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Question
Compute mean deviation from mean of the following distribution:
| Mark | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
| No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |
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Solution
Computation of mean deviation from the mean:
| Marks | Number of Students \[f_i\]
|
Midpoints \[x_i\]
|
\[f_i x_i\]
|
\[\left| x_i - X \right|\]
\[\left| x_i - 49 \right|\]
|
\[f_i \left| x_i - X \right|\]
|
| 10−20 | 8 | 15 | 120 | 34 | 272 |
| 20−30 | 10 | 25 | 250 | 24 | 240 |
| 30−40 | 15 | 35 | 525 | 14 | 210 |
| 40−50 | 25 | 45 | 1125 | 4 | 100 |
| 50−60 | 20 | 55 | 1100 | 6 | 120 |
| 60−70 | 18 | 65 | 1170 | 16 | 288 |
| 70−80 | 9 | 75 | 675 | 26 | 234 |
| 80−90 | 5 | 85 | 425 | 36 | 180 |
|
\[N = \sum^8_{i = 1} f_i = 110\]
|
\[\sum^8_{i = 1} f_i x_i = 5390\]
|
\[\sum^8_{i = 1} f_i \left| x_i - X \right| = 1644\]
|
and
\[\sum^8_{i = 1} f_i x_i = 5390\]
\[X = \frac{\sum^8_{i = 1} f_i x_i}{N}\]
\[ = \frac{5390}{110}\]
\[ = 49\]
\[\text{ Mean deviation } = \frac{\sum^8_{i = 1} f_i \left| x_i - X \right|}{N}\]
\[ = \frac{1644}{110}\]
\[ = 14 . 945\]
\[ \approx 14 . 95\]
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