to z values of (8.70 - 8.70) / 0.6 = 0 and (9.30 - 8.70) / 0.6 = 1, that is 0 and 1 are 0.5 and 0.159.

The difference is 0.341.

c). We must compute P(8.70 < x < 9.30) taking into account that x is normally distributed with mean 9.4 and SD 0.6.

The cumulative probabilities corresponding to z values of (8.70 - 9.4) / 0.6 = -1.167 and (9.30 - 9.4) / 0.6 = -0.167 are 0.123 and 0.434

The difference is 0.311.

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