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If an integer between 1 and 100 is to be chosen at random, what is the expected value?

Short Answer

Expert verified

The expected value of the integer is 45.5

Step by step solution

01

Given information

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

02

Compute the expected value

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

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Suppose that a random variable X has a continuous distribution with the p.d.f. has given in Example 4.1.6. Find the expectation of 1/X

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Suppose that X, Y , and Z are nonnegative random variables such that\({\rm P}\left( {X + Y + Z < 1.3} \right) = 1\). Show that X, Y , and Z cannot possibly have a joint distribution under which each of their marginal distributions is the uniform distribution on the interval\(\left( {0,1} \right)\).

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