What did I do wrong when calculating Standard Deviation?

Question

this my data set:

5.6, 5.2, 4.6, 4.9, 5.7, and 6.4

When calculating by hand I got .2 but the Standard deviation is .64. This is a paragraph of what I did by hand.

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The deviations are .2, -.2, -.8, -.5, .3, and 1 so the squared deviations are .04, -.04, -.64, -.25, .09 and 1. Adding all of those together is .2. So (1/[6-1]) * .2 = .04 and the square root is.2.

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What did I do wrong

Answer 1

First of all, the squared deviations should all be POSITIVE, as

any squared number is positive.  Also, adding them together is .04+.04+.64+.25+.09+1= 2.06  So the variance is (1/6)(2.06), and the standard deviation is the sqrt of this = .5859  ANSWER

Answer 2

Simple - When you multiply two negative numbers, you get a positive result. Therefore, when you square a negative number, your result should be positive. Repeat your math taking this into account and it works out properly.

Data set:

[0.2, 0.2, -0.8, -0.5, 0.3, 1]

Squared:

[0.04, 0.04, 0.64, 0.25, 0.09, 1]

Summed:

2.06

1/(6-1) = 1/5 = 0.2

2.06 * 0.2 = 0.412

(0.412)^(1/2) ~ 0.6419

(Note:  "~" means "about equal to")