Q.1.) Give the merits and demerits of quartile deviation. Calculate quartile deviation for the following data:
Farm Size (in acres) No. of farms
0— 40 394
40— 80 461
80—120 391
120—160 334
160—200 169
200—240 113
240 and over
• Solution:
Let the farm size be X.
Let the no of farms be = f .
Sum of no of farms = n.
There for n = ∑f.
Farm size: No of farms: Mead value A=140 ∑fdx =
X f of : X d = x-A f * d
0— 40 394 20 -120 394 * -120 = -47280
40— 80 461 60 -80 461 * -80 = -36880
80—120 391 100 -40 391 * -40 = -15640
120—160 334 140 0 334 * 0 = 0
160—200 169 180 40 169 * 40 = 6760
200—240 113 220 80 113 * 80 = 9040
∑n = 1862 ∑fdx = -84000
A.M. = X = A+∑xfi
n
= 140 - 84000
1862
X = 94.8872 ------------------------Ans.
Q.2.) Calculate quartile deviation and its relative measure.
Variable Frequency
20—29 306
30—39 182
40—49 144
50—59 96
60—69 42
70—79 24
• Solution:
Let the variable be X.
Let the frequency be = f .
Sum of no of frequency be = n.
There for n = ∑f.
Variable: Frequency: Mead value of : A=44.5 ∑fdx =
X f X d = x-A f * d
20— 29 306 24.5 -30 306 * -30 = -9180
30— 39 182 34.5 -20 182 * -20 = -3640
40—49 144 44.5 -10 144 * -10 = -1440
50—59 96 54.5 0 96 * 0 = 0
60—69 42 64.5 10 42 * 10 = 420
70—79 24 74.5 20 24 * 20 = 480
∑n = 794 ∑fdx = -13360
A.M. = X = A+∑xfi
n
= 44.5 - 13360
794
X = 37.674 ------------------------Ans.
Q.3.) From the following distribution, calculate the value of them MEAN
X F
300—399 14
400—499 46
500—599 58
600—699 76
700—799 68
800— 899 62
900—999 48
1000—1099 22
1100—1199 6
• Solution:
Sum of no of frequency be = n.
There for n = ∑f.
Mead value of : A=749.5 ∑fdx =
X : f: X d = x-A f * d
300—399 14 349.5 -400 14 * -400 = -5600
400— 499 46 449.5 -300 46 * -300 = -13800
500—599 58 549.5 -200 58 * -200 = -11600
600—699 76 649.5 -100 76 * -100 = -7600
700—799 68 749.5 0 68 * 0 = 0
800—899 62 849.5 100 62 * 100 = 6200
900—999 48 949.5 200 48 * 200 = 9600
1000—1099 22 1049.5 300 22 * 300 = 6600
1100—1199 6 1149.5 400 6 * 400 = 2400
∑n = 400 ∑fdx = -13800
A.M. = X = A+∑xfi
n
= 749.5 - 13800
400
X = 715 ------------------------Ans.
Q.4.) Following are the monthly sales of a firm in a year:
Months 1 2 3 4 5 6 7 8 9 10 11 12
Sales (in'000): 50 30 25 44 48 51 55 60 42 35 28 40
Find out mean deviation from the above data.
• Solution:
Let the months be X.
Let the sales be = f .
Sum of sales be = n.
There for n = ∑f.
Months Sales A=7 ∑fdx =
X : f: d = x-A f * d
1 50000 -6 50000 * -6 = -300000
2 30000 -5 30000 * -5 = -150000
3 25000 -4 25000 * -4 = -100000
4 44000 -3 44000 * -3 = -132000
5 48000 -2 48000 * -2 = -96000
6 51000 -1 51000 * -1 = -51000
7 55000 0 55000 * 0 = 0
8 60000 1 60000 * 1 = 60000
9 42000 2 42000 * 2 = 84000
10 35000 3 35000 * 3 = 105000
11 28000 4 28000 * 4 = 112000
12 40000 5 40000 * 5 = 200000
∑n = 508000 ∑fdx = -268000
A.M. = X = A+∑xfi
n
= 7 - 268000
508000
X = 6.47 ------------------------Ans.
5.) Calculate the Mean Deviation from the following data relating to heights (to the nearest inch) of 100 children:
Height (inches): 60 61 62 63 64 65 66 67 68
No. of children: 2 0 15 29 25 12 10 4 3
• Solution:
Let the height be X.
Let the no of children be = f .
Sum of no of children be = n.
There for n = ∑f.
Height No of Children A=64 ∑fdx =
X : f: d = x-A f * d
60 2 -4 2 * -4 = -8
61 0 -3 0 * -3 = 0
62 15 -2 15 * -2 = -30
63 29 -1 29 * -1 = -29
64 25 0 25 * 0 = 0
65 12 1 12 * 1 = 12
66 10 2 10 * 2 = 20
67 4 3 4 * 3 = 12
68 3 4 3 * 4 = 12
∑n = 100 ∑fdx = -11
A.M. = X = A+∑xfi
n
= 7 - 11
100
X = 63.89 ------------------------Ans.
6.) The following table gives the distribution of monthly wages of 1000 workers of a factory:
Wages ($) No. of workers
20 3
40 13
60 43
80 102
100 175
120 220
140 204
160 139
180 69
200 25
220 6
240 1
Find the mean deviation of the above group and also compute die dispersion.
• Solution:
Let the wages be X.
Let the no of workers be = f .
Sum of no of workers be = n.
There for n = ∑f.
Wages No of workers A=140 ∑fdx =
X : f: d = x-A f * d
20 3 -120 3 * -120 = -360
40 13 -100 13 * -100 = -1300
60 102 -80 102 * -80 = -3440
80 175 -60 175 * -60 = -6120
100 220 -40 220 * -40 = 7000
120 204 -20 204 * -20 = -4400
140 139 0 139 * 0 = 0
160 69 20 69 * 20 = 2780
180 26 40 26 * 40 = 2760
200 25 60 25 * 60 = 1500
220 6 80 6 * 80 = 480
240 1 100 1 * 100 = 100
∑n = 1000 ∑fdx = -15000
A.M. = X = A+∑xfi
n
= 140 - 15000
1000
X = 125 ------------------------Ans.
Q.7) Calculate the value of coefficient of mean deviation of the following data:
Marks: 10—20 20—30 30—40 40—50 50—60 60—70 70—80 80—90
No. of Students: 2 6 12 18 25 20 10 7
• Solution:
Let the marks be X.
Let the no of students be = f .
Sum of no of students be = n.
There for n = ∑f.
Marks No of students Mead value of : A=55 ∑fdx =
X : f: X d = x-A f * d
10—20 2 15 -40 2 * -40 = -80
20—30 6 25 -30 6 * -30 = -180
30—40 12 35 -20 12 * -20 = -240
40—50 18 45 -10 18 * -10 = -180
50—60 25 55 0 25 * 0 = 0
60—70 20 65 10 20 * 10 = 200
70—80 10 75 20 10 * 20 = 200
80—90 7 85 30 7 * 30 = 210
∑n = 100 ∑fdx = -700
A.M. = X = A+∑xfi
n
= 55 - 700
100
X = 54.3 ------------------------Ans.
8. Calculate Mean Deviation for the following frequency distribution.
Age (years) No. of persons
1—5 7
6—10 10
11—15 16
16—20 32
21—25 24
26—30 18
31—35 10
36—40 5
41—45 1
• Solution:
Let the age be X.
Let the persons be = f .
Sum of no of persons be = n.
There for n = ∑f.
Age: No of Persons Mead value of : A=23 ∑fdx =
X f X d = x-A f * d
1— 5 7 3 -20 7 * -20 = -140
6— 10 10 8 -15 10 * -15 = -150
11— 15 16 13 -10 16 * -10 = -160
16— 20 32 18 -5 32 * -5 = -160
21— 25 24 23 0 24 * 0 = 0
26— 30 18 28 5 18 * 5 = 90
31— 35 10 33 10 10 * 10 = 100
36— 40 5 38 15 5 * 15 = 75
41— 45 1 43 20 18 * 20 = 20
∑n = 123 ∑fdx = -325
A.M. = X = A+∑xfi
n
= 23 - 325
123
X = 20.3577 ------------------------Ans.
Q.9) Find out coefficient of Mean deviation by using mean (X) from
The following data:
Class: 0—3 3—6 6-9 9-12 12—15 15—18 18—21
Frequency: 2 7 10 12 9 6 4
• Solution:
Let the class be X.
Let the frequency be = f .
Sum of no of frequency be = n.
There for n = ∑f.
Class: Frequency: Mead value of : A=10.5 ∑fdx =
X f X d = x-A f * d
0— 3 2 1.5 -9 2 * -9 = -18
3— 6 7 4.5 -6 7 * -6 = -42
6—9 10 7.5 -3 10 * -3 = -30
9—12 12 10.5 0 12 * 0 = 0
12—15 9 13.5 3 9 * 3 = 27
15—18 6 16.5 6 6 * 6 = 36
18—21 4 19.5 9 4 * 9 = 36
∑n = 50 ∑fdx = 9
A.M. = X = A+∑xfi
n
= 10.5 + 9
50
X = 10.68 ------------------------Ans.
Q.10) Find the mean deviation
Size Frequency
1 and up to 10 1
1 and up to 20 3
1 and up to 30 6
1 and up to 40 8
1 and up to 50 10
• Solution:
Let the size be X.
Let the frequency be = f .
Sum of no of frequency be = n.
There for n = ∑f.
Size: Frequency: Mead value of : A=15.5 ∑fdx =
X f X d = x-A f * d
1— 10 1 5.5 -10 1 * -10 = -10
1— 20 3 10.5 -5 3 * -5 = -15
1—30 6 15.5 0 6 * 0 = 0
1—40 8 20.5 5 8 * 5 = 40
1—50 10 25.5 10 10 * 10 = 100
∑n = 28 ∑fdx = 115
A.M. = X = A+∑xfi
n
= 15.5 + 115
28
= 15.5 + 4.1071
X = 19.6071 ------------------------Ans.
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