The average and standard deviation o

S.D = \(\sqrt{\frac{\sum(x-m)^2}{n}}\); where x is individual element, m is average/mean, n is the number of elements

2 = \(\sqrt{\frac{(x_1-5)^2+(x_2-5)^2+(x_3-5)^2+(x_4-5)^2}{4}}\)
squaring both sides,
4 = \(\frac{(x_1-5)^2+(x_2-5)^2+(x_3-5)^2+(x_4-5)^2}{4}\)
16 =\((x_1-5)^2+(x_2-5)^2+(x_3-5)^2+(x_4-5)^2\)
16 =\({x_1}^2+25-10*x_1+{x_2}^2+25-10*x_2+{x_3}^2+25-10*x_3+{x_4}^2+25-10*x_4\)
16 =\(({x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2)+100-10*(x_1+x_2+x_3+x_4)\)
16 =\(({x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2)+100-10*(20)\);sum of the four numbers is 20
116 = \({x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2\)
\(\frac{{x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2}{4}=\frac{116}{4}\)=29
I think option D needs to be changed to 29.