JAMB 1998 · UME · Q45

Find the positive value of $x$ if the standard deviation of the numbers $1$ , $x + 1$ , $2x + 1$ is $\sqrt6$ .

Worked solution (try it first)

The three numbers add up to $3x + 3$ , so the mean is $\frac{3x + 3}{3} = x + 1$ .
The deviations are $-x$ , 0 and $x$ , so the variance is $\frac{x^2 + 0 + x^2}{3} = \frac{2x^2}{3}$ .
The standard deviation is $\sqrt6$ , so the variance is 6.
Then $\frac{2x^2}{3} = 6$ gives $x^2 = 9$ .
The positive value is $x = 3$ , option C.