JAMB 1999 · UME · Q44

If the standard deviation of the numbers 3, 6, $x$ , 7, 5 is $\sqrt2$ , find the least possible value of $x$ .

Worked solution (try it first)

Use variance $= \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$ with $n = 5$ , $\sum x = 21 + x$ and $\sum x^2 = 119 + x^2$ .
The variance is 2, so multiply through by 25: $5(119 + x^2) - (21 + x)^2 = 50$ .
Expand and collect terms: $4x^2 - 42x + 104 = 0$ , which is $2x^2 - 21x + 52 = 0$ , or $(x - 4)(2x - 13) = 0$ .
So $x = 4$ or $x = 6\frac12$ .
The least value is 4, option C.