Descriptive Statistics: Measuring the Spread of the Data (edited: Teegarden)
1.1
2008/08/12 14:17:01.928 GMT-5
2008/08/12 14:18:19.901 GMT-5
Mary
T
Teegarden
tteegard@sdccd.edu
Mary
T
Teegarden
tteegard@sdccd.edu
elementary
statistics
This module describes a number of statistical measures used to describe data, such as percentiles, spread, and skewness. Labs changed to include information on mini-tabs.
The most common measure of spread is the standard deviation. The standard deviation is a number that measures how far data values are from their mean. For example, if the mean of a set of data containing 7 is 5 and the standard deviation is 2, then the value 7 is one (1) standard deviation from its mean because 5 + (1)(2) = 7. The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line, 7 is to the right of 5. We say, then, that 7 is one standard deviation to the right of 5. If 1 were also part of the data set, then 1 is two standard deviations to the left of 5 because 5 +(-2)(2) = 1. 1=5+(-2)(2) ; 7=5+(1)(2)
Formula: value =
x
+ (#ofSTDEVs)(s)Generally, a value = mean + (#ofSTDEVs)(standard deviation), where #ofSTDEVs = the number of standard deviations.If x is a value and
x
is the sample mean, then x-
x
is called a deviation. In a data set, there are as many deviations as there are data values. Deviations are used to calculate the sample standard deviation. To calculate the standard deviation, calculate the variance first. The variance is the average of the squares of the deviations. The standard deviation is the square root of the variance. You can think of the standard deviation as a special average of the deviations (the x-
x
values). The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma) represents the population standard deviation. We use
s
2
to represent the sample variance and
σ
2
to represent the population variance. If the sample has the same characteristics as the population, then s should be a good estimate of σ.In practice, use either a calculator or computer software to calculate the standard deviation. However, please study the following step-by-step example.
In a fifth grade class, the teacher was interested in the average age and the standard deviation of the ages of her students. What follows are the ages of her students 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
The 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 s.
Data
Freq.
Deviations
Deviations
2
(Freq.)(
Deviations
2
)
x
f
(x-
x
)
(
x
-
x
¯
)
2
(
f
)
(
x
-
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
×
.275625
=
1.1025
10.5
4
10.5
-
10.525
=
-
0.025
(
-
0.025
)
2
=
0.000625
4
×
.000625
=
.0025
11
6
11
-
10.525
=
0.475
(
0.475
)
2
=
0.225625
6
×
.225625
=
1.35375
11.5
3
11.5
-
10.525
=
0.975
(
0.975
)
2
=
0.950625
3
×
.950625
=
2.851875
The sample variance,
s
2
, is equal to the last sum (9.7375) divided by the total number of data values minus one (20 - 1):
s
2
=
9.7375
20
-
1
=
0.5125
The sample standard deviation, s, is equal to the square root of the sample variance:
s
=
0.5125
=
.
0715891
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.
Verify the mean and standard deviation calculated above on your calculator or computer. Find the median and mode.
- Median = 10.5
- Mode = 11
Find the value that is 1 standard deviation above the mean. Find
(
x
¯
+
1
s
)
.
(
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
)
=
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
)
=
11.61
Explanation of the table: The deviations show how spread out the data are about the mean. The value 11.5 is farther from the mean than 11. The deviations 0.97 and 0.47 indicate that. If you add the deviations, the sum is always zero. (For this example, there are 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. The variance, then, is the average squared deviation. It is small if the values are close to the mean and large if the values are far from the mean.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.For the sample variance, we divide by the total number of data values 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. By dividing by (n-1), we get a better estimate of the population variance.Your concentration should be on what the standard deviation does, not on the arithmetic. 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 sample standard deviation, s
, is either zero or larger than zero. When s
=
0
, there is no spread. When s
is a lot larger than zero, the data values are very spread out about the mean. Outliers can make s 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.The formula for the standard deviation is at the end of the chapter.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
- aCreate a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
- bCalculate the following to one decimal place using a TI-83+ or TI-84 calculator:
- iThe sample mean
- iiThe sample standard deviation
- iiiThe median
- ivThe first quartile
- vThe third quartile
- viIQR
- cConstruct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
- a
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
- iThe sample mean = 73.5
- iiThe sample standard deviation = 17.9
- iiiThe median = 73
- ivThe first quartile = 61
- vThe third quartile = 90
- viIQR = 90 - 61 = 29
Chebyshev's TheoremThe proportion (or fraction) of any data set lying withing K standard deviations of the mean is always at least 1 - 1/K^2, where K is any positive number greater then 1. (Why is this?)
For K = 2, the proportion is 1 - 1/4 = 3/4, hence 3/4ths or 75% of the data falls within 2 standard deviations of the mean.
For K = 3, the proportion is 1 - 1/9 = 8/9, hence 8/9ths or 89% or the data falls within 3 standard deviations of the mean.
Using the data from the pre-calculus class exams and K = 2, this means that at least 75% of the scores fall between 73.5 - 2(17.9) and 73.5 + 2(17.9), or between 37.7 and 109.3.
In actual fact all but one data value falls in this range, however Chevyshev's Theorem gives the worst case scenario.
Two students, John and Ali, from different high schools, wanted to find out who had the highest G.P.A. when compared to his school. Which student had the highest G.P.A. when compared to his school?
Student
GPA
School Mean GPA
School Standard Deviation
John
2.85
3.0
0.7
Ali
77
80
10
Use the formula value = mean + (#ofSTDEVs)(stdev) and solve for #ofSTDEVs for each student (stdev = standard deviation):
#
ofSTDEVs
=
value
-
mean
stdev
:For John,
#
ofSTDEVs
=
2.85
-
3.0
0.7
=
-
0.21
For Ali,
#
ofSTDEVs
=
77
-
80
10
=
-
0.3
John has the better G.P.A. when compared to his school because his G.P.A. is 0.21 standard deviations below his mean while Ali's G.P.A. is 0.3 standard deviations below his mean.
Your concentration should be on what the standard deviation does, not on the arithmetic. 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.
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 σfor population standard deviation.
Variance
Mean of the squared deviations from the mean. Square of the standard deviation.