UPSKILL MATH PLUS
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Learn moreLet 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.