Question

Let X be a random variable that is uniformly chosen from the set of a positive odd number less than 100. The expectation E[x], is

Answer 1

Concept:

E[x] = Mean of Random variable X

X = discrete random variable

The expectation is given by the formula:

\(E\left[ x \right] = \mathop \sum \limits_{i = 1}^n {x_i}\;P\left( {{x_i}} \right)\) 

x_i = i^th random variable

P(x_i) = Probability of x_i

Calculation:

X: set off positive odd numbers less than 100, i.e.

X: {1, 3, 5, 7, 9, 11, …, 93, 95, 97, 99}

Since X is uniformly chosen, the probability of each random variable will be:

\(P\left( {{x_i}} \right) = \frac{1}{n}\) 

Where n = total number of random variables

Here, n = number of odd numbers less than 100

n = 50

\(P\left( {{x_i}} \right) = \frac{1}{{50}}\) 

\(E\left[ x \right] = \mathop \sum \limits_{i = 1}^n {x_i}\left( {P\left( {{x_i}} \right)} \right) = \mathop \sum \limits_{i = 1}^n {x_i}\left( {\frac{1}{{50}}} \right)\) 

\(E\left[ x \right] = \frac{1}{{50}}\left[ {1 + 3 + 5 + 7 \ldots + 93 + 95 + 97 + 99} \right]\) 

Clearly an AP is formed with n = 50, a = 1 and l = 99.

The sum of an AP is given by the formula:

\(S = \frac{n}{2}\;\left\{ {a + l} \right\}\) 

\(E\left[ {{x_i}} \right] = \frac{1}{{50}}\left[ {\frac{{50}}{2}\left\{ {1 + 99} \right\}} \right]\) 

\( = \frac{1}{2} \times 100\) 

E[x_i] = 50