What is the standard deviation of the following numbers: 24, 36, 33, 21, 15, 11?

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Answer 1 · 2017-12-09T10:20:34

#sigma = (19 sqrt 2)/3#

We know:

#sigma^2 = (Sigma x^2 )/ n - ( (Sigma x) / n )^2 #

#Sigma x = 24 + 36 + 33 + 21 + 15 +11 = 140 #

#Sigma x^2 = 24^2 + 36^2 +33^2 + 21^2 + 15^2 + 11^2 = 3748 #

#n# is the number data in the sample, #=> n =6 #

Hence plugging into the formula:

#sigma^2 = 3748/6 - (140/6)^2 = 722/9 #

#=> sigma = (sqrt 722) / (sqrt 9 ) = (19 sqrt 2 )/3 #

Answer 2 · 2017-12-09T10:29:03

Standard Deviation #sigma = color (purple)(8.9567)#

Mean for the given numbers
#= (24+36+33+21+15+11) / 6 = 23.33#

Variance #sigma^2 = sum(x- bar x)^2 / (n)#

#sigma^2 =sum ((23.33-24)^2 + (23.33-36)^2 + (23.33-33)^2 + (23.33-21)^2 + (23.33-15)^2 + (23.33-11)^2) / (6)#

#sigma^2 = ((-0.67)^2 + (-12.67)^2 + (-9.67)^2 + (2.33)^2 + (8.33)^2 + (12.33)^2) / 6#

#sigma ^2 = 80.2222#

Standard Deviation #sigma = 8.9567#