- #1

santoki

- 34

- 0

After filling out the table, I used it to finally calculate the standard deviation as 0.105. Just to make sure, I ran it through Microsoft Excel and got 0.942809042. I put two together and figured I'm doing something wrong after the xw column because Excel agrees that the mean is also 1.7.

Sample calculations:

d = x - <x>

d = (1.50 m) - (1.7 m)

d = -0.2 m

d

^{2}= (-0.2 m)

^{2}= 0.04

wd

^{2}= (1)(0.04) = 0.04

SD = [itex]\sqrt{[0.04+0.0225+0.09+0.0075+0+0+0.02+0+0.04]/(21-1)}[/itex]

SD = 0.1048808848