Find the positive value of x if the standard deviation of the numbers 1, x + 1 is 6
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
mean (x) = \(\frac{1 + x + 1 + 2x + 1}{3}\)
= \(\frac{3x + 3}{3}\)
= 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 = \(\sqrt{\frac{\sum(x - 7)^2}{\sum f}}\)
= \(\sqrt{(6)}^2\)
= \(\frac{2x^2}{3}\)
= 2x2
= 18
x2 = 9
∴ x = \(\pm\) \(\sqrt{9}\)
= \(\pm\)3
mean (x) = \(\frac{1 + x + 1 + 2x + 1}{3}\)
= \(\frac{3x + 3}{3}\)
= 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 = \(\sqrt{\frac{\sum(x - 7)^2}{\sum f}}\)
= \(\sqrt{(6)}^2\)
= \(\frac{2x^2}{3}\)
= 2x2
= 18
x2 = 9
∴ x = \(\pm\) \(\sqrt{9}\)
= \(\pm\)3