Question

The lengths of plate glass parts are measured to the nearest 1/10 mm. The lengths are discretely and uniformly distributed with values at every 1/10th mm, starting at 590.0 thru 590.9 mm. Determine the mean & variance of lengths.

Answer 1

@whpalmer4

Answer 2

@ybarrap @yama_aryayee

Answer 3

It says uniformly distributed are you sure its geometric?

Answer 4

I really do not know but I guess it should be done with poisson method

Answer 5

For uniform, the mean is
$$
\bar{x}=\cfrac{590.0+590.9}{2}
$$
The variance is
$$
\sigma=\sqrt{\cfrac{n^2-1}{12}}
$$
Where n=10.

Answer 6

n is 9 or 10 since there are 9 portions

Answer 7

between 590 and 590.9

Answer 8

There are 10 portions:
590.0 590.1 590.2 590.3 590.4 590.5 590.6 590.7 590.8 590.9

Answer 9

and mean is the mean value which mean a number in the middle of these two, so do not u think it should be 590.45?

Answer 10

yes

Answer 11

590.45 is correct

Answer 12

and I think variance and mean should be equal ,,

Answer 13

BTW, variance is
$$
\sigma^2=\cfrac{n^2-1}{12}
$$
NOT
$$
\sigma=\sqrt{\cfrac{n^2-1}{12}}
$$