Question

A cola-dispensing 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.)

Value should be in between

and

.

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.)

Probability

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

Answer 1

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 =(X-μ)/σx
here mean=       μ=	7.6
std deviation   =σ=	1.000
sample size       =n=	37
std error=σx̅=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)=

1-0.7852=

0.2148

3)

probability =

P(7.73<X<8.27)

=

P(-5.29<Z<-2.01)=

0.0222-0=

0.0222