Answer: [tex]\text{Mean =590.55 tenths of millimeters and Variance =0.41 tenths of millimeters}^2}[/tex]
Step-by-step explanation:
Given : The lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.2, and continuing through 590.9.
Let x be the random variable that represents the lengths of plate glass parts.
The mean and variance of the random variable X that is uniformly distributed in interval [a,b] is given by :-
[tex]\text{Mean}=\dfrac{b+a}{2}\\\\\Rightarrow\text{Variance}=\dfrac{(b-a)^2}{12}[/tex]
In the given situation, a= 590.2 and b= 590.9
Then,
[tex]\text{Mean}=\dfrac{590.2+590.9}{2}=590.55\\\\\Rightarrow\text{Variance}=\dfrac{(590.9-590.2)^2}{12}=0.0408333333333\approx0.041[/tex]
Hence, Mean = 590.55 and Variance = 0.41
