Standard Deviation of Grouped
Data
• Standard deviation can be found by
summing the square of the deviation of each
value, or,
• If the value is present more than once, the
square of the deviation can be calculated
once and multiplied by the frequency of
occurrences
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
Sum
x
f
5
2
6
3
7
6
8
2
9
1
xf
x - xbar
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
• x   xf  95  6.786
n
Sum
14
x
f
xf
5
2
10
6
3
18
7
6
42
8
2
16
9
1
9
14
95
x - xbar
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
• x   xf  95  6.786
n
Sum
14
x
f
xf
x - xbar
5
2
10
-1.786
6
3
18
-0.7857
7
6
42
0.2143
8
2
16
1.2143
9
1
9
2.2143
14
95
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
• x   xf  95  6.786
n
Sum
14
x
f
xf
x - xbar
(x – xbar)^2
5
2
10
-1.786
3.1888
6
3
18
-0.7857
0.61735
7
6
42
0.2143
0.04592
8
2
16
1.2143
1.4745
9
1
9
2.2143
4.9031
14
95
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
• x   xf  95  6.786
n
Sum
14
x
f
xf
x - xbar
(x – xbar)^2
(x – xbar) ^2 * f
5
2
10
-1.786
3.1888
6.378
6
3
18
-0.7857
0.61735
1.852
7
6
42
0.2143
0.04592
0.276
8
2
16
1.2143
1.4745
2.949
9
1
9
2.2143
4.9031
4.903
14
95
16.357
Standard Deviation of Grouped Data
• Find the sample standard deviation of the
following data:
• 7, 6, 7, 6, 7, 8, 5, 6, 7, 5, 7, 8, 9, 7
• x   xf  95  6.786
n
Sum
14
x
f
xf
x - xbar
(x – xbar)^2
(x – xbar) ^2 * f
5
2
10
-1.786
3.1888
6.378
6
3
18
-0.7857
0.61735
1.852
7
6
42
0.2143
0.04592
0.276
8
2
16
1.2143
1.4745
2.949
9
1
9
2.2143
4.9031
4.903
14
95
16.357
Standard Deviation of Grouped Data


2
 xx f
16.357
s s 

 1.122
n 1
13
2
•The calculator also gives 1.122
Sum
x
f
xf
x - xbar
(x – xbar)^2
(x – xbar) ^2 * f
5
2
10
-1.786
3.1888
6.378
6
3
18
-0.7857
0.61735
1.852
7
6
42
0.2143
0.04592
0.276
8
2
16
1.2143
1.4745
2.949
9
1
9
2.2143
4.9031
4.903
14
95
16.357
Standard Deviation of Grouped Data
• We grouped the data in the above example.
• The same process can be used when given
data in the form of a histogram or pie chart.
• Since the values of the specific data points has
been lost,assume all the data points within a
cell have the same value as the cell midpoint.
• The student is left to review Example 10 on
page 77.
Standard Deviation of Grouped Data
• Assume the histogram on the following slide
represents our data.
• Make a table of values (x values – the
midpoint of each column), including the
frequency of each column.
• Calculate the sample standard deviation of the
data represented in the histogram
Standard Deviation of Grouped Data
Frequency versus Midpoints
6
5
4
3
2
1
0
1
2
3
4
5
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
Sum
x
f
1
2
2
3
3
5
4
4
5
1
xf
x - xbar
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
n
Sum
15
x
f
xf
1
2
2
2
3
6
3
5
15
4
4
16
5
1
5
15
44
x - xbar
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
What does this
n
15
add to?
Sum
x
f
xf
x - xbar
1
2
2
-1.933
2
3
6
-0.933
3
5
15
0.067
4
4
16
1.067
5
1
5
2.067
15
44
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
What does this
n
15
add to?
Sum
x
f
xf
x - xbar
1
2
2
-1.933
2
3
6
-0.933
3
5
15
0.067
4
4
16
1.067
5
1
5
2.067
15
44
0
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
What does this
n
15
add to?
Sum
x
f
xf
x - xbar
(x – xbar) * f
1
2
2
-1.933
-3.867
2
3
6
-0.933
-2.800
3
5
15
0.067
0.3333
4
4
16
1.067
4.2667
5
1
5
2.067
2.067
15
44
0
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
n
Sum
15
x
f
xf
x - xbar
(x – xbar) * f
1
2
2
-1.933
-3.867
2
3
6
-0.933
-2.800
3
5
15
0.067
0.3333
4
4
16
1.067
4.2667
5
1
5
2.067
2.067
15
44
0
0
(x – xbar)^2
(x – xbar) ^2 * f
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
n
15
•
Sum
x
f
xf
x - xbar
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
1
2
2
-1.933
-3.867
3.738
7.476
2
3
6
-0.933
-2.800
0.871
2.613
3
5
15
0.067
0.3333
0.004
0.022
4
4
16
1.067
4.2667
1.137
4.551
5
1
5
2.067
2.067
4.271
4.271
15
44
0
0
18.933
Standard Deviation of Grouped Data
• The cell midpoints are 1, 2, 3, 4, and 5
• The frequencies are 2, 3, 5, 4, and 1
• x   xf  44  2.933
n
• s s 
2
Sum
15


2
 xx f
18 .933

 1.163
n 1
14
x
f
xf
x - xbar
(x – xbar) * f
(x – xbar)^2
(x – xbar) ^2 * f
1
2
2
-1.933
-3.867
3.738
7.476
2
3
6
-0.933
-2.800
0.871
2.613
3
5
15
0.067
0.3333
0.004
0.022
4
4
16
1.067
4.2667
1.137
4.551
5
1
5
2.067
2.067
4.271
4.271
15
44
0
0
18.933
Standard Deviation of Grouped Data
•
•
•
•
•
•
•
How can we do this in our calculator?
Put the “x” values in L1
Put the frequency in L2
Stat
Calc
1-Var Stats
2nd L1, 2nd L2, Enter
Homework
• Pg 81 & 82, # 29 – 32 all (4)