How is it that the Mean of a set of values is related to the Standard Deviation of that set?

Answer 1

If you're in a hurry, then skip to the end, but know that this explanation is very thorough, and may help you remember why the relationship is what it is. Remembering that will help you to remember the relationship itself, and will even help you to derive that relationship if you should forget it.

The Mean is the average value of a set of values, found by dividing the sum of all values in the set by the total number of values in the set.

For example, the average of the set of 5 numbers, {13, 17, 15, 16, 20}, would be (13+17+15+16+20) divided by 5, the total number of values in the set.

That would be 81 divided by 5, which would be 16.2.

Thus, the average value of the numbers in the set is 16.2, even though none of them is exactly 16.2.

To find the Standard Deviation, you need the average value of the numbers in the set (the Mean), as well as the total number of values in the set, and the values themselves.

Because the Standard Deviation is the average distance of the actual values from the Mean, we must first find the Difference between the actual values and the Mean.

We do this by subtracting the Mean from each of the actual values, but, because some of the actual values are lower than the Mean, this will produce some negative numbers, and, if you think about the Difference as Distance, then you will realize that the Differences should not be negative, as Distance cannot be negative.

13 - 16.2 = -3.2

17 - 16.2 = 0.8

15 - 16.2 = -1.2

16 - 16.2 = -0.2

20 - 16.2 = 3.8

The Standard Deviation would be artificially narrowed if you calculated the average Difference between the actual values and the Mean using negative Differences, because the negative Differences would cancel out the positive Differences, so that when you calculated the average Difference, your numerator, the sum of distances from the Mean, would be artificially small.

If we tried to treat negative Differences the same way that we treat positive distances, we would see that something goes horribly wrong. The negative Differences completely cancel out the positive Differences, leaving us with zero.

(-3.2) + (0.8) + (-1.2) + (-0.2) + (3.8) =

(0.8) + (3.8) + (-1.2) + (-0.2) + (-3.2) =

0.8 + 3.8 - 1.2 - 0.2 - 3.2 =

4.6 - 4.6 = 0

This makes sense, because the Mean is basically the point around which all actual values in a set are balanced.

The fact that some Differences are negative only confirms that some values in the set are lower than the Mean, and others are greater than the Median.

In this case, the negative signs indicate direction, but distance itself is always positive, so we need to find a way to make all of these Differences positive.

13 - 16.2 = -3.2

17 - 16.2 = 0.8

15 - 16.2 = -1.2

16 - 16.2 = -0.2

20 - 16.2 = 3.8

We find that we are able to do this by squaring the original Differences, as the square of any real number is always positive.

(-3.2)2 = 10.24

(0.8)2 = 0.64

(-1.2)2 = 1.44

(-0.2)2 = 0.04

(3.8)2 = 14.44

We now have the positive squares of the Differences, and so we can now find the average of those squares by adding them and then dividing their sum by the number of values in their set, which is the same as the number of values in the original set, which, in this case, is 5.

10.24 + 0.64 + 1.44 + 0.04 + 14.44 = 26.8

26.8 / 5 = 5.36

Remember that we had to square the Differences in order to have positive numbers to work with, so, what we have here is the average squared Difference between the values of the original set and the Mean. Therefore, in order to find the average Difference, we must take the square root of the average squared Difference.

SQRT (5.36) = 2.3152

This gives us the average distance between values in a set and the Mean of that set, and we know that distance to mean the same thing as the Standard Deviation, so, the Standard Deviation in this example is 2.3152.

In short, the relationship between the Mean and the Standard Deviation can be expressed as follows:

The Standard Deviation is the Square Root of the Average of the Squares of all the Differences between values in a set and the Mean of that set.

SQRT ( ( (V1 - Mean)2 + (V2 - Mean)2 + ... + (Vn- Mean)2 ) / (n) )

SQRT = take the Square Root of the following expression in parentheses

V = a value in a set, such as, the 1st, 2nd, and nth value in a set

n = the number of values in that set