The mean and the standard deviation for the population of products produced with the process are 2.0 and 0.3329
Start by calculating the population mean using:
[tex]\mu = \frac{L + U}{2}[/tex]
Where:
L = 1.9 and U = 2.1
So, we have:
[tex]\mu = \frac{1.9 + 2.1}{2}[/tex]
[tex]\mu = 2.0[/tex]
The z value when p = 5%, from z table of probability is:
z = -1.645
Also, we have the z-score to be:
[tex]z = \frac{x - \mu}{\sigma/\sqrt n}[/tex]
So, we have:
[tex]-1.645 = \frac{1.9 - 2.0}{\sigma/\sqrt {30}}[/tex]
[tex]-1.645 = \frac{-0.1}{\sigma/\sqrt {30}}[/tex]
[tex]16.45 = \frac{1}{\sigma/\sqrt {30}}[/tex]
Cross multiply
[tex]\sigma/\sqrt {30} * 16.45 = 1[/tex]
Take the square root of 30
[tex]\sigma/5.477* 16.45 = 1[/tex]
[tex]\sigma = 0.3329[/tex]
Hence, the mean and the standard deviation for the population of products produced with the process are 2.0 and 0.3329
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