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A machine is used to fill 1-liter bottles of a type of soft drink. We can assume that the output of the machine approximately follows a normal distribution with a mean of 1.0 liter and a standard deviation of .01 liter. The firm uses means of samples of 25 observations to monitor the output, answer the following questions: Determine the upper limit of the control chart such that it will include roughly 97 percent of the sample means when the process is in control

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Answer:

1.00434

Step-by-step explanation:

Given the following :

Given a normal distribution ;

Mean (m) = 1.0 liter

Standard deviation (σ) = 0.01 liter

Sample size (n) = 25

For 97% sample means (sm) = 0.97

Z = (m - sm) / s

Zcrit = 1 - (100% - 97%)/2

Zcrit = 1 - (0.03/2)

Zcrit = 1 - 0.015 = 0.985

The z score which corresponds to 0.985 = 2.17

Upper limit : m + z*(σ/√n)

Upper limit : 1.0 + 2.17*(0.01/√25)

Upper limit : 1. 0 + 2.17*(0.01/5)

= 1.0 + 2.17*0.002

= 1.0 + 0.00434

= 1.00434

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