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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 - Mathematics

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
 
\[N = \sum^8_{i = 1} f_i = 110\]
 
 

\[\sum^8_{i = 1} f_i x_i = 5390\]
 
 

\[\sum^8_{i = 1} f_i \left| x_i - X \right| = 1644\]
\[N = \sum^8_{i = 1} f_i = 110\]
and

\[\sum^8_{i = 1} f_i x_i = 5390\]

\[X = \frac{\sum^8_{i = 1} f_i x_i}{N}\]
\[ = \frac{5390}{110}\]
\[ = 49\]

\[\text{ Mean deviation } = \frac{\sum^8_{i = 1} f_i \left| x_i - X \right|}{N}\]

\[ = \frac{1644}{110}\]

\[ = 14 . 945\]

\[ \approx 14 . 95\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.3 | Q 3 | Page 16
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