Question

A cola-dispensing machine is set to dispense 10 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 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.

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 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.)

3. If the population mean shifts to 10.4, 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.)

Answer 1

ANSWER::

Mean=10

SD=0.8

n=31

5% is the control limit on both the sides

Hence corresponding z value is 1.645(90% probability)

a)

UCL=mean+z*SD/sqrt(n)

=10+1.645*0.8/sqrt(31)

=10+0.236

UCL=10.236

LCL=mean-z*SD/sqrt(n)

=10-0.236

LCL=9.76

b)

new mean=9.7

SD=0.8

to find probability that change will not be detected

P(9.76<x<10.24)

Z values are (9.76-9.7) and (10.24-9.7)

Z values=0.06 and 0.54

From Z table the corresponding probabilities are 0.5239 and 0.7054

The difference is our required probability (0.7054-0.5239)

=0.1815

c)

new mean=10.4

SD=0.8

to find probability that change will not be detected

P(9.76<x<10.24)

Z values are (9.76-10.4) and (10.24-10.4)

Z values= -0.64 and -0.16

From Z table the corresponding probabilities are 0.2611 and 0.4364

The difference is our required probability (0.2611-0.4364)

=0.1753

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