108 CHAPTER 3 Describing, Exploring, and Comparing Data CP EXAMPLE 3 Calculating Standard Deviation with Formula 3-5 Use Formula 3-5 to find the standard deviation of these ”Space Mountain” wait times (minutes) from Example 1: 50, 25, 75, 35, 50, 25, 30, 50, 45, 25, 20. SOLUTION Here are the components needed in Formula 3-5. n = 11 (because there are 11 values in the sample) Σx = 430 (found by adding the original sample values) Σx2 = 19,550 (found by adding the squares of the sample values, as in 502 + 252 + 752 + 352 + 502 + 252 + 302 + 502 + 452 + 252 + 202 = 19,550) Using Formula 3-5, we get s = Cn1Σx22-1Σx22 n1n-12 = C11119,5502-143022 11111-12 = C 30,150 110 = 16.6 minutes The result of s = 16.6 minutes is the same as the result in Example 2. (Because Formulas 3-4 and 3-5 are equivalent, those two results should always be the same.) YOUR TURN. Find the standard deviation in Exercise 7 “Celebrity Net Worth.” Where Are the 0.400 Hitters? The last baseball player to hit above 0.400 was Ted Williams, who hit 0.406 in 1941. There were averages above 0.400 in 1876, 1879, 1887, 1894, 1895, 1896, 1897, 1899, 1901, 1911, 1920, 1922, 1924, 1925, and 1930, but none since 1941. Are there no longer great hitters? The late Stephen Jay Gould of Harvard University noted that the mean batting average has been steady at 0.260 for about 100 years, but the standard deviation has been decreasing from 0.049 in the 1870s to 0.031, where it is now. He argued that today’s stars are as good as those from the past, but consistently better pitchers now keep averages below 0.400. T b h w l 0 T Range Rule of Thumb for Understanding Standard Deviation The range rule of thumb is a crude but simple tool for understanding and interpreting standard deviation. It is based on the principle that for many data sets, the vast majority (such as 95%) of sample values lie within 2 standard deviations of the mean. We could improve the accuracy of this rule by taking into account such factors as the size of the sample and its distribution, but here we sacrifice accuracy for the sake of simplicity. The concept of significance that follows will be enhanced in later chapters, TABLE 3-4 General Procedure for Finding Standard Deviation with Formula 3-4 Specific Example Using These Sample Values: 50, 25, 75, 35, 50, 25, 30, 50, 45, 25, 20 Step 1: Compute the mean x. The sum of 50, 25, 75, 35, 50, 25, 30, 50, 45, 25, 20 is 430; therefore: x = Σx n = 50+25+75+35+50+25+30+50+45+25+20 11 = 430 11 = 39.1 Step 2: Subtract the mean from each individual sample value. [The result is a list of deviations of the form 1x - x2.] Subtract the mean of 39.1 from each sample value to get these deviations away from the mean: 10.9, -14.1, 35.9, -4.1, 10.9, -14.1, -9.1, 10.9, 5.9, -14.1, -19.1. Step 3: Square each of the deviations obtained from Step 2. [This produces numbers of the form 1x - x22.] The squares of the deviations from Step 2 are: 118.81, 198.81, 1288.81, 16.81, 118.81, 198.81, 82.81, 118.81, 34.81, 198.81, 364.81. Step 4: Add all of the squares obtained from Step 3. The result is Σ1x - x22. The sum of the squares from Step 3 is 2740.91. Step 5: Divide the total from Step 4 by the number n - 1, which is 1 less than the total number of sample values present. With n = 11 data values, n - 1 = 10, so we divide 2740.91 by 10 to get this result: 2740.91 10 = 274.091. Step 6: Find the square root of the result of Step 5. The result is the standard deviation, denoted by s. The standard deviation is 2274.091 = 16.556. Expressing the result with one more decimal place than the original data, we get s = 16.6 minutes.
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