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2.7 Measures of the spread of the data  (Page 4/25)

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  • 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

  1. Verify the mean and standard deviation on your calculator or computer.
  2. Find the value that is one standard deviation above the mean. Find ( x ¯ + 1s).
  3. Find the value that is two standard deviations below the mean. Find ( x ¯ – 2s).
  4. Find the values that are 1.5 standard deviations from (below and above) the mean.
    • Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.
    • 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.
    • 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.
    • 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.
    • x ¯ = 10.525
    • Use Sx because this is sample data (not a population): Sx=0.715891
  2. ( x ¯ + 1s) = 10.53 + (1)(0.72) = 11.25
  3. ( x ¯ – 2 s ) = 10.53 – (2)(0.72) = 9.09
    • ( x ¯ – 1.5 s ) = 10.53 – (1.5)(0.72) = 9.45
    • ( x ¯ + 1.5 s ) = 10.53 + (1.5)(0.72) = 11.61
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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.

μ = 30.68

s = 6.09
( x ¯ + 2 s ) = 30.68 + (2)(6.09) = 42.86.

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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 [link] , 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 them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.

The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.

Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one ( n – 1). Why not divide by n ? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by ( n – 1) gives a better estimate of the population variance.

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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