Calculating Means and Medians

We have already seen an example of the concept of a probability density distribution in the context of the probability of decay in a radioactive sample. It is worth reviewing the main properties of such distributions, since they are cntral to the discussion of these pages.



Properties of Probability Density Functions:



If $ p(x) $ is a probability density distribution defined on the interval $ [a,b] $ then
\[ (1)~~~\int_a^b ~p(x)~dx ~=~1 \] \[ (2)~~~p(x) \geq 0 ~~~{\rm for}~~a<x<b \] \[ (3)~~~~~\int_c^d ~p(x)~dx ={\rm probability~that~} c<x<d \]


To paraphrase these properties, (1) means that the total probability that the random variable $ x $ takes values anywhere in its permissible range is defined to be 1. (We can think of this as a convention, or a "polite agreement".) (2) means that we do not attach any meaning to negative probabilities, and (3) is the working definition of what we mean by the probability density. Among other things, this means that we cannot really attach a meaning to the probabillity that x takes on an exact value, but we can talk about the probability that it is within some range of values.

This is slightly different from the way that discrete probabilities are handled, though the two are strongly related. If we were to talk about a discrete random event, such the tossing of a coin or a dice, we would assign an exact numerical probability to each possible outcome. For example, in a fair dice, each face value (1,2,...6) is equally likely, so that each one can be assigned a discrete probability of 1/6. The sum of the discrete probabilities (1/6+1/6 +...1/6) is 1. When we deal with continuous random variables, and probability densities, we can only consider ranges of values of the outcome of an experiment, as discussed in the following examples.



How far can a group of people jump

To illustrate the idea of a continuous random variable, suppose we measure the distances that a group of people can jump at a track and field competition. Suppose that the best jumping record so far is a distance of $ b $ meters, and the weakest jump is a mere $ a $ meters. In fact, suppose that the distance jumped is always between these two values,

$ a \leq x \leq b $
Let us define
\[ F(x) = ~{\rm fraction~of~people~ who~jump~ less~ than ~ x~ meters} \]

Then
\[ p(x) = F'(x) \]

would be the probability density distribution for jumps of a given distance, and among its properties are that
\[ \int_{h_1}^{h_2}~p(x)~dx = F(h_2)-F(h_1) \]

The above is the fraction of people that can jump a distance between $ h_1 $ and $ h_2 $ . Suppose we keep track of all these jumps and find that
\[ p(x) = c \sin(\frac{\pi}{6}x)~~~~0~\leq ~x ~ \leq ~ 6 \]

What can we say about the average performance of this group of people?
To answer this question, we first need to convert the above function to a probability density distribution by finding the value of the constant $ c $ . This process is called Normalization.

Finding the Constant (Normalization)
To find the constant we use the fact that
\[ \int_0^6 ~p(x)~dx = 1 \]

Thus:
\[ c \int_0^6~ \sin(\frac{\pi}{6}x)~dx~=~1 \] \[ c(-\frac{6}{\pi} \cos(\frac{\pi}{6}x))|_0^6 = 1 \] \[ c \frac{12}{\pi} = 1 \]

Thus the constant is $ c=\pi/12 $ , and we now have the (normalized) probability density distribution,
\[ p(x) = \frac{\pi}{12}~ \sin(\frac{\pi}{6}x) \]


Finding the average jump distance
With this probability distribution, we can now caluclulate the average jump distance. (This is also refered to as the mean and the expectation value of x. We use the notation $ {\bar x} $ to denote this average value. Then
\[ {\bar x} = \int_0^6 ~x~p(x)~dx = \frac{\pi}{12}~ \int_0^6 ~x~ \sin(\frac{\pi}{6}x)~dx \]

We calculate that
\[ {\bar x} = \ \frac{\pi}{12}~ [ - \frac{6}{\pi} x \cos( \frac{\pi}{6}x) |_0^6 + \frac{6}{\pi} \int_0^6 ~\cos(\frac{\pi}{6}x)~dx \]

We calculate that
\[ {\bar x} = \ \frac{\pi}{12}~ [ \frac{36}{\pi} + + \frac{36}{\pi^2}~\sin(\frac{\pi}{6}x)|_0^6 = 3 \]

Thus, the average jump distance is three meters. This is not too surprising given the fact that the probability density distribution is a function which is symmetric about $ x=3 $ . We will see later that for this distribution, the average (mean) coincides with another "typical value" called the median.

Probability of Jumping more than 5 meters
We can compute this probability simply by calculating the integral
\[ I=\frac{\pi}{12}~ \int_5^6 ~\sin(\frac{\pi}{6}x)~dx \]

The result is:
\[ I=\frac{\pi}{12}~ \frac{6}{\pi} \cos(\frac{\pi}{6}x)|_5^6 = \frac{1}{2}~ (\cos(\frac{5 \pi}{6}) +1) = 0.067 \]




For your consideration:




Means and Medians

There is more than one way to define a typical value in the results of an experiment (such as the jumping contest above). We have already discussed the mean or average value. Below, we will also introduce the idea of a median. Essentially, the median is a value "in the middle of the distribution" such that it is equally probable to score better or worse. We will make this idea more precise in the example below. Note that this example will involve a probability distribution that is not symmetric. (When the distribution is symmetric, as in our first example, the mean and the median are at the same place.
Consider the probability density distribution
\[ p(x) = c(1-x) \]

Then we might note that
  • The distribution has the shape of a straight line.

  • It is positive only for $ 0 \leq x \leq 1 $ . This will be the range over which we consider the distribution.

  • The constant $ c $ is not yet specified, but it must have a value such that the area under the curve of $ p(x),~~ 0 \leq x \leq 1 $ is 1.

The last constraint means that
\[ \int_0^1~p(x)~dx = c \int_0^1~ (1-x)~dx = 1 \]

You should be able to conclude that $ c=2 $ .
If we compute the mean of the distribution, we find that
\[ {\bar x}=\int_0^1~x~p(x)~dx = 2 \int_0^1~ x~(1-x)~dx = 2 \int_0^1~ (x-x^2)~dx \]

We find that
\[ {\bar x}= 2 [\frac{x^2}{2}-\frac{x^3}{3}]|_0^1 =2 [\frac{1}{2}-\frac{1}{3}] = \frac{1}{3} \]

Thus the mean is at $ {\bar x} = 1/3 $

The median is the value $ m $ such that
\[ \int_0^m~~p(x)~dx = \int_m^1~~p(x)~dx = \frac{1}{2} \]

To find the median, we calculate the above integral and then determine m as follows:
\[ \frac{1}{2}~=~\int_0^m~~p(x)~dx = \int_0^m~~2~(1-x)~dx = 2(x - \frac{x^2}{2})|_0^m = 2 ~(m- \frac{m^2}{2}) \]

Thus
\[ 2 ~(m- \frac{m^2}{2})={1}{2} \]

We need to solve this equation for tha value of m. Rearranging leads to a quadratic equation in standard form:
\[ m^2-2m+\frac{1}{2} = 0 \]

The result is then
\[ m ~=~ \frac{2 \pm \sqrt{2}}{2} = 1 \pm 0.7071 \]

Only one of these values, namely m=0.2929 is inside the required range (in the interval [0,1]), and so is the value we are looking for. The median in this case is not at the same location as the mean, but rather slightly to the left of it. It denotes a value which subdivides the area under the graph of $ p(x) $ into two equal portions, each one equal to 1/2. For a probability distribution as given here, it is equally likely to score a value smaller or larger than 0.2929.





For your consideration: