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| rev | Admin | (Created page with "'''Solution: A''' Say a number is chosen at random from the integers <math>1,\ldots,n</math> and denote the outcome by <math>X</math>. Then <math display = "block"> E[X] = \frac{1}{n}\sum_{i=1}^n i = \frac{n(n+1)}{2n} = \frac{n+1}{2} </math> and <math display = "block"> E[X^2] = \frac{1}{n}\sum_{i=1}^n i^2 = \frac{(n+1)(2n+1)}{6n}. </math> Hence the variance equals <math display = "block"> \frac{(n+1)(2n+1)}{6n} - \left(\frac{n+1}{2} \right)^2. </math> Setting...") | Jun 25'24 at 1:51 | +505 |