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 1.0 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 50, 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 9.7, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

c. If the population mean shifts to 10.5, what is the probability that the change will not be detected? (Round your intermediate calculations to 2 decimal places and final answer to 4 decimal places.)

Explanation / Answer

a)

for 5% upper and lowe control limit ; z =-/+ 1.64

therefore lower control limit =mean -z*std deviaiton =10-1.64*1=8.36

upper control limit =mean +z*std deviaiton =10+1.64*1=11.64

b)

P(8.36<X<11.64)=P((8.36-9.7)/1<Z<(11.64-9.7)/1)=P(-1.34<Z<1.94)=0.9738-0.0901 =0.8837

c)

P(8.36<X<11.64)=P((8.36-10.5)/1<Z<(11.64-10.5)/1)=P(-2.14<Z<1.14)=0.8729-0.0162 =0.8567

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