# 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?

Dec 17, 2014

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:

$0.1 \cdot 0 + 0.3 \cdot 1 + 0.4 \cdot 2 + 0.1 \cdot 3 + 0.1 \cdot 4 = 1.8$

Dec 17, 2014

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

In general, if a random variable $\xi$ takes values ${x}_{1} , {x}_{2} , \ldots , {x}_{n}$ with probabilities, correspondingly, ${p}_{1} , {p}_{2} , \ldots , {p}_{n}$, 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
$E \left(\xi\right) = {p}_{1} \cdot {x}_{1} + {p}_{2} \cdot {x}_{2} + \ldots + {p}_{n} \cdot {x}_{n}$

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.