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Data Frequency Relative Frequency Cumulative Relative Frequency
33 1 0.032 0.032
42 1 0.032 0.064
49 2 0.065 0.129
53 1 0.032 0.161
55 2 0.065 0.226
61 1 0.032 0.258
63 1 0.032 0.29
67 1 0.032 0.322
68 2 0.065 0.387
69 2 0.065 0.452
72 1 0.032 0.484
73 1 0.032 0.516
74 1 0.032 0.548
78 1 0.032 0.580
80 1 0.032 0.612
83 1 0.032 0.644
88 3 0.097 0.741
90 1 0.032 0.773
92 1 0.032 0.805
94 4 0.129 0.934
96 1 0.032 0.966
100 1 0.032 0.998 (Why isn't this value 1?)

Try it

The following data show the different types of pet food stores in the area carry.
6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;
Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.

μ = 9.3

s = 2.2

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Standard deviation of grouped frequency tables

Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: M e a n   o f   F r e q u e n c y   T a b l e = f m f
where f = interval frequencies and m = interval midpoints.

Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.

Find the standard deviation for the data in [link] .

Class Frequency, f Midpoint, m m 2 x ¯ 2 fm 2 Standard Deviation
0–2 1 1 1 7.58 1 3.5
3–5 6 4 16 7.58 96 3.5
6–8 10 7 49 7.58 490 3.5
9–11 7 10 100 7.58 700 3.5
12–14 0 13 169 7.58 0 3.5
15–17 2 16 256 7.58 512 3.5

For this data set, we have the mean, x ¯ = 7.58 and the standard deviation, s x = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, s x = f ( m x ¯ ) 2 n 1 where s x = sample standard deviation, x ¯ = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.

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Try it

Find the standard deviation for the data from the previous example

Class Frequency, f
0–2 1
3–5 6
6–8 10
9–11 7
12–14 0
15–17 2

First, press the STAT key and select 1:Edit

Input the midpoint values into L1 and the frequencies into L2

Select STAT , CALC , and 1: 1-Var Stats

Select 2 nd then 1 then , 2 nd then 2 Enter

You will see displayed both a population standard deviation, σ x , and the sample standard deviation, s x .

Comparing values from different data sets

The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.

  • For each data value, calculate how many standard deviations away from its mean the value is.
  • Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
  • # o f S T D E V s = value – mean standard deviation
  • Compare the results of this calculation.

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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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