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Question
Find the mean deviation from the mean for the data:
| Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequencies | 6 | 8 | 14 | 16 | 4 | 2 |
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Solution
We will compute the mean deviation from the mean in the following way:
| Classes | Frequency
\[f_i\]
|
Midpoints \[x_i\]
|
\[f_i x_i\]
|
\[\left| x_i - X \right|\]
\[\left| x_i - 358 \right|\]
|
\[f_i \left| x_i - X \right|\]
|
| 0−10 | 6 | 5 | 30 | 22 | 132 |
| 10−20 | 8 | 15 | 120 | 12 | 96 |
| 20−30 | 14 | 25 | 350 | 2 | 28 |
| 30−40 | 16 | 35 | 560 | 8 | 128 |
| 40−50 | 4 | 45 | 180 | 18 | 72 |
| 50−60 | 2 | 55 | 110 | 28 | 56 |
|
\[\sum^6_{i = 1} f_i = 50\]
|
\[\sum^6_{i = 1} f_i = 50\]
|
\[\sum^8_{i = 1} f_i \left| x_i - X \right| = 512\] |
\[N = \sum^6_{i = 1} f_i = 50\] and
\[X = \frac{\sum^6_{i = 1} f_i x_i}{\sum^6_{i = 1} f_i}\]
\[ = \frac{1350}{50}\]
\[ = 27\]
\[\therefore \text{ Mean deviation } = \frac{1}{N} \sum^8_{i = 1} f_i \left| x_i - X \right|\]
\[ = \frac{512}{50}\]
\[ = 10 . 24\]
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