Find the positive value of x if the standard deviation of the numbers 1, x + 1 is 6

Find the positive value of x if the standard deviation of the numbers 1, x + 1 is 6

Answer Details

mean (x) = $\frac{1+x+1+2x+1}{3}$
= $\frac{3x+3}{3}$
= 1 + x
$\begin{array}{ccc}X& (X-X)& (X-X{)}^2\\ 1& -x& x^2\\ x+1& 0& 0\\ 2x+1& x& x^2\\ & & 2x^2\end{array}$
S.D = $\sqrt{\frac{\sum (x-7{)}^2}{\sum f}}$
= ${\sqrt{\left(6\right)}}^2$
= $\frac{2x^2}{3}$
= 2x2
= 18
x2 = 9
∴ x = $\pm$ $\sqrt{9}$
= $\pm$3