PDF chapter test TRY NOW

Let us look at an example to find standard deviation of a grouped data by mean method.
Example:
Case 1: Discrete data
 
Find the standard deviation of the following data by mean method.
 
x
1
2
3
4
f
4
8
12
16
  
Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:
 
\overline x = \frac{\sum_{i = 1}^{n}  f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}
 
Where x_{i} is the midpoint of the class interval and f_{i} is the frequency.
Let us form a frequency distribution table.
 
x_{i}
f_{i}
f_{i} x_{i}
1
4
4
2
8
16
3
12
36
4
16
64
  
\sum_{i = 1}^{4} f_{i} = 40
\sum_{i = 1}^{4}  f_{i} x_{i} = 120
 
Substituting the known values in the above formula, we get:
 
Mean \overline x = \frac{120}{40} = 3
 
Thus, the meanof the given data is
 
Now, let us find the standard deviation for the given data.
 
x_{i}
f_{i}
d_{i} = x_{i} - \overline x
 
= x_{i} - 3
d_{i}^{2}
f_{i}d_{i}^{2}
1
4
-2
4
16
2
8
-1
1
8
3
12
0
0
0
4
16
1
1
16
 
\sum_{i = 1}^{4} f_{i} = 40
 
 
\sum_{i = 1}^{4}  f_{i} d_{i}^{2} = 40
The formula to calculate the standard deviation by mean method is given by:
 
\sigma = \sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}} where N = \sum_{i = 1}^{n} f_{i}
Substitute the required values in the above formula.
 
\sigma = \sqrt{\frac{40}{40}}
 
= \sqrt{1}
 
= 1
 
Therefore, the standard deviation of the given data is 1.
 
Case 2: Continuous data
 
Find the standard deviation of the following data by mean method.
 
x
0 - 2
2 - 4
4 - 6
6 - 8
f
4
8
12
16
 
Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:
 
\overline x = \frac{\sum_{i = 1}^{n}  f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}
 
Where x_{i} is the midpoint of the class interval and f_{i} is the frequency.
Let us form a frequency distribution table.
 
Class interval
Frequency
(f_{i})
Midpoint
(x_{i})
f_{i} x_{i}
0 - 2
4
1
4
2 - 4
8
3
24
4 - 6
12
5
60
6 - 8
16
7
112
  
\sum_{i = 1}^{4} f_{i} = 40
 
\sum_{i = 1}^{4}  f_{i} x_{i} = 200
 
Substituting the known values in the above formula, we get:
 
Mean \overline x = \frac{200}{40} = 5
 
Thus, the mean of the given data is 5
 
Now, let us find the standard deviation for the given data.
 
x_{i}
f_{i}
d_{i} = x_{i} - \overline x
 
= x_{i} - 5
d_{i}^{2}
f_{i}d_{i}^{2}
1
4
-4
16
64
3
8
-2
4
32
5
(12\)
0
0
0
7
16
2
4
64
 
\sum_{i = 1}^{4} f_{i} = 40
 
 
\sum_{i = 1}^{4}  f_{i} d_{i}^{2} = 160
The formula to calculate the standard deviation by mean method is given by:
 
\sigma = \sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}} where N = \sum_{i = 1}^{n} f_{i}
Substitute the required values in the above formula.
 
\sigma = \sqrt{\frac{160}{40}}
 
= \sqrt{4}
 
= 2
 
Therefore, the standard deviation of the given data is 2.