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Maximum spread within bounds

StatisticsHard

$5$ real numbers each lie in the interval $[0, 10]$ and have mean $4$ . What is the largest possible population variance (dividing by $5$ )?

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Hint

Variance equals the mean of squares minus the square of the mean; squares are convex, so push values to the ends.

Solution

The total is $5 \times 4 = 20$ , and the variance is the mean of squares minus $4^{2} = 16$ .
The sum of squares is largest when values sit at the endpoints, with at most one value strictly inside.
Taking $2$ values equal to $10$ and $3$ values equal to $0$ gives total $20$ , so this is feasible, with sum of squares $200$ .
The variance is $200 \div 5 - 4^{2} = 24$ , so the answer is $24$ .

Answer: $24$