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

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