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Gregoltsov · Answer 1 · 2023-09-18T08:55:25+0000

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find the mean and standard deviation for the diameter of the ball bearings from supplier A

For this case we need to remember the formulas for the mean and deviation given by:

$\text{Mean}=\frac{x_1+x_2+\text{.}\ldots+x_n}{n}=(16.23+16.26+16.31+16.35+16.37+16.41+16.44)/(7)=16.34$
$\text{Variance}=((16.23-16.34)^2+(16.26-16.34)^2+(16.31-16.34)^2+(16.35-16.34)^2+(16.37-16.34)^2+(16.41-16.34)^2+(16.44-16.34)^2)/(7-1)$

And we got:

Variance= 0.00588

And the deviation would be given by:

$\text{Devition}=\sqrt[]{0.00588}=0.0767$

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