Probability, Mean and Median In the last section, we considered (probability) density functions. We went on to discuss their relationship with cumulative distribution functions. The goal of this section is to take a closer look at densities, introduce some common distributions and discuss the mean and median. Recall, we define probabilities as follows: Proportion of population for Area under the graph of ()which is between and ( ) between and bapxdxxabpxab The cumulative distribution function gives the proportion of the population that has values below t. That is, Proportion of population() ( )having values of below tPt pxdxxt When answering some questions involving probabilities, both the density function and the cumulative distribution can be used, as the next example illustrates. Example 1: Consider the graph of the function p(x). 2 4 6 8 10x0.10.2px Figure 1: The graph of the function p(x) a. Explain why the function is a probability density function. b. Use the graph to find P(X < 3) c. Use the graph to find P(3 § X § 8) 1Solution: a. Recall, a function is a probability density function if the area under the curve is equal to 1 and all of the values of p(x) are non-negative. It is immediately clear that the values of p(x) are non-negative. To verify that the area under the curve is equal to 1, we recognize that the graph above can be viewed as a triangle. Its base is 10 and its height is 0.2. Thus its area is equal to 110 0.2 12. b. There are two ways that we can solve this problem. Before we get started, though, we begin by drawing the shaded region. 2 4 6 8 10x0.10.2px The first approach is to recognize that we can determine the area under the curve from 0 to 3 immediately. The shaded area is another triangle, with a base of 3 and a height of 0.1. Thus, the area is equal to 0.15. A second approach would be to find the equation of the lines that form p(x) and use the integral formula on the previous page. For the first line, notice that the line passes through the points (0, 0) and (6, 0.2). Using the point-slope formula, we see that the line is given by p(x) = (1/30)x. The second line passes through the points (6, 0.2) and (10, 0). Again, using the point-slope formula, we see that the line is given by p(x) = -(1/20)x + 1/2. So, we have that 1301120 2if 0 6() if 6 100otherwixxpx x xse    Returning to the original question, we have that P(X < 3) is given by the integral . On [0, 3), p(x) = (1/30)x. Notice that P(t) = (1/60)t2. So, we have that 30() (3) (0)pxdx P P2211 9( 3)PX(3) (0) (3) (0) 0.1560 60 60P P     . 2c. Again, we have two ways that we can approach this problem. Again, we start by drawing out the shaded region. 2 4 6 8 10x0.10.2px If we want to use triangles, it is easiest to use the fact that the area under the curve is equal to 1. The shaded region is thus equal to one minus the two triangles on the sides. In (b), we found the area of the left triangle is equal to 0.15. The area of the right triangle is equal to 0.1. So, the area of the shaded region is 1 – 0.15 – 0.1 = 0.75. If instead we were to use integrals, notice that p(x) changes functions at x = 6. Thus, in order to compute the integral 83()pxdx, we need to split into two pieces. That is, 868336() () ()pxdx pxdx pxdx. 666233311( ) 0.4530 60pxdx xdx x .888266611 1 1() 0.320 2 40 2pxdx x dx x x      . So, we see that the shaded area is equal to 0.45 + 0.3 = 0.75, which agrees with the answer we found the other way. Often times, we are concerned with finding the “average” value of a distribution. There are two common measured that are used: the mean and the median. The Mean If a quantity has a density function p(x), then we define the mean value of the quantity as ( )xpxdx. 3Example 2: Returning to the density function given in Example 1, compute its mean. Solution: Notice that p(x) changes functions at x = 6. Thus, in order to compute the integral 100()xpxdx, we will need to again split it into two pieces. Thus, we have that the mean is equal to 10 6 10006613320611()30 20 211190 60 4216 176 1690 60 301xp x dx x xdx x x dxxxx    The Median A median of a quantity x distributed through a population is a value T such that half of the population has values of x less than T and half the population has values of x greater than T. That is, T satisfies the equation 1()2Tpxdx where p(x) is the density function of the quantity. In words, we have that half the area under the graph of p(x) lies to the left of T (and half lies to the right of T.) Example 3: Returning to the density function given in Example 1, compute its median. Solution: Looking at Figure 1, notice that more than half of the area occurs in the left side of the triangle. Thus, the median will be a number between 0 and 6. 4Since we do not need to worry about the function changing (since it is the same on the interval [0, 6]), we have that 0130 2Txdx1. That is, 2160 2T. Solving for T, we see that 30T  . Note: We did not use the 30 for T, since we know that T is a positive number. There are a number of important distributions that arise in a variety of situations. Below, we list three such distributions as well as associated properties. The first important distribution we shall consider is the uniform distribution. We introduced this distribution in the previous section. The graph of the density function is constant on the interval [a, b] and zero elsewhere. a bx1b-apx Figure 2: The density of the uniform distribution on [a, b] Uniform Distribution The density of the uniform distribution is given by 1()pxba, for a § x § b The cumulative distribution function is given by () ( )tataPt pxdxba, for a § t § b Another important distribution we shall consider is the exponential distribution. The graph of the density function is characterized by an exponential decay. 5x1px Figure 3: The density of the exponential distribution for c > 0. Exponential