A coladispensing machine is set to dispense 9 ounces of cola

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 35, 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.)

B) 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.)

C) 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.)

Solution

a) SD of sample mean = 0.6 / sqrt(35) = 0.1014

z value corresponding to 95% probability is 1.65

x = m +- (z * SD) = 9 +- (1.65*0.1014) = 8.83 to 9.17

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b) we have to find P(8.83 < x < 9.17), given m = 8.7 and SD = 0.6

z = (x - m)/SD

P(8.83 < x < 9.17) = P( (8.83-8.7)/0.6 < z < (9.17-8.7)/0.6) )

= P( 0.02 < z < 0.08)

= 0.5319 - 0.5080

= 0.0239

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c) we have to find P(8.83 < x < 9.17), given m = 9.7 and SD = 0.6

z = (x - m)/SD

P(8.83 < x < 9.17) = P( (8.83-9.7)/0.6 < z < (9.17-9.7)/0.6) )

= P(-1.45 < z < -0.88)

= 0.1894 - 0.0735

= 0.1159