A coladispensing machine is set to dispense 8 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 37, 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.
1. At what value should the control limit be set? (Round z values to two decimal places. Round your answers to 2 decimal places.)

2. 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.)

3. If the population mean shifts to 8.6, 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.)
Probability 
1) here for top and bottom 5% critical value =1.64
hence Value should be in between =mean/+z*std error=8/+1.64*1/sqrt(37)=7.73 and 8.27
2)
for normal distribution z score =(Xμ)/σx  
here mean= μ=  7.6 
std deviation =σ=  1.000 
sample size =n=  37 
std error=σ_{x̅}=σ/√n=  0.1644 
probability that the change will not be detected:
probability =  P(7.73<X<8.27)  =  P(0.79<Z<4.08)=  10.7852=  0.2148 
3)
probability =  P(7.73<X<8.27)  =  P(5.29<Z<2.01)=  0.02220=  0.0222 
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