Standard Deviation - Basic Examples

Basic Examples

Consider a population consisting of the following eight values:

2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9.

These eight data points have the mean (average) of 5:

To calculate the population standard deviation, first compute the difference of each data point from the mean, and square the result of each:

\begin{array}{lll} (2-5)^2 = (-3)^2 = 9 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (7-5)^2 = 2^2 = 4 \\ (4-5)^2 = (-1)^2 = 1 && (9-5)^2 = 4^2 = 16. \\ \end{array}

Next, compute the average of these values, and take the square root:

\sqrt{ \frac{(9 + 1 + 1 + 1 + 0 + 0 + 4 + 16)}{8} } = 2.

This quantity is the population standard deviation; it is equal to the square root of the variance. The formula is valid only if the eight values we began with form the complete population. If they instead were a random sample, drawn from some larger, "parent" population, then we should have divided by 7 (which is n − 1) instead of 8 (which is n) in the denominator of the last formula, and then the quantity thus obtained would have been called the sample standard deviation. See the section Estimation below for more details.

A slightly more complicated real life example, the average height for adult men in the United States is about 70", with a standard deviation of around 3". This means that most men (about 68%, assuming a normal distribution) have a height within 3" of the mean (67"–73") — one standard deviation — and almost all men (about 95%) have a height within 6" of the mean (64"–76") — two standard deviations. If the standard deviation were zero, then all men would be exactly 70" tall. If the standard deviation were 20", then men would have much more variable heights, with a typical range of about 50"–90". Three standard deviations account for 99.7% of the sample population being studied, assuming the distribution is normal (bell-shaped).

Read more about this topic:  Standard Deviation

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