What is the standard deviation of #1, 2, 3, 4# and #5#?

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De Rono Share
Nov 5, 2015

Answer:

The standard deviation of #{1, 2, 3, 4, 5}#
#= [(5^2-1)/(12)]^(1/2) = sqrt2 #

Explanation:

Let's develop a general formula then as a particular you get standard deviation of #1, 2, 3, 4# and #5#. If we have #{1, 2,3, .... , n}# and we need to find the standard deviation of this numbers.

Note that

#"Var" (X) = 1/n sum_{i=1}^n x_i^2 - (1/n sum _(i=1)^n x_i)^2#

#implies "Var" (X) = 1/n sum_{i=1}^n i^2 - (1/n sum _(i=1)^n i)^2#

#implies "Var" (X) = 1/n * (n(n+1)(2n+1))/(6) - (1/n *(n(n+1))/2)^2#
#implies "Var" (X) = ((n+1)(2n+1))/(6)- ((n+1)/2)^2#
#implies "Var" (X) = (n+1)/(2)[(2n+1)/3- (n+1)/2]#

#implies "Var" (X) = (n+1)/(2) * (n-1)/6#
#implies "Var" (X) = (n^2-1)/(12)#

So, Standard deviation of #{1, 2,3, .... , n}# is #["Var" (X)]^(1/2) = [ (n^2-1)/(12)]^(1/2)#

In particular, your case the standard deviation of #{1, 2, 3, 4, 5}#
#= [(5^2-1)/(12)]^(1/2) = sqrt 2 #.

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