# Long Questions

Find the mean deviation and its coefficient from median of the data given below:
10, 50, 60, 40, 30, 20

Arranging given data in the ascending order: 10, 20, 30, 40, 50, 60
Total number (N) = 6
Median (Md) = $(\frac{\text{N + 1}}{2})$th item = 3.5th item
So, Md = $\frac{\text{30 + 40}}{2}$ = 35

Now, tabulating the given data,

X |D| = |X - Md|
10 25
20 15
30 5
40 5
50 15
60 25
N = 6 ∑|D| = 90
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{90}{6}$ = 15
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{15}{25}$ = 0.43
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 20 30 50 40 60 70 2 4 7 5 2 1
Here, Calculation of median by arranging the data in ascending order,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
20 2 2
30 4 4+2 = 6
40 5 5+6 = 11
50 7 7+11=18
60 2 2+18 = 20
70 1 1+20 = 21
N = 21

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 11th item

Corresponding value of Cumulative frequency (c.f.) equal to 11 is 40.
So, Median = 40

Finally, Calculation of mean deviation from median,

Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
20 |20-40| = 20 2 40
30 |30-40| = 10 4 40
40 |40-40| = 0 5 0
50 |50-40| = 10 7 70
60 |60-40| = 20 2 40
70 |70-40| = 30 1 30
N = 21 ∑f|D| 220
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{220}{21}$ = 10.47
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10.47}{40}$ = 0.26
Compute mean deviation from median of the following data.
 Marks obtained No. of students 5 10 15 20 25 30 4 5 6 7 3 2
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
5 4 4
10 5 4+5 = 9
15 6 9+6 = 15
20 7 15+7 = 22
25 3 22+3 = 25
30 2 25+2 = 27
N = 27

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 14th item

Cumulative frequency (c.f.) just greater than 14 is 15. Its corresponding value is 15.
So, Median = 15

Finally, Calculation of mean deviation from median,

Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
5 |5-15| = 10 4 40
10 |10-15| = 5 5 25
15 |15-15| = 0 6 0
20 |20-15| = 5 7 35
25 |25-15| = 10 3 30
30 |30-15| = 15 2 30
N = 27 ∑f|D| 160
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{160}{27}$ = 5.925
Compute the mean deviation from median of the following data.
 Marks obtained No. of students 20 30 40 50 60 70 4 7 12 2 4 6
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
20 4 4
30 7 4+7 = 11
40 15 11+12 = 23
50 2 23+2 = 25
60 4 25+4 = 29
70 6 29+6 = 35
N = 35

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 18th item

Cumulative frequency (c.f.) just greater than 18 is 23. Its corresponding value is 40.
So, Median = 40

Finally, Calculation of mean deviation from median,

Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
20 |20-40| = 20 4 80
30 |30-40| = 10 7 70
40 |40-40| = 0 12 0
50 |50-40| = 10 2 20
60 |60-40| = 20 4 80
70 |70-40| = 30 6 180
N = 35 ∑f|D| 430
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{430}{35}$ = 12.285
Calculate the mean deviation of the data given below from median.
 Marks obtained No. of students 10-20 20-30 30-40 40-50 50-60 5 4 5 4 2
Here, Calculation of median,
 Marks obtained (X) Number of students (f) Cumulative frequency (c.f.) mid-value (m) 10-20 15 5 5 20-30 25 4 5+4 = 9 30-40 35 5 9+5 = 14 40-50 45 4 14+4 = 18 50-60 55 2 18+2 = 20 N = 20

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 10th item

Cumulative frequency (c.f.) which is greater than 10 is 14. Its corresponding value is 30-40
So, Median lies in class (30-40).
L = 30, f = 5, c.f. = 9 and i = 10
Now, Median = L + $\frac {i}{f}(\frac N2 - c.f.)$ = 30 + $\frac {10}{5}(10 - 9)$
= 32

Finally, Calculation of mean deviation from median,

Marks obtained (m) |m - Median| = |D| Number of students (f) f|D|
15 |15-32| = 17 5 85
25 |25-32| = 7 4 2
35 |35-32| = 3 5 15
45 |45-32| = 13 4 52
55 |55-32| = 23 2 46
N = 20 ∑f|D| = 226
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{226}{20}$ = 11.3
Compute the mean deviation from median of the following data.
 Marks obtained No. of students 25 35 45 55 65 5 7 8 6 4
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
25 5 5
35 7 5+7 = 12
45 8 12+8 = 20
55 6 20+6 = 26
65 4 26+4 = 30
N = 30

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 15.5th item

Cumulative frequency (c.f.) just greater than 15.5 is 20. Its corresponding value is 45.
So, Median = 45

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
25 |25-45| = 20 5 100
35 |35-45| = 10 7 70
45 |45-45| = 0 8 0
55 |55-45| = 10 6 60
65 |65-45| = 20 4 80
N = 30 ∑f|D| 310
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{310}{30}$ = 10.33
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 55 65 75 85 95 4 5 6 3 2
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
55 4 4
65 5 4+5 = 9
75 6 9+6 = 15
85 3 15+3 = 18
95 2 18+2 = 20
N = 20

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 10.5th item

Cumulative frequency (c.f.) just greater than 10.5 is 15. Its corresponding value is 75.
So, Median = 75

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
55 |55-75| = 20 4 80
65 |65-75| = 10 5 50
75 |75-75| = 0 6 0
85 |85-75| = 10 3 30
95 |95-75| = 20 2 40
N = 20 ∑f|D| 200
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{200}{20}$ = 10
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{75}$ = 0.133
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 10 20 30 40 50 2 3 6 5 4
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
10 2 2
20 3 2+3 = 5
30 6 5+6 = 11
40 5 11+5 = 16
50 4 16+4 = 20
N = 20

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item
= value of 10.5th item

Cumulative frequency (c.f.) just greater than 10.5 is 11. Its corresponding value is 30.
So, Median = 30

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
10 |10-30| = 20 2 40
20 |20-30| = 10 3 30
30 |30-30| = 0 6 0
40 |40-30| = 10 5 50
50 |50-30| = 20 4 80
N = 20 ∑f|D| 200
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{200}{20}$ = 10
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{30}$ = 0.33
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 5 10 15 20 25 3 4 6 2 1
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
5 3 3
10 4 3+4 = 7
15 6 7+6 = 13
20 2 13+2 = 15
25 1 15+1 = 16
N = 16

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 8.5th item

Cumulative frequency (c.f.) just greater than 10.5 is 13. Its corresponding value is 15.
So, Median = 15
Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
5 |5-15| = 10 3 30
10 |10-15| = 5 4 20
15 |15-15| = 0 6 0
20 |20-15| = 5 2 10
25 |25-15| = 10 1 10
N = 16 ∑f|D| 70
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{70}{16}$ = 4.375
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{4.375}{15}$ = 0.29
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 10 30 50 20 40 60 2 10 8 5 12 3
Here, Calculation of median by arranging the data in ascending order,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
10 2 2
20 5 2+5 = 7
30 10 7+10 = 17
40 12 17+12 = 29
50 8 29+8 = 37
60 3 37+3 = 40
N = 40

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 20.5th item

Cumulative frequency (c.f.) just greater than 20.5 is 29. Its corresponding value is 40.
So, Median = 40

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
10 |10-40| = 30 2 60
20 |20-40| = 20 5 100
30 |30-40| = 10 10 100
40 |40-40| = 0 12 0
50 |50-40| = 10 8 80
60 |60-40| = 20 3 60
N = 40 ∑f|D| 400
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{400}{40}$ = 10
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{10}{40}$ = 0.25
Compute the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 200 400 300 350 250 4 7 6 1 3
Here, Calculation of median by arranging the data in ascending order,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
200 4 4
250 3 4+3 = 7
300 6 7+6 = 13
350 1 13+1 = 14
400 7 14+7 = 21
N = 21

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 11th item

Cumulative frequency (c.f.) just greater than 11 is 13. Its corresponding value is 300.
So, Median = 300

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
200 |200-300| = 100 4 400
250 |250-300| = 50 3 150
300 |300-300| = 0 6 0
350 |350-300| = 50 1 50
400 |400-300| = 100 7 700
N = 21 ∑f|D| 1300
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{1300}{21}$ = 61.9
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{61.9}{300}$ = 0.206
Find the mean deviation from median of the following data.
 Marks obtained No. of students 20 10 30 50 40 3 4 5 2 1
Here, Calculation of median by arranging the data in ascending order,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
10 4 4
20 3 4+3 = 7
30 5 7+5 = 12
40 1 12+1 = 13
50 2 13+2 = 15
N = 15

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 8th item

Cumulative frequency (c.f.) just greater than 8 is 12. Its corresponding value is 30.
So, Median = 30

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
10 |10-30| = 20 4 80
20 |20-30| = 10 3 30
30 |30-30| = 0 5 0
40 |40-30| = 10 1 10
50 |50-30| = 20 2 40
N = 15 ∑f|D| 160
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{160}{15}$ = 10.67
Find the mean deviation and its coefficient from median of the following data.
 Marks obtained No. of students 10 20 30 40 50 60 2 3 9 21 11 5
Here, Calculation of median,
Marks obtained (X) Number of students (f) Cumulative frequency (c.f.)
10 2 2
20 3 2+3 = 5
30 9 5+9 = 14
40 21 14+21 = 35
50 11 35+11 = 46
60 5 46+5 = 51
N = 51

Now, median = value of $(\frac{\text{N + 1}}{2})^{th}$ item = value of 26th item

Cumulative frequency (c.f.) just greater than 26 is 35. Its corresponding value is 40.
So, Median = 40

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - Median| = |D| Number of students (f) f|D|
10 |10-40| = 30 2 60
20 |20-40| = 20 3 60
30 |30-40| = 10 9 90
40 |40-40| = 0 21 0
50 |50-40| = 10 11 110
60 |60-40| = 20 5 100
N = 51 ∑f|D| 420
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{420}{51}$ = 8.235
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Median}}$ = $\frac{8.235}{40}$ = 0.21
Calculate the mean deviation and its coefficient from the mean of the data given below.
 Marks obtained No. of students 20 25 30 35 40 2 5 6 5 2
Here, Calculation of median,
Marks obtained (X) Number of students (f) fX
20 2 40
25 5 125
30 6 180
35 5 175
40 2 80
N = 2 $\sum fX$ = 600

Now, mean = $\frac{\sum fX}{N}$ = $\frac{600}{2}$ = 30

Finally, Calculation of mean deviation from median,
Marks obtained (X) |X - mean| = |D| Number of students (f) f|D|
20 |20-30| = 10 2 20
25 |25-30| = 5 5 25
30 |30-30| = 0 6 0
35 |35-30| = 5 5 25
40 |40-30| = 10 2 20
N = 20 ∑f|D| 90
Hence, Mean deviation from median = $\frac{\sum f|D|}{N}$ = $\frac{90}{20}$ = 4.5
Also, coefficient of mean deviation from median = $\frac{\text{Mean deviation}}{\text{Mean}}$ = $\frac{4.5}{30}$ = 0.15