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Answer 1 · 2017-08-06T18:56:15

#sqrt2~~1.414#

The Standard Deviation #sigma# of #n# observations

#{x_i}_(i=1)^(i=n),# is given by,

#sigma^2=1/n{sum_(i=1)^(i=n)(x_i-barx)^2},# where,

#barx=1/n{sum_(i=1)^(i=n)x_i}.#

#=1/5(1+2+3+4+5)=15/5=3.#

Thus, we have,

#x_i : 1,2,3,4,5.#

#(x_i-barx) : -2,-1,0,1,2.#

#(x_i-barx)^2 : 4,1,0,1,4.#

#sum_(i=1)^(i=n)(x_i-barx)^2 : 10.#

# rArr sigma^2=10/5=2.#

# sigma=sqrt2~~1.414#