A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a st

Question

A cola-dispensing machine is set to dispense 8 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 38, 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 7.6, 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 8.4, 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

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 8
standard Deviation ( sd )= 0.8/ Sqrt ( 38 ) =0.1298
sample size (n) = 38

a.
control limit if 5% less and 5%greater
Z for 5% less= (7.6-8)/0.8/ Sqrt ( 38 )
Z =-3.08
Z greater 5%
Z = (8.4-8)/0.8/ Sqrt ( 38 )
Z = 3.08
control limit -3.08 to 3.08

b.
If the population mean shifts to 7.6
the probability that the change will not be detected
P(X < 7.6) = (7.6-8)/0.8/ Sqrt ( 38 )
= -0.4/0.1298= -3.08
= P ( Z <-3.0822) From Standard NOrmal Table
= 0.001
c.
If the population mean shifts to 8.4
the probability that the change will not be detected
P(X > 8.4) = (8.4-8)/0.8/ Sqrt ( 38 )
= 0.4/0.13= 3.08
= P ( Z >3.0822) From Standard Normal Table
= 0.001

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