Solution
five consecutive positive integers = 1,2,3,4,5
Mean = 3
Standard deviation = $$\sqrt{\frac{\left(\ \left(3-1\right)^2+\left(3-2\right)^2+\left(3-3\right)^2+\left(3-4\right)^2+\left(3-5\right)^2\right)}{5}}$$ = $$\sqrt{\ 2}$$
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