An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation.
The standard deviation is a number that measures how far data values are from their mean.
- provides a numerical measure of the overall amount of variation in a data set
- can be used to determine whether a particular data value is close to or far from the mean
First we will investigate what the standard deviation tells us about data; then we will learn to calculate the standard deviation.
The standard deviation is always positive or 0. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.
Suppose that we are studying waiting times at the checkout line for customers at supermarket A and supermarket B; the average wait time at both markets is 5 minutes. At market A, the standard deviation for the waiting time is 2 minutes; at market B the standard deviation for the waiting time is 4 minutes.
Because market B has a higher standard deviation, we know that there is more variation in the waiting times at market B. Overall, wait times at market B are more spread out from the average; wait times at market A are more concentrated near the average.
Suppose that Rosa and Binh both shop at Market A. Rosa waits for 7 minutes and Binh waits for 1 minute at the checkout counter. At market A, the mean wait time is 5 minutes and the standard deviation is 2 minutes.
- 7 is 2 minutes longer than the average of 5; 2 minutes is equal to one standard deviation.
- Rosa's wait time of 7 minutes is 2 minutes longer than the average of 5 minutes.
- Rosa's wait time of 7 minutes is one standard deviation above the average of 5 minutes.
- A wait time that is only one standard deviation from the average is considered close to the average.
- 1 is 4 minutes less than the average of 5; 4 minutes is equal to two standard deviations.
- Binh's wait time of 1 minute is 4 minutes less than the average of 5 minutes.
- Binh's wait time of 1 minute is two standard deviations below the average of 5 minutes.
- A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than 2 standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than 2 standard deviations. (We will learn more about this in later chapters.)
- In general, a value = mean + (#ofSTDEVs)(standard deviation)
- where #ofSTDEVs = the number of standard deviations
- 7 is one standard deviation more than the mean of 5 because: 7=5+(1)(2)
- 1 is two standard deviations less than the mean of 5 because: 1=5+(−2)(2)
If xx is a data value, then the difference "xx - mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the data is for a population, in symbols a deviation is
x−μx−μ .
For sample data, in symbols a deviation is
x-x-
x¯
x
.
The procedure to calculate the standard deviation depends on whether the data is for the entire population or comes from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is a population or a sample. The lower case letter ss represents the sample standard deviation and the Greek letter σσ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then ss should be a good estimate of σσ.
To calculate the standard deviation, we need to calculate the variance first. The variance is an average of the squares of the deviations (the x-x-
x¯
x
values for a sample, or the x−μx−μ values for a population). The symbol
σ
2
σ
2
represents the population variance; the population standard deviation σσ is the square root of the population variance. The symbol
s
2
s
2
represents the sample variance; the sample standard deviation ss is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the data is from a population, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population. If the data is from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n-1, one less than the number of items in the sample. You can see that in the formulas below.
- s=s= size 12{s={}} {}
Σ
(
x
−
x
¯
)
2
n
−
1
Σ
(
x
−
x
¯
)
2
n
−
1
or
s=s= size 12{s={}} {}
Σ
f
·
(
x
−
x
¯
)
2
n
−
1
Σ
f
·
(
x
−
x
¯
)
2
n
−
1
- For the sample standard deviation, the denominator is n-1, that is the sample size MINUS 1.
-
σ=σ= size 12{σ={}} {}
Σ
(
x
−
μ
¯
)
2
N
Σ
(
x
−
μ
¯
)
2
N
or
σ=σ= size 12{σ={}} {}
Σ
f
·
(
x
−
μ
¯
)
2
N
Σ
f
·
(
x
−
μ
¯
)
2
N
- For the population standard deviation, the denominator is N, the number of items in the population.
In these fomulas, ff represents the frequency with which a value appears. For example, if a value appears once, ff is 1. If a value appears three times in the data set, ff is 3.
In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83,83+,84+ calculator, you need to select the appropriate standard deviation σσ or ss from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean.
At an elementary school, a teacher was interested in the average age and the standard deviation of the ages of the students in the fifth grade. The following data are the ages for a SAMPLE of n=20 fifth grade students. The ages are rounded to the nearest half year:
9 ; 9.5 ; 9.5 ; 10 ; 10 ; 10 ; 10 ; 10.5 ; 10.5 ; 10.5 ; 10.5 ; 11 ; 11 ; 11 ; 11 ; 11 ; 11 ; 11.5 ; 11.5 ; 11.5
x
¯
=
9
+
9.5
×
2
+
10
×
4
+
10.5
×
4
+
11
×
6
+
11.5
×
3
20
=
10.525
x
¯
=
9
+
9.5
×
2
+
10
×
4
+
10.5
×
4
+
11
×
6
+
11.5
×
3
20
=10.525
(1)The sample average age is 10.53 years, rounded to 2 places.
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating ss.
Table 1
| Data |
Freq. |
Deviations |
Deviations
2
Deviations
2
|
(Freq.)(
Deviations
2
Deviations
2
) |
| xx |
ff |
(x-x¯)(x-
x
) |
(
x
-
x
¯
)
2
(
x
-
x
¯
)
2
|
(
f
)
(
x
-
x
¯
)
2
(
f
)
(
x
-
x
¯
)
2
|
| 99 |
11 |
9
-
10.525
=
-
1.525
9-10.525=-1.525 |
(
-
1.525
)
2
=
2.325625
(
-
1.525
)
2
=2.325625 |
1
×
2.325625
=
2.325625
1
×
2.325625
=
2.325625
|
| 9.59.5 |
22 |
9.5
-
10.525
=
-
1.025
9.5-10.525=-1.025 |
(
-
1.025
)
2
=
1.050625
(
-
1.025
)
2
=1.050625 |
2
×
1.050625
=
2.101250
2
×
1.050625
=
2.101250
|
| 1010 |
44 |
10
-
10.525
=
-
0.525
10-10.525=-0.525 |
(
-
0.525
)
2
=
0.275625
(
-
0.525
)
2
=0.275625 |
4
×
.275625
=
1.1025
4
×
.275625
=
1.1025
|
| 10.510.5 |
44 |
10.5
-
10.525
=
-
0.025
10.5-10.525=-0.025 |
(
-
0.025
)
2
=
0.000625
(
-
0.025
)
2
=0.000625 |
4
×
.000625
=
.0025
4
×
.000625
=
.0025
|
| 1111 |
66 |
11
-
10.525
=
0.475
11-10.525=0.475 |
(
0.475
)
2
=
0.225625
(
0.475
)
2
=0.225625 |
6
×
.225625
=
1.35375
6
×
.225625
=
1.35375
|
| 11.511.5 |
33 |
11.5
-
10.525
=
0.975
11.5-10.525=0.975 |
(
0.975
)
2
=
0.950625
(
0.975
)
2
=0.950625 |
3
×
.950625
=
2.851875
3
×
.950625
=
2.851875
|
The sample variance
s
2
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
=
9.7375
20
-
1
=
0.5125
s
2
=
9.7375
20
-
1
=0.5125
The sample standard deviation ss is equal to the square root of the sample variance:
s
=
0.5125
=
0.715891
s=
0.5125
=0.715891 Rounded to two decimal places, s
=
0.72
s=0.72
Typically, you do the calculation for the standard deviation on a calculator or computer. Also note that we only rounded at the end of the calculation. The intermediate results are not rounded; this is done for accuracy.
Verify the mean and standard deviation calculated above using a calculator or computer.
- For the TI-83,83+,84+, enter data into the list editor.
- Put the data values in list L1 and the frequencies in list L2.
- STAT CALC 1-VarStats L1, L2
-
x¯
x
=10.525
- Use Sx because this is sample data (not a population): Sx=.715891
- For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation)
- For a sample:
xx =
x¯
x
+ (#ofSTDEVs)(s)
- For a population:
xx =
μ
μ + (#ofSTDEVs)( σσ)
- For this example, use
xx =
x¯
x
+ (#ofSTDEVs)(s) because the data is from a sample
Find the value that is 1 standard deviation above the mean. Find
(
x
¯
+
1
s
)
(
x
¯
+
1
s
)
.
(
x
¯
+
1
s
)
=
10.53
+
(
1
)
(
0.72
)
=
11.25
(
x
¯
+
1
s
)
=10.53+(1)(0.72)=11.25
Find the value that is two standard deviations below the mean. Find
(
x
¯
-
2
s
)
(
x
¯
-
2
s
)
.
(
x
¯
-
2
s
)
=
10.53
-
(
2
)
(
0.72
)
=
9.09
(
x
¯
-
2
s
)
=10.53-(2)(0.72)=9.09
Find the values that are 1.5 standard deviations from (below and above) the mean.
-
(
x
¯
-
1.5
s
)
=
10.53
-
(
1.5
)
(
0.72
)
=
9.45
(
x
¯
-
1.5
s
)
=10.53-(1.5)(0.72)=9.45
-
(
x
¯
+
1.5
s
)
=
10.53
+
(
1.5
)
(
0.72
)
=
11.61
(
x
¯
+
1.5
s
)
=10.53+(1.5)(0.72)=11.61
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. The deviations 0.97 and 0.47 indicate that. A positive deviation occurs when the data value is greater than the mean. A negative deviation occurs when the data value is less than the mean; the deviation is -1.525 for the data value 9. If you add up all the deviations, the sum is always zero; the positive and negative deviations offset each other. (For this example, there are n=20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, they become 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-1n-1). Why not divide by nn? 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)(n-1) gives a better estimate of the population variance.
Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.
The standard deviation, ss or σσ, is either zero or larger than zero. When the standard deviation is 0, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation
is a lot larger than zero, the data values are very spread out about the mean; outliers can make ss or σσ very large.
The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data.
Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
- a. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
- b. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:
- i. The sample mean
- ii. The sample standard deviation
- iii. The median
- iv. The first quartile
- v. The third quartile
- vi. IQR
- c. Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
- a.
Table 2
| 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?) |
- b.
- i. The sample mean = 73.5
- ii. The sample standard deviation = 17.9
- iii. The median = 73
- iv. The first quartile = 61
- v. The third quartile = 90
- vi. IQR = 90 - 61 = 29
- c. The x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram; number of intervals is 5 for the histogram so the width of an interval is (100.5 - 32.5) divided by 5 which is equal to 13.6. Endpoints of the intervals: starting point is 32.5, 32.5+13.6 = 46.1, 46.1+13.6 = 59.7, 59.7+13.6 = 73.3, 73.3+13.6 = 86.9, 86.9+13.6 = 100.5 = the ending value; No data values fall on an interval boundary.
The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73-33=40) than the spread in the upper 50% (100-73=27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR=29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.
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, it can be misleading to compare the data values directly.
- For each data value, calculate how many standard deviations the value is away from its mean.
- Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
-
#
ofSTDEVs
=
value
-
mean
standard deviation
#ofSTDEVs=
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:
Table 3
| Sample |
xx =
x¯
x
+ z s |
z
=
x
-
x¯
s
z=
x
-
x
s
|
| Population |
xx =
μ
μ + z σσ |
z
=
x
-
μ
σ
z=
x
-
μ
σ
|
Two students, John and Ali, from different high schools, wanted to find out who had the highest grade point average (GPA) when compared to his school. Which student had the highest GPA when compared to his school?
Table 4
| 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.
#
ofSTDEVs
=
value
-
mean
standard deviation
#ofSTDEVs=
value
-
mean
standard deviation
;
z
=
x
-
μ
σ
z=
x
-
μ
σ
For John,
z=#
ofSTDEVs
=
2.85
-
3.0
0.7
=
-
0.21
z=#ofSTDEVs=
2.85
-
3.0
0.7
=-0.21
For Ali, z=
#
ofSTDEVs
=
77
-
80
10
=
-
0.3
z=#ofSTDEVs=
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 mean while Ali's GPA is 0.3 standard deviations below his 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.
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.
- At least 75% of the data is within 2 standard deviations of the mean.
- At least 89% of the data is within 3 standard deviations of the mean.
- At least 95% of the data is within 4 1/2 standard deviations of the mean.
- This is known as Chebyshev's Rule.
- Approximately 68% of the data is within 1 standard deviation of the mean.
- Approximately 95% of the data is within 2 standard deviations of the mean.
- More than 99% of the data is within 3 standard deviation of the mean.
- This is known as the Empirical Rule.
- It is important to note that this rule only applies when the shape of the distribution of the data is mound-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
- Standard Deviation:
A number that is equal to the square root of the variance and measures how far data values are from their mean. Notations: s for sample standard deviation and σσfor population standard deviation.
- Variance:
Mean of the squared deviations from the mean. Square of the standard deviation.
