Mathematics The text defines standard deviation as a measure of how spread out a data set is and then discusses the following aspects of standard deviation: Calculating standard deviation Interpreting the standard deviation of normally distributed data Relating standard deviation to the concavity of the normal curve Here are two reasons statisticians don’t generally use mean absolute deviation—that is, just finding the average deviation from the mean: Mean absolute deviation involves the use of absolute value, which is difficult to work with in calculus. Squaring differences from the mean and then taking a square root at the end is computationally easier than using absolute value. Mean absolute deviation is more suitable if the median is used as the “central value” instead of the mean.

Discussing the Reference Pages You may want to have students read over this material as part of their homework. Define standard deviation using the example from the reference pages: the numbers 5, 8, 10, 14, and 18. (You may want to copy and hand out these pages so students can write on them.) Calculation of Standard Deviation Have students follow along with the steps and example in the student book as you demonstrate how to calculate the standard deviation. Find the mean. Find the difference between each data item and the mean. Square each of the differences. Find the average (mean) of these squared differences. Take the square root of this average. The computation of the mean is shown below the table to the left. Students may choose to ignore the sign of each difference in step 2, in effect using the absolute value of the difference rather than the difference itself. Since the differences are squared in step 3, their signs do not affect the final result. You may want to bring this out to emphasize the similarity between standarddeviation and meanabsolutedeviation. Below the table to the right, step 4 of the computation of standard deviation is broken down into substeps: (1) adding the squares of the differences and (2) dividing by the number of data items. x x - x ˉ size 12{ { bar {x}}} {} (x - x ˉ size 12{ { bar {x}}} {} )² 5 –6 36 8 –3 9 10 1 1 14 3 9 18 7 49

sum of data items = 55 number of data items = 5 sum of data squared differences = 104 mean of the squared difference = 2 - .8 x ˉ = size 12{ { bar {x}} {} 55 ÷ 5 = 11 σ(standard deviation) = 20 . 8 = 4 . 6 size 12{s \( s"tan""dard" ital "deviation" \) = sqrt {"20" "." 8} =4 "." 6} {}

If students are comfortable with the summation notation presented in Kai and Mai Spread Data in the formula for mean absolute deviation, show them that the definition of standard deviation involves just two changes to that expression: Replacing ∣xi−xˉ∣ size 12{ lline x rSub { size 8{i} } - { bar {x}} rline } {} with xi−xˉ2 size 12{ left (x rSub { size 8{i} } - { bar {x}} right ) rSup { size 8{2} } } {} Taking the square root of the final expression Thus, the formula for standard deviation is ∑ i = 1 n x i − x ˉ 2 n size 12{ sqrt { { { Sum cSub { size 8{i=1} } cSup { size 8{n} } { left (x rSub { size 8{i} } - { bar {x}} right )} rSup { size 8{2} } } over {n} } } } {} The symbol usually used for standard deviation is the lowercase form of the Greek letter sigma, written σ. Mention this symbol, as students will be looking for it on their calculators. Remind the class that the uppercase Greek sigma, , is the summation symbol. (The issue of the distinction between σ and s—the sample standard deviation—is discussed later.) Standard Deviation and the Normal Distribution Explain that the following facts hold true whenever a set of data is normally distributed: Approximately 68% of all results will be within one standard deviation of the mean. Approximately 95% of all results will be within two standard deviations of the mean. Illustrate these facts using the areas shown in the next diagram, which also appears in the student book.

The darkly shaded area stretches from one standard deviation below the mean to one standard deviation above the mean and is approximately 68% of the total area under the curve. The two lightly shaded areas represent data between one and two standard deviations from the mean. The total shaded area stretches from two standard deviations below the mean to two standard deviations above the mean and comprises approximately 95% of the total area under the curve. Explain that the standard deviation provides a good rule of thumb for deciding whether something is “rare.”