Question

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 590.9. 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).

Answer 1

Answer:  $\text{Mean =590.55 tenths of millimeters and Variance =0.41 tenths of millimeters}^2}$

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 :-

$\text{Mean}=\dfrac{b+a}{2}\\\\\Rightarrow\text{Variance}=\dfrac{(b-a)^2}{12}$

In the given situation, a= 590.2 and b= 590.9

Then,

$\text{Mean}=\dfrac{590.2+590.9}{2}=590.55\\\\\Rightarrow\text{Variance}=\dfrac{(590.9-590.2)^2}{12}=0.0408333333333\approx0.041$

Hence, Mean = 590.55 and Variance = 0.41