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Page No 277:

Question 33:

The following are the scores made by two batsmen A and B in a series of innings:

A : 12 115 6 73 7 19 119 36 84 29
B : 47 12 76 42 4 51 37 48 13 0

Who is better as a run-getter ? Who is more consistent?
 

Answer:

For Batsman A

S. No. Score
(XA)
Deviation from mean
xA=XA-XA
xA2
1
2
3
4
5
6
7
8
9
10
12
115
6
73
7
19
119
36
84
29
−38
65
−44
23
−43
−31
69
−14
34
−21
144
4225
1936
529
1849
961
4761
196
1156
441
N = 10 ΣxA = 500   ΣxA2=17498

X=ΣXN=50010=50SD σA=ΣxA2N=1749810=41.83Coefficient of Variation=σAXA×100=41.8350×100=83.66%

For Batsman B
S. No. Score
XB
Deviation from Mean
xB=XB-XB
xB2
1
2
3
4
5
6
7
8
9
10
47
12
76
42
4
51
37
48
13
0
14
−21
43
9
−29
18
4
15
−20
−33
196
441
1849
81
841
324
16
225
400
1089
N = 10 ΣxB = 330   ΣxB2=5462

XB=ΣXN=33010=33SD σB=ΣxB2N=546210=546.2=23.37Coefficient of Variation=σBXB×100=23.3733×100=70.82%
Since average score of A is more than B, he is better as run-getter but since, coefficient of variance is less for B, therefore B is more consistent.



Page No 280:

Question 1:

The daily wages of ten workers are given below. Find out range and its coefficient.

No. of Workers : A B C D E F G H I J
Wages in (₹) : 175 50 50 55 100 90 125 145 70 60

Answer:

Range = Largest item (L) − Smallest item (S)
          =175 − 50 = 125

Coefficient of Range=L-SL+S=175-50175+50                             =125225=0.55

Page No 280:

Question 2:

Following are the marks obtained by students in Sec . A and Sec . B . Compare the range of marks of students in two sections:

Marks ( Section A) : 20 25 28 45 15 30
Marks ( Section B) : 45 52 36 42 28 25
 

Answer:

For Section A
Range = Largest item (L) − Smallest item (S)
          = 45 − 15 = 30
Coefficient of Range=L-SL+S=45-1545+15                              =3060=0.5

For Section B
Range = Largest item (L) − Smallest item (S)
          = 52 − 25 = 27
Coefficient of Range=L-SL+S                              =52-2552+25=2777                              =0.351
Thus, range of marks of students in section A is greater than range of marks of students in section B.

Page No 280:

Question 3:

Find range and coefficient of range of the following:
(a) Per day earning of seven agricultural labourers in â‚¹:

60, 72, 36, 85, 35, 52, 72

(b)  Mean temperature deviation from normal (2002).
Jan. Feb. Mar. Apr. May June
+1.5 +2.4 +3.1 −1.5 −0.4 +3.3
July Aug. Sep. Oct. Nov. Dec.
−0.1 −0.6 −1.5 −0.6 −1.9 −6.1
 

Answer:

(a) Range = Largest item (L) − Smallest item (S)

= 85 − 35 = 50
Coefficient of Range=L-SL+S=85-3585+35                                     =50120=0.42

(b) Range = Largest item (L) − Smallest item (S)
= +3.3 − (−6.1)
= +3.3 + 6.1 = +9.4
Coefficient of Range=L-SL+S                              =+3.3--6.1+3.3+-6.1                              =+3.3+6.1+3.3-6.1=+9.4-2.8                              =-3.36

Page No 280:

Question 4:

Calculate range and coefficient of range for the following data:

Income (₹) : 50 70 80 90 100 120 130 150
No of workers : 2 8 12 7 4 3 8 6
 

Answer:

Range = Largest item (L) − Smallest item (S)
= 150 − 50 = 100
Coefficient of Range=L-SL+S=150-50150+50                              =100200=0.5



Page No 281:

Question 5:

Calculate range and coefficient of range of the following data:

X : 10 15 20 30 40 50
f : 4 12 7 3 5 2
 

Answer:

Range = Largest Value (L) − Smallest Value (S)
= 50 − 10 = 40
Coefficient of Range=L-SL+S=50-1050+10                              =4060=0.67

Page No 281:

Question 6:

Find the range and coefficient of range of the following:

Age in Years : 5-10 10-15 15-20 20-25
Frequency : 10 15 20 5

Answer:

Range = Largest item (L) − Smallest item (S)
          = 25 − 5 = 20

Coefficient of Range=L-SL+S=25-525+5                              =2030=0.67

Page No 281:

Question 7:

Calculate range and coefficient of range for the following distribution:

Class : 1-5 6-10 11-15 16-20 21-25 26-30
Frequency : 2 8 15 35 20 10
 

Answer:

The given series is an inclusive series. For the calculation of range the given series must first be converted into exclusive series as follows:
 

Inclusive class Exclusive class Frequency
1−5      0.5−5.5 2
6−10   5.5−10.5 8
11−15 10.5−15.5 15
16−20 15.5−20.5 35
21−25 20.5−25.5 20
26−30 25.5−30.5 10

Range = Largest item (L) − Smallest item (S)
          = 30.5 − 0.5 = 30

Coefficient of Range=L-SL+S=30.5-0.530.5+0.5                              =3031=0.97

Page No 281:

Question 8:

The following table gives the height of 100 persons . Calculate dispersion by Range Method.

Height
(in centimetres)
No. of Persons
Below 162 2
Below 163 8
Below 164 19
Below 165 32
Below 166 45
  Below 167 58
Below 168 85
Below 169 93
Below 170 100
 

Answer:

Height Cumulative Frequency Frequency
161−162 2 2
162−163 8 8−2 = 6
163−164 19 19−8 = 11
164−165 32 32−19 = 13
165−166 45 45−32 = 13
166−167 58 58−45 = 13
167−168 85 85−58 = 27
168−169 93 93−85 = 8
169−170 100 100−93 = 7

For the calculation of range, the given data must be converted in continuous series as done above.

Range = Largest item (L) − Smallest item (S)
          = 170 − 161 = 9

Coefficient of Range=L-SL+S=170-161170+161                              =9331=0.03

Page No 281:

Question 9:

Calculate Quartile Deviation and its Coefficient of Rajesh's daily income.

Months : 1 2 3 4 5 6 7 8 9 10 11
Income (₹) : 239 250 251 251 257 258 260 261 262 262 273
 

Answer:

Month Income
(Rs)

 
Month Income
(Rs)

 
1 239 7 260
2 250 8 261
3 251 9 262
4 251 10 262
5 257 11 273
6 258    

N = 11
Q1=Size of N+14th item    =Size of 11+14th item=3rd item    =Size of 3rd item     =251Q3=Size of 3 N+14th item    =Size of 3 3rd item    =Size of 9th item    =Size of 9th item    =262Quartile deviation=Q3-Q12                           =262-2512                           =112=5.5Coefficient of Q. D.=Q3-Q1Q3+Q1                           =262-251262+251                           =11513=0.0214
Thus, the quartile deviation and coefficient of quartile deviation are 5.5 and 0.021 respectively.

Page No 281:

Question 10:

Find the Quartile Deviation and its Coefficient from the following data relating to the daily wages of seven workers:

Daily Wages ( in â‚¹) : 50 90 70 40 80 65 60

Answer:

S. No. Daily Wages
1 40
2 50
3 60
4 65
5 70
6 80
7 90

For the calculation of quartile deviation, we first arrange the given data in ascending order as above. The quartile deviation is then calculated in the following manner. 
Q1=Size of N+14th item    =Size of 7+14th item=Size of 2nd item    =50Q3=Size of 3 N+14th item    =Size of 6th item=80Quartile deviation=Q3-Q12=80-502=302=15Coefficient of Q. D.=Q3-Q1Q3+Q1                           =80-5080+50=30130=0.23
Thus, the quartile deviation and coefficient of quartile deviation are 15 and 0.23 respectively.

Page No 281:

Question 11:

Find out Quartile Deviation and Coefficient of Quartile Deviation of the following items:

145 130 200 210 198
234 159 160 178 257
260 300 345 360 390
 

Answer:

S. No. Item
1 130
2 145
3 159
4 160
5 178
6 198
7
8
9
10
11
12
13
14
15
200
210
234
257
260
300
345
360
390
   
   

We first arrange the given data in ascending order as above.

Q1=Size of N+14th item    =Size of 15+14th item    =Size of 4th item=160Q3=Size of 3 N+14th item    =Size of 12th item    =300Quartile Deviation=Q3-Q12=300-1602                          =1402=70Coefficient of Q. D.=Q3-Q1Q3+Q1=300-160300+160                           =140460=0.304
Thus, the quartile deviation and coefficient of quartile deviation are 70 and 0.304 respectively.



Page No 282:

Question 12:

Find out Quartile Deviation , Coefficient of Quartile Deviation of the following data:

X : 10 15 20 25 30 35 40 45
f : 6 17 29 38 25 14 9 1
 

Answer:

X f Cumulative Frequency
10 6 6
15 17 23
20 29 52
25 38 90
30 25 115
35 14 129
40 9 138
45 1 139
    N = 139

Q1=Size of N+14th item    =Size of 139+14th item    =Size of 35th item    
The 35th item corresponds to 52 in the cumulative frequency.
Q1=20

Q3=Size of 3 N+14th item    =Size of 105th item    =30
The 105th item corresponds to 115 in the cumulative frequency.
Q3=30
Coefficient of Q. D.=Q3-Q1Q3+Q1=30-2030+20=1050=0.2
Thus, coefficient of quartile deviation is 0.2.

Page No 282:

Question 13:

Calculate the Quartile Deviation and Coefficient of Quartile Deviation of the following data:

Marks : 5-9 10-14 15-19 20-24 25-29 30-34 35-39
Students : 1 3 8 5 4 2 2
 

Answer:

Marks Students
(f)

 
Cumulative Frequency
5−9 1 1
10−14 3 1 + 3 = 4
15−19 8 4 + 8 = 12
20−24 5 12 + 5 = 17
25−29 4 17 + 4 = 21
30−34 2 21 + 2 = 23
35−39 2 23 + 2 = 25
  N = Æ©f = 25  

Q1=Size of N4th item    =Size of 254th item    =6.25th item
This corresponds to the class interval (15−19).
Q1=l1+N4-c.ff×i    =15+6.25-48×4    =15+2.258×4    =15+98    =15+1.125    =16.125
 Q1=16.125

Q2=Size of N2th item    =Size of 252th item    =12.5th item
This corresponds to the class interval (20−24).

Q2=l1+N2-c.ff×i    =20+12.5-125×4    =20+25    =20+0.4    =20.04
 Q2=20.04

Q3=Size of 3 N4th item    =Size of 18.75th item
This corresponds to the class interval (25−29).

Q3=l1+3N4-c.ff×i    =25+18.75-174×4    =25+1.75    =26.75
 Q3=26.75
Coefficient of Quartile Deviation=Q3-Q1Q3+Q1                                               =26.75-16.12526.75+16.125                                               =10.62542.875                                               =0.25

Page No 282:

Question 14:

Calculate the Semi-interquartile Range and its Coefficient of the following data:

Marks : 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of Students : 4 8 10 14 10 9 5
 

Answer:

Marks No. of Students
(f)
c.f
0−10 4 4
10−20 8 12
20−30 10 22
30−40 14 36
40−50 10 46
50−60 9 55
60−70 5 60
  N = 60  
The first quartile class is given by the size of the N4thitem, i.e. the 604thitem, which is the 15th item.
This corresponds to the class interval 20−30, so this is the first quartile class.

Q1=l1+N4-c.ff×i    =20+15-1210×10    =23
The third quartile class is given by the size of the 3N4thitem, i.e. the 3604thitem, which is the 45th item.
This corresponds to the class interval 40−50, so this is the third quartile class.

Q3=l1+3N4-c.ff×i    =40+45-3610×10    =40+9    =49

Quartile Deviation=Q3-Q12=49-232=13Coefficient of Q.D.=Q3-Q1Q3+Q1=49-2349+23=2672=0.361

Page No 282:

Question 15:

Calculate the Interquartile Range for the data given below:

Class : 35-40 30-35 25-30 20-25 15-20 10-15 5-10 0-5
Frequency : 1 4 9 11 10 6 5 4
 

Answer:

Class Frequency
(f)
c.f.
0−5 4 4
5−10 5 9
10−15 6 15
15−20 10 25
20−25 11 36
25−30 9 45
30−35 4 49
35−40 1 50
  N = 50  

The first quartile class is given by the size of the N4thitem, i.e. the 504thitem, which is the 12.5th item.
This corresponds to the class interval 10−15, so this is the first quartile class.
Q1=l1+N2-c.ff×i    =10+12.5-96×5    =10+3.5×56    =10+2.92    =12.92
The third quartile class is given by the size of the 3N4thitem, i.e. the 3504thitem, which is the 37.5th item.
This corresponds to the class interval 25−30, so this is the third quartile class.

Q3=l1+3 N4-c.f.f×i    =25+37.5-369×5    =25+1.5×59    =25.83

 Interquartile range=Q3-Q1                               =25.83-12.92                               =12.91

Page No 282:

Question 16:

Calculate the Mean deviation from median of the following data:

Marks : 41 66 59 38 54 21 32 49 68

Answer:

Marks
(X)

 
Deviation from Median
D= |XM|
(M = 49)
21 28
32 17
38 11
41 8
49 0
54 5
59 10
66 17
68 19
  ΣD = 115

For the calculation of mean deviation, the given data is first arranged in the ascending order as done above. Then median is calculated in the following manner.
Median=Size of N+12th item           =Size of 9+12th item           =Size of 5th item           =49 Mean deviation from Median=ΣDN=1159=12.77
Thus, mean deviation from median is 12.77.

Page No 282:

Question 17:

Find  Coefficient of Mean deviation from median :

Price (Rs) : 25 28 32 32 36 48 44 45 50 50

Answer:

Price Deviation from Median
D = |XM|
M = 40
25 15
28 12
32 8
32 8
36 4
44 4
45 5
48 8
50 10
50 10
  ΣD = 84

Median=Size of N2 item+Size of N2+1th item2           =Size of 102 item+Size of 102+1th item2           =Size of 5th item+Size of 6th item2           =36+442=802=40
Mean deviation from Median=ΣDN=8410=8.4 Coefficient of MD=Mean deviation from MedianMedian                           =8.440                           =0.2
Thus, coefficient of mean deviation from median is 0.2.

Page No 282:

Question 18:

Calculate the Mean Deviation from mean and median of the following data:

X : 54 71 57 52 49 45 72 57 47

Answer:

X Deviation from Median
DM =|X−M|
M = 54
Deviation from Mean
DX¯ =X-X
 X=56
45 9 11
47 7 9
49 5 7
52 2 4
54 0 2
57 3 1
57 3 1
71 17 15
72 18 16
ΣX = 504 Σ|X−Md| = 64 ΣX-X=66

Median = Size of N+12th item = 9+12th = 5th item = 54Mean = ΣXN = 5049 = 56
Mean deviation from Mean=Σ X-XN=669=7.33Mean deviation from Median=ΣX-MN=649=7.11

Page No 282:

Question 19:

Calculate  Mean deviation from (i) arithmetic mean, (ii) Median:

Marks : 7 4 10 9 15 12 7 9 7

Answer:

Marks
(X)

 
Deviation from Mean
Dx¯ =X-X
Deviation from Median
DM =|X−M|
4 4.89 5
7 1.89 2
7 1.89 2
7 1.89 2
9 0.11 0
9 0.11 0
10 1.11 1
12 3.11 3
15 6.11 6
ΣX = 80 ΣX-X=21.11 Σ|X−Md| = 21

X=ΣXN=809=8.89Median=Size of N+12th item           =Size of 9+12th item           =Size of 5th item           =9 Mean deviation from Mean=X-XN=21.119=2.34 Mean deviation from Median=X-MN=219=2.32

Page No 282:

Question 20:

Find out the average deviation from median for the following distribution:

Values : 6 12 18 24 30 36 42
Frequency : 4 7 9 18 15 10 5
 

Answer:

Value
(X)
Frequency
(f)
c.f Deviation from Median
|DM|=|XM|
(M = 24)
f|DM|
6 4 4 18 72
12 7 11 12 84
18 9 20 6 54
24 18 38 0 0
30 15 53 6 90
36 10 63 12 120
42 5 68 18 90
  Σf = N = 68     Σf|dm| = 510

Median=Size of N+12th item           =Size of 68+12th item           =34.5th item           
This corresponds to value 24. Thus, median is 24.
Mean deviation from Median=ΣfDMΣf=51068=7.5

Page No 282:

Question 21:

Calculate Mean Deviation from median:

No. of tomatoes per plant : 0 1 2 3 4 5 6 7 8 9 10
No. of Plants : 2 5 7 11 18 24 12 8 6 4 3
 

Answer:

Tomatoes Frequency
(f)

 
c.f Deviation from Median
|DM|=|X−M|
(M = 5)
f|DM|
0 2 2 5 10
1 5 7 4 20
2 7 14 3 21
3 11 25 2 22
4 18 43 1 18
5 24 67 0 0
6 12 79 1 12
7 8 87 2 16
8 6 93 3 18
9 4 97 4 16
10 3 100 5 15
  Σf = N = 100     Σf|dm|=168

Median=Size of N+12th item           =Size of 100+12th item           =50.5th item
This corresponds to 5. Thus, median is 5.
Mean deviation from Median=ΣfDMΣf=168100=1.68



Page No 283:

Question 22:

Calculate (a) Median Coefficient of dispersion from the following data:

Size of item : 4 6 8 10 12 14 16
Frequency : 2 4 5 3 2 1 4
 

Answer:

Calculation of MD From Median
 
Size of Item
(X)
Frequency
(f)
c.f Deviation from Median
|DM|=|X−M|
(M = 8)
f|DM|
4 2 2 4 8
6 4 6 2 8
8 5 11 0 0
10 3 14 2 6
12 2 16 4 8
14 1 17 6 6
16 4 21 8 32
  Σf = N = 21     Σf|dm| = 68

Median=Size of N+12th item           =Size of 21+12th item           =11th item
This corresponds to size 8. Thus, median is 8.
Mean deviation from Median=ΣfDMΣf=6821=3.24Coefficient of MD from Median=MD from MedianMedian                                              =3.248=0.405
 
Calculation of MD From Mean
 
Size
(X)
Frequency
(f)
fX Deviation from Mean DX
X=9.71 
fDX
4 2 8 5.71 11.42
6 4 24 3.71 14.84
8 5 40 1.71   8.55
10 3 30  0 .29    0.87
12 2 24 2.29   4.58
14 1 14 4.29   4.29
16 4 64 6.29 25.16
  Σf = 21 ΣfX = 204   ΣfdX=69.71

X=ΣfXΣf=20421=9.71Mean deviation from Mean=ΣfDXΣf=69.7121=3.32Coefficient of Mean deviation from Mean=MD from MeanMean                                                           =3.329.71=0.342

Page No 283:

Question 23:

Calculate Mean and Mean Deviation and coefficient  of M.D. for the following distribution:

Weekly wages : 20-40 40-60 60-80  80-100
Workers : 20 40 30 10

Answer:

Weekly wages
(X)
Mid value
m
f fm DX=m-XX=56 fDX
20−40 30 20 600 26 520
40−60 50 40 2000 6 240
60−80 70 30 2100 14 420
80−100 90 10 900  34 340
    Σf = 100 Σfm = 5600   ΣfDX=1520

X=ΣfmΣf=5600100=56Mean deviation from Mean=ΣfDXΣf=1520100=15.2Coefficient of MD=MD from MeanMean                           =15.256                           =0.27

Page No 283:

Question 24:

Find Mean Deviation from median of the marks secured by 100 students in a class-test as given below:

Marks : 60-63 63-66 66-69 69-72 72-75
No. of Students : 5 18 42 27 8
 

Answer:

Marks Mid Value f c.f Deviation from
Median
DM = |m−M|
(M= 67.93)
f|DM|
60−63 61.5 5 5 6.43 32.15
63−66 64.5 18 23 3.43 61.74
66−69 67.5 42 65  0 .43 18.06
69−72 70.5 27 92 2.57 69.39
72−75 73.5 8 100 5.57 44.56
    Σf = 100     Σf|DM| = 225.9
The median class is given by the size of the N2thitem, i.e. the 1002thitem, which is the 50th item.
This corresponds to the class interval 66−69, so this is the median class.

M=l1+N2-c.ff×i  =66+50-2342×3  =66+1.93  =67.93Mean deviation from Median=Σf DMΣf=225.9100=2.259=2.26

Page No 283:

Question 25:

Using Mean Deviation from median of the income group of 5 and 7 members given below, compare which of the group has more variability?

Group A : 4000 4200 4400 4600 4800                 
Group B : 3000 4000 4200 4400 4600 4800 5800
 

Answer:

For Group A

Income
(X)
Deviation from median
DM=|XM|
(M=4400)
4000 400
4200 200
4400 0
4600 200
4800 400
  Σ|DM|=1200

Mean deviation from Median=ΣDMN=12005=240Coefficient of MD from median= MD from MedianMedian=2404400=.054                                              

For Group B
Income
 
Deviation from median
DM=|XM|
(M=4400)
3000 1400
4000 400
4200 200
4400 0
4600 200
4800 400
5800 1400
  Σ|DM|=4000

Mean deviation from Median=ΣDMN=40007=571.4Coefficient of MD from median=MD from MedianMedian=571.44400=0.13

In case of group B, as the coefficient of MD is more, so group B has greater variation.

Page No 283:

Question 26:

Calculate the standard Deviation of wage earner's daily earnings: 

Week : 1 2 3 4 5 6 7 8 9 10
Earnings (in â‚¹) : 54 62 63 65 68 71 73 78 82 84
 

Answer:

S.No.
(Week)
Farming
(X)
Deviation X-X
X=70
Square of deviation
X-X2
1
2
3
4
5
6
7
8
9
10
54
62
63
65
68
71
73
78
82
84
−16
−8
−7
−5
−2
1
3
8
12
14
256
64
49
25
4
1
9
64
144
196
  ΣX = 700   ΣX-X2=812

The mean of the given series can be calculated in the following manner.
X=ΣXN=70010=70
To calculate standard deviation, we use the following formula.
SD σ=ΣX-X2NPutting the vaules we get,SD σ=81210=81.2=9.01
Thus, standard deviation of wage earner's daily earnings is 9.01.

Page No 283:

Question 27:

Calculate Standard Deviation of the following two series . Which series has more variability:

A: 58 59 60 65 66 52 75 31 46 48
B: 56 87 89 46 93 65 44 54 78 68

Answer:

For A

S. No. (X) Deviation
x=X-XX=56
x2
1
2
3
4
5
6
7
8
9
10
58
59
60
65
66
52
75
31
46
48
2
3
4
9
10
−4
19
−25
−10
−8
4
9
16
81
100
16
361
625
100
64
  ΣX = 560   Σx2 = 1376

X=ΣXN=56010=56SD σ=Σx2N=137610=137.6=11.7Coefficient of variation=σX×100                                 =11.756×100                                 =20.89%

For B
S. No. Y Deviation
y=Y-yy=68
y2
1
2
3
4
5
6
7
8
9
10
56
87
89
46
93
65
44
54
78
68
−12
19
21
−22
25
−3
−24
−14
10
0
144
361
441
484
625
9
576
196
100
0
  ΣY = 680   Σy2 = 2936

Y¯=ΣYN=68010=68SD σ=Σy2N=293610=293.6=17.1Coefficient of variation=σY×100                               =17.168×100=171068                               =25.14%
Since the coefficient of variation of series B is more, it is said to have more variability.

Page No 283:

Question 28:

Calculate Mean and Standard Deviation of income of 8 employees of  a firm:

Income (₹): 100 120 140 120 180 140 120 150

Answer:

S. No.
(X)
Deviation
x=X-XX=133.75
x2
100
120
140
120
180
140
120
150
−33.75
−13.75
6.25
−13.75
46.25
6.25
−13.75
16.25
1139.06
189.06
39.06
189.06
2139.06
39.06
189.06
264.06
ΣX= 1070   Σx2 = 4187.48

X=ΣXN=10708=133.75SD=Σx2N=4187.488=523.43=22.87
Thus, mean standard deviation is 22.87.

Page No 283:

Question 29:

Calculate Mean and Standard Deviation from the following data:

Marks ( Above) : 0 10 20 30 40 50 60 70
No. of Students : 100 90 75 50 25 15 5 0
 

Answer:

As we are given marks (above) series, we can convert it into classes and frame the following table.

Marks Mid Value
(m
)
cf Frequency
(f)
fm Deviation from mean
x=m-XX=31
x2 fx2
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
5
15
25
35
45
55
65
75
100
90
75
50
25
15
5
0
10
15
25
25
10
10
5
0
50
225
625
875
450
550
325
0
-26
-16
-6
4
14
24
34
44
676
256
36
16
196
576
1156
1936
6760
3840
900
400
1960
5760
5780
0
      Σf = 100 Σfm = 3100     Σfx2=25400

X=ΣfmΣf=3100100=31SD=Σfx2Σf=25400100       =254       =15.93



Page No 284:

Question 30:

Calculate Mean and Variance for the following frequency distribution:

X : 1 2 3 4 5 6 7 8 9 10
f : 2 5 8 16 21 18 13 10 4 3
 

Answer:

X f fX Deviation
x=X-XX=5.5
x2 fx2
1
2
3
4
5
6
7
8
9
10
2
5
8
16
21
18
13
10
4
3
2
10
24
64
105
108
91
80
36
30
−4.5
−3.5
−2.5
−1.5
−.5
.5
1.5
2.5
3.5
4.5
20.25
12.25
6.25
2.25
.25
.25
2.25
6.25
12.25
20.25
40.5
61.25
50
36
5.25
4.5
29.25
62.5
49
60.75
  Σf = 100 ΣfX = 550     Σfx2 = 399

X=ΣfXΣf=550100=5.5Variance σ2=Σfx2Σf=399100=3.99
Thus, mean and variance are 5.5 and 3.99 respectively.

Page No 284:

Question 31:

Calculate  Mean, Standard Deviation and Variance:

Variable : 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequency : 2 5 7 13 21 16 8 3
 

Answer:

Variable Mid Value (m) Frequency
(f)
fm Deviation Mean
x=m-XX=21.9
x2 fx2
0 − 5
5 − 10
10 − 15
15 − 20
20 − 25
25 − 30
30 − 35
35 − 40
2.5
7.5
12.5
17.5
22.5
27.5
32.5
37.5
2
5
7
13
21
16
8
3
5
37.5
87.5
227.5
472.5
440
260
112.5
19.4
14.4
9.4
4.4
0.6
5.6
10.6
15.6
376.36
207.36
88.36
19.36
0.36
31.36
112.36
243.36
752.72
1036.8
618.52
251.68
7.56
501.76
898.88
730.08
    Σf = 75 Σfm = 1642.5     Σfx2 = 4798

X=ΣfmΣf=1642.575=21.9SD=Σfx2Σf=479875=63.97=7.99Variance σ2=Σfx2Σf=479875=63.97

Page No 284:

Question 32:

Calculate the Coefficient of Variation for the following distribution of the wages of 200 workers in a factory:

Wages ( in â‚¹) : 40-49 50-59 60-69 70-79 80-89 90-99 100-109 110-119
No. of workers : 11 23 40 60 35 16 9 6
 

Answer:

In order to calculate coefficient of variation, first we will convert the given inclusive series into exclusive series as follows:

Wages Mid Point
(m)
Frequency
(f)
fm Deviation from
x=m-XX=74.45
x2 fx2
39.5 − 49.5
49.5 − 59.5
59.5 − 69.5
69.5 − 79.5
79.5 − 89.5
89.5 − 99.5
99.5 − 109.5
109.5 − 119.5
44.5
54.5
64.5
74.5
84.5
94.5
104.5
114.5
11
23
40
60
35
16
9
6
489.5
1253.5
2580
4470
2957.5
1512
940.5
687
29.95
19.95
9.95
.05
10.05
20.05
30.05
40.05
897
398
99
.002
101
402
903
1604
9867
9154
3960
0.12
3535
6432
8127
9624
    Σf = 200 Σfm = 14890     Σfx2 = 50699.12

X=ΣfmΣf=14890200=74.45SD σ=Σfx2Σf=50699.12200=253.4956=15.921Coefficient of variation=σX×100                                 =15.92174.45×100=1592.174.45                                 =21.38%

Page No 284:

Question 34:

The index number of prices of cotton and coal shares in 1998 were as under:
Index Number of Prices:

Cotton : 188 178 173 164 172 183 184 185 211 217 232 240
Coal : 131 130 130 129 129 129 127 127 130 137 140 142

Which of these two shares do you consider more variable in prices:
 

Answer:

For Cotton
S. No. Price
X
Deviation from mean
x=X-XX=193.9
x2
1
2
3
4
5
6
7
8
9
10
11
12
188
178
173
164
172
183
184
185
211
217
232
240
5.9
15.9
20.9
29.9
21.9
10.9
9.9
8.9
17.1
23.1
38.1
46.1
34.81
252.81
436.81
894.01
479.61
118.81
98.01
79.21
292.41
533.61
1451.61
2125.61
  ΣX = 2327   Σx2 = 6796.92

X=ΣXN=232712=193.9SD=Σx2N=6796.9212=566.41=23.80Coefficient of Variation=SDX×100=23.8193.9×100=12.27%
 
For Coal
S. No. Price
X
Deviation from mean
x=X-X
x2
1
2
3
4
5
6
7
8
9
10
11
12
131
130
130
129
129
129
127
127
130
137
140
142
−.75
−1.75
−1.75
−2.75
−2.75
−2.75
−4.75
−4.75
−1.75
5.25
8.25
10.25
.56
3.06
3.06
7.56
7.56
7.56
22.56
22.56
3.06
27.56
68.06
105.06
  ΣX = 1581   Σx2= 278.22

X=ΣXN=158112=131.75SD σ=Σx2N=278.2212=4.815Coefficient of Variation=σX×100=4.815131.75×100=3.65%
Since coefficient of variance is more of cotton than that of coal. So, prices of cotton shares are more variable.

Page No 284:

Question 35:

Calculate the arithmetic mean, standard deviation and variance from the following distribution:

Class : 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequency : 2 5 7 13 21 16 8 3
 

Answer:

Size Mid Value
(m)
(f) fm Deviation from mean
x
x2 fx2
0 − 5
5 − 10
10 − 15
15 − 20
20 − 25
25 − 30
30 − 35
35 − 40
2.5
7.5
12.5
17.5
22.5
27.5
32.5
37.5
2
5
7
13
21
16
8
3
5
37.5
87.5
227.5
472.5
440
260
112.5
19.4
14.4
9.4
4.4
.6
5.6
10.6
15.6
376.36
207.36
88.36
19.36
.36
31.36
112.36
243.36
752.72
1036.8
618.52
251.68
7.56
501.76
898.88
730.08
    Σf = 75 Σfm = 1642.5     Σfx2 = 4798

X=ΣfmΣf=1642.575=21.9SD=Σfx2Σf=479875=63.97=7.99Variance σ2=Σfx2Σf=479875=63.97

Page No 284:

Question 36:

Calculate arithmetic mean, standard deviation and variance from the following series:

Marks : 70-80 60-70 50-60 40-50 30-40 20-30
No of students : 7 11 22 0 15 5
 

Answer:

Marks Mid Value
(m)
f fm Deviation from mean
x=m-XX=51.67
x2 fx2
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
25
35
45
55
65
75
5
15
0
22
11
7
125
525
0
1210
715
525
26.67
16.67
6.67
3.33
13.33
23.33
711.29
277.89
44.49
11.09
177.69
544.29
3556.45
4168.35
0
243.98
1954.59
3810.03
    Σf = 60 Σfm = 3100     Σfx2 = 13733.4

X=ΣfmΣf=310060=51.67SD=Σfx2N=13733.460=228.89=15.13Variance=SD2=228.89



Page No 285:

Question 37:

You are given the following data about height of boys and girls:

  Boys Girls
Number 72 38
Average height( in inches) 68 61
Variance of distribution (in inches) 9 4

(a) Calculate Coefficient of Variation.
(b) Decide whose height is more variable.

Answer:

(a)
For Boys
Standard Deviation=Variance=9=3Coefficient of variance=SDX×100                                =368×100=4.41%

For Girls
Standard deviation=Variance=4=2Coefficient of variance=SDX×100                                =261×100=3.27%
(b) Since coefficient of variance of boys is more than that of girls, thus the height of boys is more variable.

Page No 285:

Question 38:

Lives of two models of refrigerators in a recent survey are as under:

Life No fo Years : 0-2 2-4 4-6 6-8 8-10 10-12
No of Refrigerators :            
Model A : 5 16 13 7 5 4
Model B : 2 7 12 19 9 1

What is the average life of each model of refrigerators? Which model has more uniformity?

Answer:

Model A:

Years Mid Value
(m)
f fm Deviation from mean
xA=m-XX=5.12
xA2 fxA2
0 − 2
2 − 4
4 − 6
6 − 8
8 − 10
10 − 12
1
3
5
7
9
11
5
16
13
7
5
4
5
48
65
49
45
44
−4.12
−2.12
−.12
1.88
3.88
5.88
16.97
4.49
0.01
3.53
15.05
34.57
84.85
71.84
0.13
24.71
75.25
138.28
    Σf = 50 Σfm = 256     fxA2=395.06

XA=ΣfmΣf=25650=5.12SD=ΣfxA2Σf=395.0650=7.90=2.81Coefficient of variation=SDXA=2.815.12×100=54.9%
Average life of model A is 5.12 year.

Model B:
Years Mid Value
(m)
f fm Deviation from mean
xB=m-XX=6.16
xB2 fxB2
0 − 2
2 − 4
4 − 6
6 − 8
8 − 10
10 − 12
1
3
5
7
9
11
2
7
12
19
9
1
2
21
60
133
81
11
−5.16
−3.16
−1.16
0.84
2.84
4.84
26.62
9.98
1.34
0.70
8.06
23,42
53.24
69.86
16.08
13.3
72.54
23.42
    Σf = 50 Σfm = 308     fxB2=248.44

XB=ΣfmΣf=30850=6.16SD=ΣfxB2Σf=248.4450=4.96=2.23Coefficient of variation=SDXB=2.236.16×100=36.2%
Average life of model B is 6.16 year.

Since the coefficient of variation of model B is less than that of A, thus model B has more uniformity.

Page No 285:

Question 39:

Two brands of Tyres are tested with the following results:

Life ( '000 miles) : 20-25 25-30 30-35 35-40 45-50 50-55
Brand X   8 15 12 18 13 9
Brand Y   6 20 32 30 12 0
Which brand has a greater variation?

Answer:

For Brand 'X'


Life
Mid value
m
f fm Deviation
from mean
x=m-X
X=35.17
x2 fx2
20−25 22.5 8 180 −12.67 160.53 1284.24
25−30 27.5 15 412.5   −7.67 58.83 882.45
30−35 32.5 12 390   −2.67 7.13 85.56
35−40 37.5 18 675     2.33 5.43 97.74
40−45 42.5 13 552.5     7.33 53.73 698.49
45−50 47.5 9 427.5   12.33 152.03 1368.27
    Σf =75 Σfm =2637.5     Σfx2= 4416.75

X=ΣfmΣf=2637.575=35.17SDX=Σfx2Σf=4416.7575=58.89=7.67Coefficient of variation=SDXX×100                                 =7.6735.17×100                                 =21.81%

For Brand 'Y'

Life
Mid value
m
f fm Deviation
from mean
x=m-X
X=33.60
x2 fx
20−25 22.5 6 135 −11.1 123.21 739.26
25−30 27.5 20 550   −6.1   37.21   744.2
30−35 32.5 32 1040   −1.1     1.21   38.72
35−40 37.5 30 1125     3.9   15.21   456.3
40−45 42.5 12 510     8.9   79.21 950.52
45−50 47.5 0 0   13.9 193.21 0
    Σf=100 Σfm=3360     Σfx2=2929

X=ΣfmΣf=3360100=33.60SDY=ΣfxB2Σf=2929100=29.29=5.41Coefficient of variation=SDYX×100                                 =5.4133.6×100                                 =16.101

As the coefficient of variation is more for brand X, it is said to be more variable than brand Y.

Page No 285:

Question 40:

From the data given below state which series is more consistent:

Variable : 10-20 20-30 30-40 40-50 50-60 60-70
Series A : 10 18 32 40 22 18
Series B : 18 22 40 32 18 10
 

Answer:

For Series A

Variable Mid value
(m)
f fm Deviation
from mean
x=m-XA
XA=42.14
xA2 fxA2
10−20 15 10   150 −27.14 736.58  7365.8
20−30 25 18   450 −17.14 293.78 5288.04
30−40 35 32 1120   −7.14   50.98 1631.36
40−50 45 40 1800     2.86     8.18     327.2
50−60 55 22 1210   12.86 165.38 3638.36
60−70 65 18 1170   22.86 522.58  9406.44
    Σf=140 Σfm=5900     fxA2=27657.2
XA=ΣfmΣf=5900140=42.14SDA=Σfx2Σf=27657.2140=14.06Coefficient of variation=SDAXA×100                                 =14.0642.14×100=33.36%

For Series B

Variable
Mid value
(m)
f fm Deviation
from mean
x=m-XB
XB=37.86
xB2 fxB2
10−20 15 18   270 −22.86 522.58 9406.44
20−30 25 22   550 −12.86 165.38 3638.36
30−40 35 40 1400   −2.86     8.18    327.2
40−50 45 32 1440     7.14   50.98 1631.36
50−60 55 18   990   17.14  293.78 5288.04
60−70 65 10   650   27.14  736.58  7365.8
    Σf=140 Σfm=5300     ΣfxB2=27657.2

XB=ΣfmΣf=5300140=37.86SDB=ΣfxB2Σf=27657.2140=197.55=14.06Coefficient of variation=SDBXB×100                                 =14.0637.86×100=37.14%

As coefficient of variation for series 'A' is less than series 'B', therefore, series A is more consistent.

Page No 285:

Question 41:

The following table gives the distribution of wages in the two branches of a factory:

Monthly wages ( in â‚¹) : 100-150 150-200 200-250 250-300 300-350 Total
Number of workers :            
Branch A : 167 207 253 205 168 1000
Branch B : 63 93 157 105 82 500
Find the mean and standard deviation for the two branches for the wages separately.
(a) Which branch pays higher average wages?
(b) Which branch has greater variability in wages in relation to the average wages?
(c) What is the average monthly wage for the factory as a whole?
(d) What is the variance of wages of all the workers in the two branchesA and B taken together?

Answer:

For Branch A

Monthly Wages Mid Value
(m)
f fm Deviation from mean
xA=m-XAXA=225
xA2 fxA2
100 − 150
150 − 200
200 − 250
250 − 300
300 − 350
125
175
225
275
325
167
207
253
205
168
20875
36225
56925
56375
54600
−100
−50
0
50
100
10000
2500
0
2500
10000
1670000
517500
0
512500
1680000
    Σf = 1000 Σfm = 225000     ΣfxA2=4380000
 
XA=ΣfmΣf=2250001000=225SDA=ΣfxA2Σf=43800001000=4380=66.19Coefficient of variation=SDAXA×100                                 =66.19225×100                                 =29.42%


For Branch B
Monthly wage Mid value
(m)
f fm Deviation from mean
xB=m-XBXB=230
xB2 fxB2
100 − 150
150 − 200
200 − 250
250 − 300
300 − 350
125
175
225
275
325
63
93
157
105
82
7875
16275
35325
28875
26650
−105
−55
−5
45
95
11025
3025
25
2025
9025
694575
281325
3925
212625
740050
    Σf = 500 Σfm = 115000     ΣfxB2=1932500

XB=ΣfmΣf=115000500=230SDB=ΣfxB2Σf=1932500500=3865=62.16Coefficient of variation=SDBXB×100=62.16230×100=27.02%

(a) Branch B pays higher average monthly wages i.e. Rs 230.
(b) Branch A has greater variability since its coefficient of variation is higher than that of B.
(c)
Combined mean X¯AB=XA NA+XB NBNA+NB                                  =225×1000+230×5001000+500                                  =225000+1150001500                                  =3400001500                                  =226.67
Thus, the average monthly wage for the factory as a whole is Rs 226.67.
                            
(d)
σA,B=NAσA2+NB σB2+NAdA2+NBdB2NA+NBdA2=XA-XA,B2=225-226.672   =2.79dB2=XB-XA,B2=230-226.672    =11.09σA,B =1000×4381.12+500×3863.86+1000×2.79+1000×11.091000+500       =4381120+1931930+2790+110901500       =63269301500=4217.95Variance=SD2            =4217.952             =4217.95
Thus, the variance of wages of all the workers in the two branches- A and B taken together is Rs 4217.95.



Page No 286:

Question 42:

Goals scored by two teams A and B in football matches were as follows:

No. of Goals in a match : 0 1 2 3 4
No.of Matches:     A : 27 9 8 5 4
                              B : 17 9 6 5 3
Find the team which is more consistent in its performance.

Answer:

For Team A
X f fX Deviation from mean
xA=X-XA
xA2 fxA2
0
1
2
3
4
27
9
8
5
4
0
9
16
15
16
−1.06
−0.06
0.94
1.94
2.94
1.12
0.0036
0.88
3.76
8.64
30.24
0.03
7.04
18.8
34.56
  Σf = 53 ΣfX = 56     ΣfxA2=90.68

XA=ΣfXΣf=5653=1.06SDA=ΣfxA2Σf=90.6853=1.71      =1.31Coefficient of variation=SDAXA×100=1.311.06×100=123.6%
 
For Team B
X f fX Deviation from mean
xB=X-XB
xB2 fxB2
0
1
2
3
4
17
9
6
5
3
0
9
12
15
12
−1.2
−0.2
0.8
1.8
2.8
1.44
0.04
0.64
3.24
7.84
24.48
0.36
3.84
16.2
23.52
  Σf = 40 ΣfX= 48     ΣfxB2=68.4

XB=ΣfXΣf=4840=1.2SDB=ΣfxB2Σf=68.440=1.71=1.31Coefficient of variation=SDBXB×100                                 =1.311.2×100=109.1%
Since, coefficient of variation is lesser for team B, it is more consistent.

Page No 286:

Question 43:

Find mean and the standard deviation of the following two groups taken together:

Group Number Mean S.D.
A 113 159 22.4
B 121 149 20.0
 

Answer:

Given,XA=159, σA=22.4, NA=113XB=149, σB=20, NB=121Combined MeanXA,B=XANA+XBNBNA+NB     =159×113+149×121113+121     =17967+18029234     =35996234     =153.83
Now, to calculate combined standard deviation, we need to find,

dA2=XA-XA,B2=159-153.832                            =5.172=26.73dB2=XB-XA,B2=149-153.832                          =-4.832=23.33

Combined Standard Deviation
σA,B=NA σA2+NBσB2+NAdA2+NBdB2NA+NB      =11322.42+121202+11326.73+12123.33113+121      =56698.88+48400+3020.49+2822.93234      =110942.3234      =474.112      =21.774 or 21.8 approx.
Thus, mean and standard deviation of the two groups-A and B taken together are 23.33 and 21.8 respectively.

Page No 286:

Question 44:

The number examined , the mean weight and standard deviations in each group of examination by two medical examiners are given below . Calculate mean and standard deviation of both the groups taken together.

Medical Examiner Number Examined Mean Weight
(lbs)
Standard deviation
(lbs)
 A 50 113 6.5
B 60 120 8.2
 

Answer:

Given:XA=113, σA=6.5, NA=50XB=120, σB=8.2, NB=60Combined MeanXA,B=XA NA+XB NBNA+NB     =113×50+120×6050+60=5650+7200110     =12850110     =116.82

Combined Standard Deviation
dA2=XA-XA,B2=113-116.822                          =-3.822=14.59dB2=XB-XA,B2=120-116.822=3.182                          =10.11σA,B=NA σA2+NB σB2+NAdA2+NBdB2NA+NB    =506.52+608.22+5014.59+6010.1150+60   =2112.5+4034.4+729.5+606.6110   =7483110=68.02   =8.25
Thus, mean and standard deviation of both the groups taken together are 116.82 and 8.25 respectively.

Page No 286:

Question 45:

The following data gives arithmetic means, standard deviations of three sub-groups . Calculate arithmetic mean  and standard deviation of the whole group.

Sub-group No. of Men Average Wage
(in â‚¹)
Standard deviation 
(in â‚¹)
A 50 61.0 8.0
B 100 70.0 9.0
C 120 80.5 10.0
 

Answer:

Given,XA=61, σA=8, NA=50XB=70, σB=9, NB=100XC=80.5, σC=10, NC=120Combined MeanXA,B,C=NAXA+NBXB+NCXCNA+NB+NC=50×61+100×70+120×80.550+100+120       =3050+7000+9660270=19710270       =73
Combined Standard Deviation
dA2=XA-XA,B,C=61-732=-122=144dB2=XB-XA,B,C=70-732=-32=9dC2=XC-XA,B,C=80.5-73=7.52=56.25σA,B,C=NA σA2+NB σB2+NCσC2+NAdA2+NBdB2+NCdC2NA+NB+NC       =5082+10092+120102+50144+1009+12056.2550+100+120      =3200+8100+12000+7200+900+6750270      =38150270=141.296      =11.9

Thus, arithmetic mean and standard deviation of the whole group are 73 and 11.9 respectively.



Page No 287:

Question 46:

A sample of 35 values has mean 80 and Standard Deviation 4 . A second sample of 65 values from same population has mean 70 and standard deviation 3. Find the mean and standard deviation of combined sample of 100 values.

Answer:

Given,
N1=35, X1=80, σ1=4N2=65, X2=70, σ2=3Combined MeanX1,2=X1N1+X2N2N1+N2=8035+706535+65    =2800+4550100=7350100     =73.5
Combined Standard Deviation
d12=X1-X1,22=80-73.52=6.52=42.25d22=X2-X1,22=70-73.52=-3.52=12.25σ1,2=σ12N1+σ22N2+N1d12+N2d22N1+N2     =4235+3265+3542.25+6512.2535+65    =560+585+1478.75+796.25100    =3420100=34.20   =5.84
Thus, mean and standard deviation of combined sample of 100 values are 73.5 and 5.84 respectively.

Page No 287:

Question 47:

For a group of 50 male workers , the mean and standard deviation of their weekly wages are â‚¹ 63 and â‚¹ 9 respectively. For a group of 40 female workers , these are â‚¹ 54 and â‚¹ 6 respectively. Find mean and standard deviation for a combined group of 90 workers.

Answer:

Given,
N1=50, X1=63, σ1=9N2=40, X2=54, σ2=6Combined MeanX1,2=X1N1+X2N2N1+N2    =6350+544050+40    =3150+216090    =531090    =59

Combined Standard Deviation
d12=X1-X1,22=63-592=42=16d22=X2-X1,22=54-592=-52=25σ1,2=N1σ12+N2σ22+N1d12+N2d22N1+N2     =5081+4036+5016+402550+40    =4050+1440+800+100090   =729090   =81   =9
Thus, mean and standard deviation for a combined group of 90 workers are 59 and 9 respectively.

Page No 287:

Question 48:

Coefficient of variation of two series are 58% and 69% and their standard deviation are 21.2 and 15.6 . What are their means?

Answer:

Let the first series be A.
Now,
Coefficient of variation of A = 58%
σA = 21.2
Coefficient of variation of A=SDAXA58%=21.2XA58100=21.2XA XA=21.2×10058 =36.55 

Let the second series B.
Now,
Coefficient of variation of B= 69%
σB = 21.2
Coefficient of variation of B=SDBXB69%=15.6XB69100=15.6XB XB=15.6×10069=22.608
Thus, means of series A and series B are 36.55 and 22.608 respectively.

Page No 287:

Question 49:

If the coefficient of variation of X series is 14.6%  and that of Y series is 36.9% and their means are 101.2 and 101.25 respectively. Find their standard deviation.

Answer:

For first series,
Coefficient of variation of A = 14.6%
XA=101.2Coefficient of variation of A=SDAXA14.6%=SDAXA SDA=14.6×101.2100=14.775

For second series,
Coefficient of variation of B= 36.9%
XB=101.25Coefficient of variation of B=SDBXB36.9%=SDB101.25 SDB=36.9×101.25100=37.36
Thus, standard deviations of first and second series are 14.775 and 37.36 respectively.

Page No 287:

Question 50:

Draw a Lorenz curve from the following:

Income (₹ '000) : 20 40 60 100 160
No of Persons ('000)            
Class A : 10 20 40 50 80
Class B : 16 14 10 6 4
 

Answer:

Income Cumulative
Income
Cumulative
Percentage


Person of
Class A

Cumulative

Cumulative
Percentage
Person of
Class B
Cumulative
Persons

 
Cumulative
Percentage
20
40
60
100
160
20
60
120
220
380
5.26
15.78
31.57
57.89
100
10
20
40
50
80
10
30
70
120
200
5
15
35
60
100
16
14
10
6
4
16
30
40
46
50
32
60
80
92
100

Page No 287:

Question 51:

Following is the frequency distribution of marks obtained by students in Economics and Statistics . Analyse the data by drawing a Lorenz Curve.

Marks (mid-values) : 5 15 25 35 45 55 65 75 85 95
No.of students (Economics) : 10 12 13 14 22 27 20 12 11 9
No. of students (Statistics) : 1 2 26 50 59 40 10 8 3 1
 

Answer:

Marks Cumulative Cumulative
Percentage
Students in
Economics
Cumulative students Cumulative
Percentage
Students in Statistics Cumulative
Students
Cumulative Percentage
5
15
25
35
45
55
65
75
85
95
5
20
45
80
125
180
245
320
405
500
1
4
9
16
25
36
49
64
81
100
10
12
13
14
22
27
20
12
11
9
10
22
35
49
71
98
118
130
141
150
6.67
14.67
23.33
32.67
47.33
65.33
78.67
86.67
94
100
1
2
26
50
59
40
10
8
3
1
1
3
29
79
138
178
188
196
199
200
0.5
1.5
14.5
39.5
69
89
94
98
99.5
100



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