The control limits that include 90% of the sample means be,
(24.9671 , 25.0329).
Given, twenty-five engine mounts are sampled each day and found to have an average width of 2 inches, with a standard deviation of 0.1 inches.
we have to find the confidence interval under confidence of 90%.
As, Mean = 2 and Standard Deviation = 0.1
Subtract 1 from your sample size. 25 – 1 = 24. This gives you degrees of freedom.
Subtract the confidence level from 1, then divide by two.
(1 – 0.9) / 2 = 0.05
For 24 degrees of freedom (df) and α = 0.05, the result is 1.645.
Divide your sample standard deviation by the square root of your sample size.
0.1 / √(25) = 0.02
Now, 0.02 × 1.645 = 0.0329
The lower end of the range
25 – 0.0329 = 24.9671
The upper end of the range
25 + 0.0329 = 25.0329
Hence, the confidence interval be, (24.9671 , 25.0329)
Learn more about Confidence Interval here https://brainly.com/question/26658887
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