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Operations Research Models and Methods
Computation Section
Subunit Continuous Distributions
 - Beta

The Beta distribution is important because it has a finite range, from 0 to 1, making it useful for modeling phenomena that cannot be above or below given values. The distribution has two parameters, and , that determine its shape. When and are equal, the distribution is symmetric. Increasing the values of the parameters decreases the variance. The symmetric case is illustrated in the figure below.

The Uniform distribution is the special case of the Beta distribution with and both equal to 1.



  When is less than the distribution is skewed to the right as shown below. When is greater than the distribution is skewed to the left. When = 1 and = 2, the Beta distribution is a triangular distribution with the mode at 0.


Generalized Beta

A linear transformation of a Beta random variable provides a random variable with an arbitrary range. The distribution is often used when an expert provides a lower bound, a, upper bound, b, and most likely value, m, for the time to accomplish a task. A transformed Beta variable could be used to represent the time to complete a task in a project. The transformed distribution is called, the Generalized Beta. It's mean, variance and mode are shown at the left.




Example: Consider the following hypothetical situation. Historical grades in a senior engineering class indicate that on average 27% of the students received an A. There is variation among classes, however, and the proportion must be considered a random variable. From past data we have measured a standard deviation of 15%. The current class has 50% A grades. What is the probability that a proportion of 50% or higher would be obtained from the population of all exams?

We model the proportion of A grades with a Beta distribution because proportions naturally fall within the range [0,1].

The data estimates the mean proportion as 0.27. The standard deviation is specified as 0.15 (variance 0.0225). We use this data to estimate and . We solve for as a function of and . Substituting = 0.27 into this expression, we obtain = 2.7037.


To solve this problem we created the Beta model as shown at the ABOVE. We use a mathematical expression in the cell for . We then use the Excel Solver to find the value of that results in a variance of 0.0225. The form shows the results of the Solver analysis. The probability that the proportion is 50% or higher is about 8%. The current class is exceptional.

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Operations Research Models and Methods
by Paul A. Jensen
Copyright 2004 - All rights reserved

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