Find the mean deviation from the mean for the data:
Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
Advertisement Remove all ads
Solution
We will compute the mean deviation from the mean in the following way:
Classes |
\[f_i\]
|
Midpoints \[x_i\]
|
\[f_i x_i\]
|
\[\left| x_i - X \right|\]
\[\left| x_i - 358 \right|\]
|
\[f_i \left| x_i - X \right|\]
|
0−100 | 4 | 50 | 200 | 308 | 1232 |
100−200 | 8 | 150 | 1200 | 208 | 1664 |
200−300 | 9 | 250 | 2250 | 108 | 972 |
300−400 | 10 | 350 | 3500 | 8 | 80 |
400−500 | 7 | 450 | 3150 | 92 | 644 |
500−600 | 5 | 550 | 2750 | 192 | 960 |
600−700 | 4 | 650 | 2600 | 292 | 1168 |
700−800 | 3 | 750 | 2250 | 392 | 1176 |
\[\sum^8_{i = 1} f_i = 50\]
|
\[\sum^8_{i = 1} f_ix_i= 17900\]
|
\[\sum^8_{i = 1} f_i \left| x_i - X \right| = 7896\] |
\[N = \sum^6_{i = 1} f_i = 50\] and
\[\sum^6_{i = 1} f_i x_i = 17900\]
\[\bar{ X } = \frac{\sum^{8} _{i = 1} f_i x_i}{\sum ^8_{i = 1} f_i} = \frac{17900}{50} = 358\]
\[\therefore \text{ Mean deviation } = \frac{1}{N} \sum^8_{i = 1} f_i \left| x_i - X \right|\]
\[ = \frac{7896}{50}\]
\[ = 157 . 92\]
Concept: Mean Deviation
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads