The lengths of plate glass parts are measured to the nearest tenth of a millimeter. The lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.2, and continuing through 590.8. Determine the mean and variance of the lengths. (a) mean (in tenths of millimeters) Round your answer to two decimal places (e.g. 98.76). (b) variance (in tenths of millimeters2) Round your answer to three decimal places (e.g. 98.765).
Accepted Solution
A:
Answer: [tex]\text{Mean length}=590.5\ mm\\\\\text{Variance of the lengths}=0.03\ mm[/tex]Step-by-step explanation:The mean and variance of a continuous uniform distribution function with parameters m and n is given by :-[tex]\text{Mean=}\dfrac{m+n}{2}\\\\\text{Variance}=\dfrac{(n-m)^2}{12}[/tex]Given : [tex] m=590.2\ \ \ n=590.80[/tex] [tex]\text{Then, Mean=}\dfrac{590.2+590.8}{2}=590.5\ mm\\\\\text{Variance}=\dfrac{(590.8-590.2)^2}{12}=0.03\ mm[/tex]