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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.

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

Use your calculator to find the standard deviation.

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.

#ofSTDEVs is often called a " z -score"; we can use the symbol z . In symbols, the formulas become:

Sample x = x + zs z = x     x ¯ s
Population x = μ + z = x     μ σ

Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?

Student GPA School Mean GPA School Standard Deviation
John 2.85 3.0 0.7
Ali 77 80 10

For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.

z = # of STDEVs = value  mean standard deviation = x + μ σ

For John, z = # o f S T D E V s = 2.85 3.0 0.7 = 0.21

For Ali, z = # o f S T D E V s = 77 80 10 = 0.3

John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.

John's z -score of –0.21 is higher than Ali's z -score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.

Try it

Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?

Swimmer Time (seconds) Team Mean Time Team Standard Deviation
Angie 26.2 27.2 0.8
Beth 27.3 30.1 1.4

For Angie: z = 26 .2 – 27 .2 0 .8 = –1.25

For Beth: z = 27 .3 30 .1 1. 4 = –2

The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.

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Source:  OpenStax, Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24. OpenStax CNX. Oct 24, 2015 Download for free at http://legacy.cnx.org/content/col11891/1.8
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