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.