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?

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Carl Swisher · Accepted Answer

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

nghodlokl · Accepted Answer

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 ξ takes values x1,x2,…,xn with probabilities, correspondingly,  p1,p2,….,pn, 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(ξ)=p1⋅x1+p2⋅x2+…+pn⋅xn  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.

You have studied the number of people waiting in line at your bank on Friday afternoon at 3 pm for m

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