Question

You have studied the number of people waiting in line at your bank on Friday afternoon at 3 pm for many years, and have created a probability distribution for 0, 1, 2, 3, or 4 people in line. The probabilities are 0.1, 0.3, 0.4, 0.1, and 0.1, respectively. What is the expected number of people (mean) waiting in line at 3 pm on Friday afternoon?

Answer 1

The expected number in this case can be thought of as a weighted average. It is best arrived at by summing the probability of a given number by that number. So, in this case:

`0.1*0 + 0.3*1 + 0.4*2 + 0.1*3 + 0.1*4 = 1.8`

Answer 2

The mean (or expected value or mathematical expectation or, simply, average) is equal to
`P=0.1*0+0.3*1+0.4*2+0.1*3+0.1*4=1.8`

In general, if a random variable `xi` takes values `x_1, x_2,...,x_n` with probabilities, correspondingly, `p_1, p_2,...,p_n`, its mean or mathematical expectation or, simply, average is defined as a weighted sum of its values with weights equal to probabilities it takes these values, that is
`E(xi)=p_1*x_1+p_2*x_2+...+p_n*x_n`

The above is a definition for discrete random variable taking a finite number of values. More complex cases with infinite number of values (countable or uncountable) require involvement of more complex mathematical concepts.

A lot of useful information on this subject can be found on the Web site Unizor by following the menu item Probability.