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What level of course is your son taking? Here is a useful site which contains information about a lot of a lot of topics.

http://www.statsoftinc.com/textbook/stathome.html

Here is an elementary discussion.

http://math.about.com/gi/dynamic/off...criptive2.html

If your son is just beginning his study, and wants to understand what the variance and standard deviation are all about, that's easy.

Here is a "population" of five numbers: 9, 10, 10 10, 11. The average (arithmetic mean) is 10. So far so good.

Here is another population: 0, 5, 10, 15, 20. The average of this population is 10, too.

So how are the two populations different, if they both have the same average? Well, in the first population, all the numbers are close together. They are all very close to the mean. In the second population they are spread apart. Some of them are a long way from the average. Thevariancemeasures, "how far from the mean are these numbers?"

So for each number, we look and see how far from average it is. The numbers 9, 10, 10, 10, 11 are, respectively, -1 unit, 0 units, 0 units, 0 units, and 1 unit away from the average of 10. Because we don't really care about the + or - sign, we square these differences, then take the average. That's the variance:

variance = [(-1)^2 + 0^2 + 0^2 + 0^2 + 1^2] / 5 = (1+0+0+0+1)/5 = 0.4

Now to make up for having squared these numbers, we have to take the square root to get the units to come out right (inches instead of square inches, etc.) This gives thestandard deviation:

Standard deviation (sigma) = square root (0.4) = 0.63.

We say informally that the typical number in this list is about 10 +/- 0.63.

So the formula is

st. dev. (denoted by "sigma") = Square root [((x1-mean)^2 + (x2 - mean)^2 + ... + (xn-mean)^2) / N]

where N is the size of the population, and thevarianceis the same thing without the square root.

For the second population, the numbers are 0, 5, 10, 15, 20. Comparing them to the mean, they are off by -10, -5, 0, 5, and 10, respectively. Square these numbers, add them up, divide by 5, take the square root and we get a standard deviation of7.07.

So now we have quantified the difference between the two populations. Although they both have the same average (10), the standard deviation is 0.63 for the first, showing that they are clustered closely about the mean, while for the second population the standard deviation is 7.07. This says that the typical member of the second population is about 7 units away from the average.

Application: If the data arenormally distributedthen we expect about 68% of the data to be within plus or minus one standard deviation of the mean, and we expect about 95% of the data to be within +/- 2 standard deviations of the mean. So for our first population we expect about 68% of the data to be in the range 10 +/- 0.63 -- that is, between 9.37 and 10.63 -- and we expect 95% of the data to be in the range 10 +/- (2*0.63), or within the interval 8.74 and 11.26.

All of this supposes that we have the whole population of numbers before us. In practice, we do not have access to every number in the whole population, and even if we did, that would be too many numbers to work with. So we estimate the statistical features of the whole population by taking a sample. Like when we take a political poll of 500 likely voters to try to predict how the whole population of 40 million are going to vote.

To compute the mean, variance and standard deviation of asampleeverything is the same except that we divide by n-1 instead of by n when we compute the variance and standard deviation. The reason for this is quite subtle and has to do with the concept of bias and with "degrees of freedom" -- basically, we lose a little accuracy whenever we use a sample statistic to estimate a real population statistic in the real world, so we have to take that into account in our formulas.

So if the first set of numbers, {0, 5, 10, 15, 20} is not the whole population but just a small sample of five things chosen at random from a larger set, then the standard deviation is

s = square root (((x1-mean)^2 + (x2 - mean)^2 + ... + (xn-mean)^2) /n-1)

= square root ((100+25+0+25+100)/4)

= 7.9,

instead of 7.07. In practice it is almost always a sample that we are working with, but on the other hand except for very small samples the difference between the two formulas is not worth bothering about.

The standard deviation has many uses besides just as a descriptive measure of variation. For instance, if you want to estimate the mean of a population by knowing the mean of a sample, a95% confidence intervalfor the true mean is given by the formula,

true mean = sample mean +/- 1.96 sigma/(square root of n),

where sigma is the standard deviation.

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