Find the positive value of x if the standard deviation of the numbers 1, x + 1, 2x + 1 is \(\sqrt6\)
- 1
- 2
-
3
- 4
The correct answer is: C
Explanation
mean (x̄) = \(\frac{1 + x + 1 + 2x + 1}{3}\)
= \(\frac{3x + 3}{3}\)
= 1 + x
\(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline 1 & -x & x^2 \\ x + 1 & 0 & 0\\2x + 1 & x & x^2\\ \hline & & 2x^2\end{array}\)
S.D = \(\sqrt{\frac{\sum(x - x)^2}{n}}\) = \(\sqrt{(6)}\) ( given )
\(\sqrt{(6)}\) = \(\sqrt{\frac{2x^2}{3}}\)
cancel square roots and cross multiply
\(2x^2\) = 18
divide both sides by 2
\(x^2\) = 9
∴ x = \(\pm\) \(\sqrt{9}\)
= \(\pm\)3