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Question
Calculate the standard deviation for the following data:
| Class: | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
| Frequency: | 9 | 17 | 43 | 82 | 81 | 44 | 24 |
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Solution
| Class |
\[f_i\]
|
Midpoint
\[\left( x_i \right)\]
|
\[y_i = \frac{x_i - 105}{30}\]
|
\[{y_i}^2\]
|
\[f_i y_i\]
|
\[f_i {y_i}^2\]
|
| 0−30 | 9 | 15 | −3 | 9 | −27 | 81 |
| 30−60 | 17 | 45 | −2 | 4 | −34 | 68 |
| 60−90 | 43 | 75 | −1 | 1 | −43 | 43 |
| 90−120 | 82 | 105 | 0 | 0 | 0 | 0 |
| 120−150 | 81 | 135 | 1 | 1 | 81 | 81 |
| 150−180 | 44 | 165 | 2 | 4 | 88 | 176 |
| 180−210 | 24 | 195 | 3 | 9 | 72 | 216 |
|
\[\sum f_i = N = 300\]
|
\[\sum f_i y_i = 137\]
|
\[\sum f_i {y_i}^2 = 665\]
|
Mean,
\[\sigma^2 = \frac{h^2}{N^2}\left[ N\sum f_i {y_i}^2 - \left( \sum f_i y_i \right)^2 \right]\]
\[ = \frac{900}{90000}\left[ 300 \times 665 - 18769 \right]\]
\[ = \frac{1}{100}\left[ 199500 - 18769 \right]\]
\[ = \frac{180731}{100} = 1807 . 31\]
\[SD, \sigma = \sqrt{1807 . 31} = 42 . 51\]
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