Tutorial: Measures of Dispersion

8.4: Measures of Dispersion

Previous tutorial: Part A: Variance and Standard Deviation of a Set of Scores

This tutorial: Part B: Variance and Standard Deviation of a Random Variable

Based on Section 8.4 in Finite Mathematics and Finite Mathematics and Applied Calculus

Variance and Standard Deviation of a Random Variable

In Part A of this tutorial, We calculated the population variance by taking the mean of the quantities (x_i - )², where the x_i are the values of a random variable X. Put another way, the population variance of a set of data can be written as the mean of all the values of (X - )², which is the same as the expected value of (X - )². We can calculate this expected value as follows:

Variance and Standard Deviation of a Random Variable

If X is a finite random variable taking on values x₁, x₂, . . . , x_n, then the variance of X is

(x₁-)²P(X = x₁) + (x₂-)²P(X = x₂) + . . . + (x_n-)²P(X = x_n)

The standard deviation of X is then the square root of the variance,, and is written as

The best way to compute the variance and standard deviation) from the probability distribution of X is to use a tabular approach similar to the one we used in the preceding tutorial. Here is an example that illustrates this method:

Example

Let us start with the distribution we looked at in the tutorial for Section 8.3:

x	0	1	2	3	4
P(X = x)	.1	.2	.4	.2	.1

First, we get the expected value as before:

x	0	1	2	3	4
P(X = x)	.1	.2	.4	.2	.1
x.P(X = x)	0.1 = 0	1.2 = .2	2.4 = .8	3.2 = .6	4.1 = .4

So, = Sum of numbers in the third row = 0 + .2 + .8 + .6 + .4 = 2. (So far, we have done nothing new.)

Now comes the new part: we need to find the expected value of the square distance from the mean, so we add three new rows:

x	0	1	2	3	4
P(X = x)	.1	.2	.4	.2	.1
x.P(X = x)	0	.2	.8	.6	.4
x-	0 - 2 = -2	1 - 2 = -1	2 - 2 = 0	3 - 2 = 1	4 - 2 = 2
(x- )²	(-2)² = 4	(-1)² = 1	0² = 0	1² = 1	2² = 4
(x- )².P(X = x)	4 .1 = .4	1 .2 = .2	0 .4 = 0	1 .2 = .2	4 .1 = .4

Then ² = Sum of numbers in the bottom row = .4 + .2 + 0 + .2 + .4 = 1.2. The standard deviation is its square root: 1.0954.

Note that we got the bottom row by multiplying the probabilities (Row 2) by the square deviations from the mean (Row 5).

Here is one for you to do. It is really a continuation of the one you did in the tutorial for Section 8.3.

Distribution of X:

x	50	150	250	350	450
P(X = x)	.1	.3	.2	.4	0

You already computed the the mean in the earlier tutorial, so we are giving you that part and asking you to fill out the rest of the table yourself.

Note: For instance, 10,000 should be entered as 10000.

Q Why didn't we divide by n or n-1 as we did for sets of scores?
A Look at how we computed the expected value . Instead of adding and dividing by n -- which is wrong unless each of the values of X is equally likely -- we multipled each value of X by its probability and then added the results. Similarly, to get the variance, which is the expected value of the square deviation from the mean, we multiply each square deviation by its probability and add.

Q Do the Empirical Rule and Chebyshev's Rule still hold for random variables?
A Yes. For instance, in the first example above, the distribution looks bell shaped and symmetric (draw its graph to see) and so, by the Empirical Rule, there is an approximately .68 probability that X will lie within one standard deviation of the mean. That is, between

Indeed, if you add up the probabilities for these val;ue of X, you get .2 + .4 + .2 = .8. This is off by about .1. In general, we cannot expect to get accurate results from the Empirical Rule unless there are lots of values of X. Put another way, how can we expect to get a bell curve from just 5 values?

For notes on the expected value and standard deviation of a binomial random variable, consult the textbook.

Now try some of the exercises in Section 8.4 of Finite Mathematics and Finite Mathematics and Applied Calculus.

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