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Question
Find the mean deviation from the mean for the data:
| Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |
| Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |
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Solution
We will compute the mean deviation from the mean in the following way:
| Classes | Frequency
\[f_i\]
|
Midpoints \[x_i\]
|
\[f_i x_i\]
|
\[\left| x_i - X \right|\]
\[\left| x_i - 128 . 58 \right|\]
|
\[f_i \left| x_i - X \right|\]
|
| 95−105 | 9 | 100 | 900 | 28.58 | 257.22 |
| 105−115 | 13 | 110 | 1430 | 18.58 | 241.54 |
| 115−125 | 16 | 120 | 1920 | 8.58 | 137.28 |
| 125−135 | 26 | 130 | 3380 | 1.42 | 36.92 |
| 135−145 | 30 | 140 | 4200 | 11.42 | 342.6 |
| 145−155 | 12 | 150 | 1800 | 21.42 | 257.04 |
|
\[\sum^6_{i = 1} f_i = 106\]
|
\[\sum^6_{i = 1} f_i x_i = 13630\]
|
\[\sum^8_{i = 1}| f_i x_i |- \bar{x}= 1272.6\] |
\[\sum^6_{i = 1} f_i x_i = 13630\]
\[\therefore X = \frac{\sum^6_{i = 1} f_i x_i}{\sum^6_{i = 1} f_i}\]
\[ = \frac{13630}{106}\]
\[ = 128 . 58\]
\[\therefore \text{ Mean deviation } = \frac{1}{N} \sum^8_{i = 1} f_i \left| x_i - X \right|\]
\[ = \frac{1272 . 6}{106}\]
\[ = 12 . 005\]
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