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 590.1, and continuing through 590.8. Determine the mean of the lengths.

590.45

0.0825 mm^2

Step-by-step explanation:

Let us introduce the discrete uniform random variable Y with the parameters:  

a = 0, b = 9  

Calculate the mean and the variance:

E(Y)= 0 + 9/ 2  = 4.5  

Var(Y)=(9-0+1)^2-1/12=8.25

Now observe that the random variable in the exercise - denoted X - is in fact:

X = 590 + Y/10

Now calculate:  

E(X) = E (590.1 + Y/10) = 590.1 +E(Y)/10=590.45

Var(X) = E ((X — E(X))^2) = E ( (590 + Y/10-590 - E(Y)^2/10)

           =E ((Y – E(Y))^2)/10

           = Var(X)/100

           = 0.0825 mm^2

is there a question here?

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answer: can you show the options ?

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