value 10.00 points A cola-dispensing machine is set to dispense 9 ounces of cola

Question

value 10.00 points A cola-dispensing machine is set to dispense 9 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 48, 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 8.74 and 9.26 b. If the population mean shifts to 8.6, 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 9.5, 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 References eBook & Resources Worksheet

Explanation / Answer

so the mean of the population mu=9

sd=1 and n=48 is given

while the upper and lower limit of 5% is also given

which gives us z value = -1.65& +1.65 for lower and upper tail respectively from Z table

As we knwo the equation is

Z=(X'-mu)/(sd/sqrt(n))

And to get X' the equation will be

X'=Z*(sd/sqrt(n))+mu

Substituting the available values we get

a) X'=+/-1.65*1/sqrt(48) + 9 = 8.76 & 9.24 which matches with your ans

b) now if the mean shifts to 8.6 we get new X'= 8.361843 & 8.838157

and so beyond 9.24 we should have detected the change which we will be able to dectect here as the upper limit has reduced to 8.83 and we will be able to detect the same but from 8.36 to 8.76 we won't be able to detect the change and that makes (8.76-8.36)/8.76 =0.0456621 approx 4.5% of the errors leave undetected and so we can say that in this case we will beable to detect 95.5% probabilities

c)now if the mean shifts to 9.5 we get new X'= 9.261843 & 9.738157

and so below 9.24 we should have detected the change which we will be able to dectect here as the lower limit has scaled up to 9.26 and we will be able to detect the same but from 9.24 to 9.74 we won't be able to detect the change and that makes (9.74-9.24)/9.24 =0.05411255 approx 5.4% of the errors leave undetected and so we can say that in this case we will beable to detect 94.6% probabilities

Hope the above explaination has helped you in understanding the problem Pls upvote the ans if it has really helped you. Good Luck!!

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