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 0.5 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.

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 ___ and ___.

b. 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 is ___.           

c. If the population mean shifts to 10.4, 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 is ___.           

Explanation / Answer

Mean ( u ) =10
Standard Deviation ( sd )= 0.5 / Sqrt(31) = 0.0898
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 10/0.0898 ) = 0.05
That is, ( x - 10/0.0898 ) = -1.64
--> x = -1.64 * 0.0898 + 10 = 9.8523                  
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 10/0.0898) = 0.05
That is, ( x - 10/0.0898) = 1.6449
--> x = 1.6449 * 0.0898+10 = 10.1477                  
control limits are 9.8523, 10.148
b.
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 9.5/0.0898 ) = 0.05
That is, ( x - 9.5/0.0898 ) = -1.64
--> x = -1.64 * 0.0898 + 9.5 = 9.3523                  
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 9.5/0.0898) = 0.05
That is, ( x - 9.5/0.0898) = 1.6449
--> x = 1.6449 * 0.0898+9.5 = 9.6477                  
control limits are 9.3523, 9.6477
c.
P ( Z < x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is -1.645
P( x-u/s.d < x - 10.4/0.0898 ) = 0.05
That is, ( x - 10.4/0.0898 ) = -1.64
--> x = -1.64 * 0.0898 + 10.4 = 10.2523                  
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 10.4/0.0898) = 0.05
That is, ( x - 10.4/0.0898) = 1.6449
--> x = 1.6449 * 0.0898+10.4 = 10.5477                  
control limits are 10.2523, 10.5477  

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