# Data freq deviations deviations 2 freq deviations 2 x

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Data Freq. Deviations Deviations 2 (Freq.)( Deviations 2 ) x f ( x ) ( x ) 2 ( f )( x ) 2 9 1 9 – 10.525 = –1.525 (–1.525) 2 = 2.325625 1 × 2.325625 = 2.325625 9.5 2 9.5 – 10.525 = –1.025 (–1.025) 2 = 1.050625 2 × 1.050625 = 2.101250 10 4 10 – 10.525 = –0.525 (–0.525) 2 = 0.275625 4 × 0.275625 = 1.1025
Data Freq. Deviations Deviations 2 (Freq.)( Deviations 2 ) 10.5 4 10.5 – 10.525 = –0.025 (–0.025) 2 = 0.000625 4 × 0.000625 = 0.0025 11 6 11 – 10.525 = 0.475 (0.475) 2 = 0.225625 6 × 0.225625 = 1.35375 11.5 3 11.5 – 10.525 = 0.975 (0.975) 2 = 0.950625 3 × 0.950625 = 2.851875 The total is 9.7375 Table2.29 The sample variance, s 2 , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s 2 = 9.737520−1 =0.5125 s2=9.737520−1=0.5125 The sample standard deviation s is equal to the square root of the sample variance: s=0.5125−−−−−√=0.715891, s=0.5125=0.715891, which is rounded to two decimal places, s = 0.72. Typically, you do the calculation for the standard deviation on your calculator or computer . The intermediate results are not rounded. This is done for accuracy. For the following problems, recall that value = mean + (#ofSTDEVs) (standard deviation) . Verify the mean and standard deviation or a calculator or computer. For a sample: x = x¯ + (#ofSTDEVs)( s ) For a population: x = μ + (#ofSTDEVs)( σ ) For this example, use x = x¯ + (#ofSTDEVs)( s ) because the data is from a sample a. Verify the mean and standard deviation on your calculator or computer. b. Find the value that is one standard deviation above the mean. Find ( x¯ + 1s). c. Find the value that is two standard deviations below the mean. Find ( x¯ – 2s). d. Find the values that are 1.5 standard deviations from (below and above) the mean. Solution 2.32 A. USING THE TI-83, 83+, 84, 84+ CALCULATOR o Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2. o Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down. o Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
o Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER. o x¯ = 10.525 o Use Sx because this is sample data (not a population): Sx=0.715891 b. ( x¯ + 1s) = 10.53 + (1)(0.72) = 11.25 c. ( x¯ – 2 s ) = 10.53 – (2)(0.72) = 9.09 d. o ( x¯ – 1.5 s ) = 10.53 – (1.5)(0.72) = 9.45 o ( x¯ + 1.5 s ) = 10.53 + (1.5)(0.72) = 11.61 TRY IT 2.32 On a baseball team, the ages of each of the players are as follows: 21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40 Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean. Explanation of the standard deviation calculation shown in the table The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero . (For Example 2.32 , there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make
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