- #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
d2 = (-0.2 m)2 = 0.04
wd2 = (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