Revision as of 00:51, 25 June 2024 by 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...")
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Exercise


ABy Admin
Jun 25'24

Answer

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]] E[X] = \frac{1}{n}\sum_{i=1}^n i = \frac{n(n+1)}{2n} = \frac{n+1}{2} [[/math]]

and

[[math]] E[X^2] = \frac{1}{n}\sum_{i=1}^n i^2 = \frac{(n+1)(2n+1)}{6n}. [[/math]]

Hence the variance equals

[[math]] \frac{(n+1)(2n+1)}{6n} - \left(\frac{n+1}{2} \right)^2. [[/math]]

Setting [math]n=10 [/math] gives 8.25.

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