91Ó°ÊÓ

An integer is chosen randomly, and we have the expected value of the chosen integer.

Let X be the random variable that a number is chosen at random. Here, X has the uniform distribution over the interval \(\left[ {1,100} \right]\) where the probability mass function f(x) is given by

\(\begin{array}{l}f(x) = \frac{1}{{100}},\,1 \le x \le 100\\\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,{\rm{otherwise}}\end{array}\)

The expected value of X denoted by \(E(X)\) is given by

\(\begin{array}{l}E(X) = \sum\limits_x {xf(x)} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_x {x\,\frac{1}{{100}}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{100}} \times \frac{{100(100 - 1)}}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45.5\end{array}\) [ The sum of first n natural numbers is \(\frac{{n(n - 1)}}{2}\) ]

So, the expected value of the integer is 45.5