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Page No 10.83:

Question 1:

Calculate figures and coefficient  of range of the following series, which gives the monthly expenditure (in ₹) of seven students: 22, 35, 32, 45, 42, 48, 39

Answer:

Given:
Highest Value (H) =48
Lowest Value (L) =22

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 48 − 22 = 26

Coefficient of Range = H- LH + L=48-2248+22=2670=0.37                                                                    
Thus, range is 26 and coefficient of range is 0.37

Page No 10.83:

Question 2:

Find range and coefficient of range from the weekly wage (in ₹) of  10 workers of a factory: 310, 350, 420, 105, 115, 290, 245, 450, 300, 375.

Answer:


Given:
Highest Value (H) = 450
Lowest Value (L) = 105

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 450 − 105 = 345

Coefficient of Range = H-LH + L=450-105450+105=345555=0.62                                                                          

Thus, range is 345 and coefficient of range is 0.62

Page No 10.83:

Question 3:

Find the range and coefficient of range of the following:

Size of shoes 6 7 8 9 10 11 12 13
No. of shoes sold 8 12 14 20 15 8 7 10

Answer:

Size of Shoes No. of Shoes Sold
6 8
7 12
8 14
9 20
10 15
11 8
12 7
13 10

Highest value (H) = 13
Lowest value (L) = 6
Range = Highest value − Lowest value
i.e. R = HL
or, R = 13 − 6=7
Coefficient of Range=H-LH+L=13-613+6=719=0.368=0.37 approx

Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.

Page No 10.83:

Question 4:

From the following data calculate range and coefficient of range:

Marks 10 20 30 40 50 60 70
No. of Studends 8 12 7 30 10 5 2

Answer:

Marks No. of Students
10 8
20 12
30 7
40 30
50 10
60 5
70 2

Highest value (H) = 70
Lowest value (L) = 10

Range = Highest value − Lowest value

i.e. R = H−L

or, R = 70 − 10= 60 marks

Coefficient of Range=H-LH+L=70-1070+10=6080=0.75

Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.


 



Page No 10.84:

Question 5:

Find out range and coefficient range.

Marks 10−20 20−30 30−40 40−50 50−60
No. of students 12 18 14 63 19

Answer:

Marks No. of Students
10 − 20 12
20 − 30 18
30 − 40 14
40 − 50 63
50 − 60 19

Range = Upper Limit of the Highest Class Interval - Lower Limit of the Lowest Class Intervalor, Range = 60 - 10 = 50

Coefficient of Range = H-LH+L=60-1060+10=5070=0.715                                                                          

Hence, the range is 50 marks and coefficient of range is 0.715

Page No 10.84:

Question 6:

Following are the marks obtained by 25 students of class XI in an exam. Find out range and coefficient of range of the marks.

Marks 5−9 10−14 15−19 20−24 25−29 30−34
No. Students 4 6 3 2 6 4

Answer:

In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.
 

Class Interval Exclusive Class Interval Frequency
5 − 9
10 −14
15 −19
20 −24
25 −29
30− 34
4.5 − 9.5
9.5 − 14.5
14.5 − 19.5
19.5 − 24.5
24.5 − 29.5
29.5 − 34.5

4
6
3
2
6
4
 

Range = Upper Limit of the Highest Class Interval - Lower Limit of the Lowest Class Intervalor, Range = 34.5 - 4.5= 30

Coefficient of Range = H- LH + L=34.5-4.534.5+4.5=3039=0.769                                                                      

Page No 10.84:

Question 7:

Calculate interquartile range, quartile deviation and coefficient of quartile deviation from the following data:

Family A B C D E F G
Income (in ₹) 1,200 1,400 1,500 1,700 2,000 2,100 2,200

Answer:

Family S. No. Income
Rs
A 1 1200
B 2 1400
C 3 1500
D 4 1700
E 5 2000
F 6 2100
G 7 2200
  N = 7  

Q1=Size of  N+14th item=Size of  7+14th item=Size of 2nd item=1400Q3=Size of 3 N+14th item=Size of 3 7+14th item=Size of 6th item=2100Inter Quartile Range=Q3-Q1=2100-1400=700Quartile deviation=Q3-Q12=7002=350

Coefficient of Q.D.=Q3 - Q1Q3 + Q1=2100-14002100+1400=7003500=0.2

Page No 10.84:

Question 8:

find out the value of quartile deviation and its coefficient from the following data:

Roll No. 1 2 3 4 5 6 7
Marks 20 28 40 12 30 15 50

Answer:

Roll No. Marks in Ascending Order
1 12
2 15
3 20
4 28
5 30
6 40
7 50
N = 7  

Q1=Size of  N+14th item=Size of  7+14th item=Size of 2nd item=15Q3=Size of 3 N+14th item=Size of 3 7+14th item=size of 6th item=40Quartile deviation=Q3-Q12=40-152=252=12.5

Coefficient of Q.D.=Q3 - Q1Q3 + Q1=40-1540+15=2555=0.45

Page No 10.84:

Question 9:

 Form the following figures, find the quartile deviation and its coefficient:

Height (cm) 150 151 152 153 154 155 156 157 158
No. of students 15 20 32 35 33 22 20 12 10

Answer:

Height
(c.m)
No. of Students Cumulative Frequency
150 15 15
151 20 15 + 20 = 35
152 32 35 + 32 = 67
153 35 67 + 35 = 102
154 33 102 + 33 = 135
155 22 135 + 22 = 157
156 20 157 + 20 = 177
157 12 177 + 12 = 189
158 10 189 + 10 = 199
  N = 199  

Q1=Size of  N+14th item=Size of  199+14th item=Size of 50th itemNow, we need to locate this item in the column of Cumulative Frequency.The item just exceeding 50th item is 67( in the c.f. column), which is corresponding to 152Hence, Q1=152Similarly,Q3=Size of 3 N+14th item=Size off 3 199+14thitem=Size of 150th item=155Quartile deviation=Q3-Q12=155-1522=1.5

Coefficient of Q.D.=Q3 - Q1Q3 + Q1=155-152155+152=3307=0.0098

Page No 10.84:

Question 10:

Compute coefficient of Quartile deviation form the following data:

Marks 10 20 30 40 50 60
No. of students 4 7 15 8 7 2

Answer:

Marks No. of Students Cumulative Frequency
10 4 4
20 7 7 + 4 = 11
30 15 11 + 15 = 26
40 8 26 + 8 = 34
50 7 34 + 7 = 41
60 2 41 + 2 = 43
  N = 43  

Q1=Size of  N+14th item=Size of  43+14th item=Size of 11th itemNow, we need to locate this item in the column of Cumulative Frequency.This item is corresponding to value 20Hence, Q1=20Similarly,Q3=Size of 3 N+14th item=Size of 3 43+14thitem=Size of 33rd item =40

Coefficient of Q.D.=Q3 - Q1Q3 + Q1=40-2040+20=2060=0.333

Page No 10.84:

Question 11:

Calculate Quartile Deviation and its Coefficient from the following data:

Size 5−10 10−15 15−20 20−25 25−30 30−35 35−40 40−45 45−50
Frequency 6 10 18 30 15 12 10 6 4

Answer:

Size Frequency Cumulative Frequency
5−10 6 6
10−15 10 16
15−20 18 34
20−25 30 64
25−30 15 79
30−35 12 91
35−40 10 101
40−45 6 107
45−50 4 111
  Σf=N = 111  


Q1 = Size of N4th item
     = Size of 1114th item
     = Size of 27.75th item


27.75 item lies in group 15−20 and falls within 34thc.f. of the series

Q1=l1+N4-c. f.f×iHere,l=Lower limit of the class intervalN= Sum total of the frequenciesc.f.=Cumulative frequency of the class preceding the first quartile classf= Frequency of the quartile classi= Class intervalThus,Q1=15 + 27.75-1618×5or, Q1=15+11.75×518or, Q1=15+58.7518or, Q1=15+3.26Q1=18.26


Like wise,
Q3 = Size of 3N4th item
      = Size of 31114th item
      = Size of 83.25th  item

83.25th  item lies in the group 30-35 within the 33rd c.f. of the series.


Thus,Q3=l1+3 N4-c. ff×ior, Q3 =30 + 83.25-7912×5or, Q3=30+4.25×512or, Q3=30+21.2512or, Q3=30+1.77Q3=31.77

Quartile Deviation=Q3 - Q12=31.77-18.262=6.75
Coefficient of Q.D.=Q3 - Q1Q3+Q1=31.77-18.2631.77+18.26=13.5150.03=0.27



Page No 10.85:

Question 12:

Calculate coefficient of quartile deviation from the following data:

X (Less than) 200 300 400 500 600
Frequency 8 20 40 46 50

Answer:

Converting less than cumulative frequency into normal frequency distribution:
 

X c.f. Frequency
100−200 8 8
200−300 20 20−8=12
300−400 40 40−20=20
400−500 46 46−40=6
500−600 50 50−46=4
    N=50

Q1 = Size of N4th item
     = Size of 504th item
     = Size of 12.5th item

12.5th item lies in group  200−300 and falls within 20thc.f. of the series.

Q1=l1+ N4-c. f.f×i

Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval

Thus,Q1=200 + 12.5-812×100or, Q1=200+45012or, Q1=200+37.5Q1=237.5

Likewise,
Q3 = Size of 3N4th item
     = Size of 3504th item
     = Size of 37.5th item

37.5th item lies in group 300−400 and fall within the 40th c.f. of the series

Q3=l1+3 N4-c. f.f×ior, Q3 =300 + 37.5-2020×100or, Q3 =300+175020Q3 =300+87.5= 387.5

Coefficient of Q.D.=Q3 - Q1Q3+Q1=387.5-237.5387.5+237.5=150625=0.24

Page No 10.85:

Question 13:

Estimate an appropriate measure of the following data:

Wages (₹) Less than 25 25−30 30−35 35−40 Above 40
No. of workers 2 10 26 16 7

Answer:

Wages No. of worker
(Frequency)

 
Cumulative
Frequency
Less than 25 2 2
25−30 10 12
30−35 26 38
35−40 16 54
Above 40 7 61
  N=61  


Q1 = Size of N4th item
     = Size of 614th item
     = Size of 15.25th item


15.25th item lies in group 30−35 and falls within 38thc.f. of the series.


Q1=l1+N4-c. ff×i

Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval

Thus,Q1=30 + 15.25-1226×5or, Q1=30+16.2526or, Q1= 30+0.625Q1 = 30.625

Likewise,
Q3 = Size of 3N4th item
     = Size of 3614th item
     = Size of 45.75th item

47.75th item lies in group 35−40 and fall within the 54thc.f. of the series

Q3=l1+3 N4-c. f.f×ior, Q3=35 + 45.75-3816×5or, Q3=35+7.75×516Q3=35+2.42=37.42

Quartile Deviation=Q3 - Q12=37.42-30.6252=6.82= Rs 3.40

Coefficient of Q.D.=Q3 - Q1Q3+Q1=37.42-30.62537.42+30.625=6.868.045=.099=0.1 approx

Page No 10.85:

Question 14:

Calculate the mean deviaiton from median and its coefficient from the data: 100, 150, 80, 90, 160, 200, 140

Answer:

 

 Sr. No   Values
    (X)
Deviation from Median
                dM=X-M
             M= 140
1
2
3
4
5
6
7
80
90
100
140
150
160
200
60
50
40
0
10
20
60
     N=7                 Σdm=240


(M) Median = Size of N+12th item
= Size of 7+12th item
= Size of 4th item
= 140

Mean deviation from Median MDM=ΣdMN=2407=34.28                                  Coefficient of MDM=MDMMedian=34.28140=0.245                      

Page No 10.85:

Question 15:

Compute Mean deviation and its coefficient by mean from the data given below:

X 210 220 225 225 225 240 250 270 280  

Answer:

Sr. No Values
(X)
Deviation from Mean
dX¯=X-X¯
1 210 28
2 220 18
3 225 13
4 225 13
5 225 13
6 235 3
7 240 2
8 250 12
9 270 32
10 280 42
N=10 ΣX = 2380 ΣdX¯ = 176

Mean,  X¯=ΣXN=238010=238Thus, calculate deviation of the values from 238Mean deviation from Mean MDX¯=ΣdX¯N=17610=17.6                               Coefficient of Mean deviation=MDX¯X¯=17.6238=0.074                       

Page No 10.85:

Question 16:

Following are the marks of the students. Find mean deviation and coefficient mean deviation from mean.

Marks 5 10 15 20 25 30 35 40
No. of students 16 32 36 44 28 18 12 14

Answer:

Marks
(X)
Frequency
(f)
fX Deviation from Mean
dX¯=X-X¯X-20.2
f×dX=fdX
5 16 80 15.2 243.2
10 32 320 10.2 326.4
15 36 540 5.2 187.2
20 44 880 0.2 8.8
25 28 700 4.8 134.4
30 18 540 9.8 176.4
35 12 420 14.8 177.6
40 14 560 19.8 277.2
ΣX = 180 Σf = 200 ΣfX=4040   ΣfdX=1531.2

Mean X¯=ΣfXΣf=4040200=20.2Thus, calculate the deviation of the values from 20.2Mean deviation from Mean MDX¯=ΣfdX¯Σf=1531.2200=7.65Coefficient of Mean deviation=MDX¯X¯=7.6520.2=0.379=0.38 approx.                             

Page No 10.85:

Question 17:

Find out the mean deviation from the median and its coefficient.

Marks 10 11 12 13 14
No. of students 3 12 18 12 3

Answer:

Marks
(X)
Frequency
(f)
Cumulative frequency
(c.f.)
Deviation from Median
dM=X-MMedian=12
f×dM=fdM
10 3 3 2 6
11 12 15 1 12
12 18 33 0 0
13 12 45 1 12
14 3 48 2 6
  N=Σf = 48     Σf|dM| =36

Median=Size of  N+12th item=Size of 48+12thitem= Size of 492thitem=Size 24.5th item

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 24.5 th item is 33 (in the c.f. column), which is corresponding to 12.

Hence, median is 12.
Thus we calculate  the deviation of the values from 12. 

Mean deviation from Median M.DM=ΣfdMΣf=3648=0.75 marks                           Coefficien of Mean deviation=M.DMMedian=0.7512=0.0625

Page No 10.85:

Question 18:

Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students:

Scores 140−150 150−160 160−170 170−180 180−190 190−200
Frequency 4 6 10 18 9 3

Answer:

Mean Deviation from Median
Scores
(X)
Frequency
(f)
Cumulative Frequency
(c.f.)
Mid -Values
(m)
Deviation from Median
dM=m-MdM=m-172.78
f×dM=fdM
140−150 4 4 145 27.78 111.12
150−160 6 10 155 17.78 106.68
160−170 10 20 165 7.78 77.8
170−180 18 38 175 2.22 39.96
180−190 9 47 185 12.22 109.98
190−200 3 50 195 22.22 66.66
  N=Σf = 50       Σf|dM|=512.2

Median class is given by the size of N2thitem, i.e. 502th   item, which is 25th item.

This corresponds to the class interval of 170-180, so this is the median class.

Median=l1+N2-c.f.f×ior, Median=170+25-2018×10or, Median=170+5018Median=172.78Mean deviation from Median M.DM=ΣfdMΣf=512.250=10.244 marks
 
Mean Deviation from Mean
Scores
(X)
Frequency
(f)
Mid -Values
(m)
fm Deviation from Mean
dX=m-XdX=m-171.2
 
f×dX=fdX
140−150 4 145 580 26.2 104.8
150−160 6 155 930 16.2 97.2
160−170 10 165 1650 6.2 62
170−180 18 175 3150 3.8 68.4
180−190 9 185 1665 13.8 124.2
190−200 3 195 585 23.8 71.4
  Σf = 50   Σfm=8560   ΣfdX=528

Mean (X¯)=ΣfmΣf=856050=171.2Mean deviation M.DX¯=ΣfdXΣf=52850=10.56 marks



Page No 10.86:

Question 19:

Find the mean deviation from mean and its coefficient for the given data:

X 0−10 10−20 20−30 30−40 40−50 50−60
F 3 5 7 2 9 4

Answer:

Values
(X)
Frequency
(f)
Mid-Values
(m)
fm dX¯=m-X¯ f×dX¯=fdX¯
0−10 3 5 15 27 81
10−20 5 15 75 17 85
20−30 7 25 175 7 49
30−40 2 35 70 3 6
40−50 9 45 405 13 117
50−60 4 55 220 23 92
  Σf = 30   Σfm=960   ΣfdX¯=430

Mean, X¯=ΣfmΣf=96030=32Mean deviation from Mean,  M.DX¯=ΣfdX¯Σf=43030=14.33Coefficient of M.D.=M.DX¯X¯=14.3332=0.44

Page No 10.86:

Question 20:

Calculate mean deviation and its coefficient from the following figures:

Class-Interval Frequency
Less than 10 5
Less than 20 12
Less than 30 20
Less than 40 35
Less than 50 54
Less than 60 60

Answer:

Converting less than cumulative frequency distribution into simple frequency distribution:
 

Class Interval
Cumulative
Frequency
(c.f.)


 
Frequency
(f)
Mid-Values
(m)
dM=m-M=m-36.67 f×dM=fdM
0−10 5 5 5 31.67 158.35
10−20 12 7 15 21.67 151.69
20−30 20 8 25 11.67 93.36
30−40 35 15 35 1.67 25.65
40−50 54 19 45 8.33 158.27
50−60 60 6 55 18.33 109.98
    Σf=N=60     Σf|dM|=696.7


Median class is given by the size of  N2th item, i.e. 602th item, which is the 30th item.

This corresponds to the class interval of 30-40, so this is the median class.

Median (M)=l1+N2-c.f.f×ior, M=30+30-2015×10or, M=30+10015M=36.67

Mean deviation from Median=ΣfdMΣf=696.760=11.61Coefficient of M.D.=M.DMMedian=11.6136.67=0.32

Page No 10.86:

Question 21:

 Calculate the Standard deviation from the following values:8 , 9, 15, 23, 5, 11, 19, 8, 10, 12.

Answer:

Values
(X)
x=X-X¯ x2
8 −4 16
9 −3 9
15 3 9
23 11 121
5 −7 49
11 −1 1
19 7 49
8 −4 16
10 −2 4
12 0 0
ΣX=120   Σx2=274

Mean, X¯=ΣXN=12010=12Standard deviation σ=Σx2Nor,  σ =27410or,  σ=27.4  σ=5.23

Hence, the standard deviation of the above series is 5.23

Page No 10.86:

Question 22:

Find the standard deviation for the following data: 3, 5, 6, 7, 10, 12, 15, 18.

Answer:

Values
(X)
x=X-X¯ x2
3 −6.5 42.25
5 −4.5 20.25
6 −3.5 12.25
7 −2.5 6.25
10 .5 .25
12 2.5 6.25
15 5.5 30.25
18 8.5 72.25
ΣX=76   Σx2=190

Mean, X¯=ΣXN=768=9.5Standard deviation σ=Σx2Nor, σ=1908or, σ=23.75 σ=4.87

Hence, standard deviation of the above series is 4.87

Page No 10.86:

Question 23:

Find out the standard deviation of the height of 10 men given below: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.

Answer:

Values
(X)
x=X-X¯ x2
160 −3 9
160 −3 9
161 −2 4
162 −1 1
163 0 0
163 0 0
163 0 0
164 1 1
164 1 1
170 7 49
ΣX=1630   Σx2=74

Mean, X¯=ΣXN=163010=163Standard deviation σ=Σx2Nor, σ=7410or, σ=7.4σ=2.72

Hence, standard deviation of the above series is 2.72

Page No 10.86:

Question 24:

Calculate standard deviation of the given below:

Size 3 4 5 6 7 8 9
Frequency 3 7 22 60 85 32 8

Answer:

Size
(X)
Frequency
(f)
fX x=X-X¯ x2 fx2
3 3 9 3.59 12.89 38.67
4 7 28 2.59   6.71 46.97
5 22 110 1.59   2.53 55.66
6 60 360  0.59  0.35   21
7 85 595  0.41   0 .17 14.45
8 32 256 1.41 1.99 63.68
9 8 72 2.41 5.81 46.48
  N=Σf=217 ΣfX=1430     Σfx2=286.91

Mean,  X¯=ΣfXΣf=1430217=6.59Standard deviation, (σ)=Σfx2Nor,  (σ) =286.91217or,  (σ) =1.32 (σ)  =1.149

Hence, standard deviation of the above series is 1.149

Page No 10.86:

Question 25:

Find the value of standard deviation and coefficient of variation from the following:

Variables 10 20 30 40 50 60 70
Frequency 6 8 16 15 32 11 12

Answer:

Size
(X)
Frequency
(f)
fX x=X-X¯ x2 fx2
10 6 60 −34 1156 6936
20 8 160 −24 576 4608
30 16 480 −14 196 3136
40 15 600 −4 16 240
50 32 1600 +6 36 1152
60 11 660 16 256 2816
70 12 840 26 676 8112
  Σf=100 ΣfX=4400     Σfx2=27000

Mean, X¯=ΣfXΣf=4400100=44Standard deviation, σ=Σfx2Σfor, σ=27000100or, σ=270σ=16.43Coefficient of S.D.=σX¯×100=16.4344×100=37.34

Page No 10.86:

Question 26:

Measurements are made to the nearest cm. of the heights of 10 children. Calculate mean and standard deviation.

Heights (cms) 60 61 62 63 64 65 66 67 68
No. of children 2 0 15 29 25 12 10 4 3

Answer:

Height
(X)
Frequency
(f)
fX x=X-X¯ x2 fx2
60 2 120 −3.89 15.13 30.26
61 0 0 −2.89 8.35 0
62 15 930 −1.89 3.57 53.55
63 29 1827 −0.89 0.79 22.91
64 25 1600 0.11 0.01 0.25
65 12 780 1.11 1.23 14.76
66 10 660 2.11 4.45 44.5
67 4 268 3.11 9.67 38.68
68 3 204 4.11 16.89 50.67
  Σf=100 ΣfX=6389     Σfx2=255.58

Mean, X¯=ΣfXΣf=6389100=63.89Standard deviation, σ=Σfx2Σf=255.58100=1.598=1.6 cm approx
Hence mean of the above series is 63.89 cms and standard deviation is 1.6 cms.



Page No 10.87:

Question 27:

Calculate standard deviation for the given data:

Age (in yrs) 20−25 25−30 30−35 35−40 40−45 45−50 50−55
No. of workers 17 11 8 5 4 3 2

Answer:

Age
(X)
No. of Workers
(f)
Mid-Values
(m)
fm m2 f×m2=fm2
20−25 17 22.5 382.5 506.25 8606.25
25−30 11 27.5 302.5 756.25 8318.75
30−35 8 32.5 260 1056.25 8450
35−40 5 37.5 187.5 1406.25 7031.25
40−45 4 42.5 170 1806.25 7225
45−50 3 47.5 142.5 2256.25 6768.75
50−55 2 52.5 105 2756.25 5512.5
  Σf=50   Σfm=1550   Σfm2=51912.5

Mean,  X¯=ΣfmΣf=155050=31Standard deviation, σ=Σfm2Σf-X¯2or, σ=51912.550-312or, σ=1038.25-961or, σ=77.25 σ=8.79 years

Hence, standard deviation of the above series is 8.79 years.

Page No 10.87:

Question 28:

Calculate Standard deviation from the following series:

Class 0−10 10−20 20−30 30−40 40−50 50−60 60−70
Frequency 2 4 6 8 6 4 2

Answer:

Class Interval
(X)

Frequency
(f)

 
Mid-Values
(m)
fm m2 f×m2=fm2
0−10 2 5 10 25 50
10−20 4 15 60 225 900
20−30 6 25 150 625 3750
30−40 8 35 280 1225 9800
40−50 6 45 270 2025 12150
50−60 4 55 220 3025 12100
60−70 2 65 130 4225 8450
  Σf=32   Σfm=1120   Σfm2=47200

Mean,  X=ΣfmΣf=112032=35Standard deviation, σ=Σfm2Σf-X¯2or, σ=4720032-352or, σ=1475-1225or, σ=250σ=15.81

Hence, standard deviation of the above series is 15.81

Page No 10.87:

Question 29:

From the following figures, find the standard deviation and the coefficient of variation:

Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
No. of persons 5 10 20 40 30 20 10 4

Answer:


Marks
(X)

 
No. of Persons
(f)
Mid-Values
(m)
fm m2 fm2
0−10 5 5 25 25 125
10−20 10 15 150 225 2250
20−30 20 25 500 625 12500
30−40 40 35 1400 1225 49000
40−50 30 45 1350 2025 60750
50−60 20 55 1100 3025 60500
60−70 10 65 650 4225 42250
70−80 4 75 300 5625 22500
  Σf=139   Σfm=5475   Σfm2=249875

Mean, X¯=ΣfmΣf=5475139=39.39Standard deviation, (σ)=Σfm2Σf-X¯2or, (σ)=249875139-39.392or, (σ)=1797.66-1551.57or, (σ)=246.09(σ)=15.69 marksCofficient of Variation=(σ)X¯×100=15.6939.39×100=39.83%

Page No 10.87:

Question 31:

The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3 and the standard deviations are 8 and 7. Find the mean and the standard deviation of the sample of size 150 obtained by combining the two samples.

Answer:

Calculating combined standard deviation,

1, 2  σ=N1σ12+N2 σ22+N1d12+N2d22N1+N2where,N1=50           N2=100σ1=8             σ2=7X¯1=54.1        X¯2=50.3

To calculate combined standard deviation, we need to find the combined mean of the observations.

Thus, Combined mean is

X¯12=N1 X¯1+N2 X¯2N1+N2=50×54.1+100×50.3150=2705+5030150=7735150=51.57d12=X¯1-X¯122=54.1-51.572=2.532=6.40d22=X2-X¯1,22=50.3-51.572=-1.272=1.61Thus, Combined standard deviation will be:σ1,2=50 82+10072+50 6.4+100 1.61150or, σ1,2=3200+4900+320+161150or, σ1,2=8581150or, σ1,2=57.20σ1,2=7.56

Hence, the combined standard deviation is 7.56

Page No 10.87:

Question 32:

For a group of 100 males, mean and standard deviation of their daily wages are ₹ 36 and ₹ 9 respectively. For a group of 50 females, it is ₹ 45 and ₹ 6. Find the standard deviation for the whole group.

Answer:

Calculating combined standard deviation,

1, 2  σ=N1σ12+N2 σ22+N1d12+N2d22N1+N2where,N1=100           N2=50X¯1=36        X¯2=45σ1=9             σ2=6

To calculate combined standard deviation, we need to find the combined mean of the observations.

Thus, Combined mean is

X¯12=N1 X¯1+N2 X¯2N1+N2=100 36+50 45150=3600+2250150= Rs 39d12=X¯1-X¯122=36-392=-32=9d22=X¯2-X¯122=45-392=62=36Thus, Combined standard deviation will be:1,2σ=N1σ12+N2 σ22+N1d 12+N2d 22N1+N2or, σ1,2=100 92+5062+100 9+50 36150or, σ1,2=8100+1800+900+1800150or, σ1,2=12600150or, σ1,2=84 σ1,2=9.16

Hence, the combined standard deviation is Rs 9.16

Page No 10.87:

Question 33:

The profit of two business concerns for 5 years are as given below. Draw Lorenz Curves to show the distribution.

Year 2001 2002 2003 2004 2005
Firm A 15 30 45 60 50
Firm B 20 30 45 60 45

Answer:

Years Firm A Cumulative Profits of Firm A Cumulative Percentage Profits of Firm A
   (x)
Firm B Cumulative Profits of Firm B Cumulative Percentage Profits of Firm B
    (y)
Coordinates of Lorenz Curve
   (x,y)
2001 15 15 15200×100=7.5 20 20 20200×100=10 (7.5,10)
2002 30 45 45200×100=22.5 30 50 50200×100=25 (22.5, 25)
2003 45 90 90200×100=45 45 95  95200×100=47.5 (45, 47.5)
2004 60 150 150200×100=75 60 155 155200×100=77.5 (75,77.5)
2005 50 200 200200×100=100 45 200 200200×100=100 (100,100)
  ΣN1=200     ΣN2=200      


Page No 10.87:

Question 34:

The given table shows the daily income of workers of two factories. Draw the Lorenz Curves for both the factories.

Daily Income (₹) 0−100 100−200 200−300 300−400 400−500
Factory A 8 7 5 3 2
Factory B 15 6 2 1 1

Answer:

For Factory A
Daily Income Mid Value Cumulative
Mid Value

 
%
Cumulative
Mid Value
No. of worker
(f)
(c.f.) %
Cumulative
Frequency
0−100 50 50 4 8 8 32
100−200 150 200 16 7 15 60
200−300 250 450 36 5 20 80
300−400 350 800 64 3 23 92
400−500 450 1250 100 2 25 100
 
For Factory B
Daily Income Mid Value Cumulative
Mid Value

 
%
Cumulative
Mid Value
No. of worker
(f)
(c.f.) %
Cumulative
Frequency
0−100 50 50 4 15 15 60
100−200 150 200 16 6 21 84
200−300 250 450 36 2 23 92
300−400 350 800 64 1 24 96
400−500 450 1250 100 1 25 100

Page No 10.87:

Question 35:

Find the mean deviation from mean and its coefficient for the given data:

Marks (more than) 0 10 20 30 40 50 60
No. of students 200 180 150 100 40 15 5

Answer:

Marks
(X)
Frequency
(f)
Mid-Values
(m)
fm dX¯=m-X¯=m-29.5 f×dX¯=fdX¯
0−10 20 5 100 24.5 490
10−20 30 15 450 14.5 435
20−30 50 25 1250 4.5 225
30−40 60 35 2100 5.5 330
40−50 25 45 1125 15.5 387.5
50−60 10 55 550 25.5 255
60−70 5 65 325 35.5 177.5
  Σf=200   Σfm=5900   ΣfdX¯=2300

Mean, X¯=ΣfmΣf=5900200=29.5 marksMean deviation M.D.X¯=ΣfdX¯Σf=2300200=11.5 marksCoefficient of M.D.=M.D.X¯X¯=11.529.5=0.39
             
             



Page No 10.88:

Question 36:

Calculate the Mean and Standard Deviation from the following distribution.

Age (years) 15−19 20−24 25−29 30−34 35−39 40−44
No. of Persons 4 20 38 24 10 4

Answer:

In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below:
 

Age
(X)
Frequency
(f)
Mid-Values
(m)
fm m2 fm2
14.5−19.5 4 17 68 289 1156
19.5−24.5 20 22 440 484 9680
24.5−29.5 38 27 1026 729 27702
29.5−34.5 24 32 768 1024 24576
34.5−39.5 10 37 370 1369 13690
39.5−44.5 4 42 168 1764 7056
  Σf=100   Σfm=2840   Σfm2=83860


Mean, X¯=ΣfmΣf=2840100=28.4 yearsStandard deviation, σ=Σfm2Σf-X¯2or, σ=83860100-28.42or, σ=838.6-806.56or, σ=32.04σ=5.66 years

Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.

Page No 10.88:

Question 37:

The following table gives the weights of one hundred persons. Copute the coefficient of dispersion by the method of limits.

Class-interval 40−45 45−50 50−55 55−60 60−65 65−70 70−75 75−80 80−85 85−90
No. of persons 4 13 8 14 9 16 17 9 8 2

Answer:

Given:

Upper limit of the Highest Class Interval (H) = 90
Lower limit of the Lowest Class Interval (L) = 40

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 90 − 40= 50

Coefficeint of Range=H-LH+L=90-4090+40=50130=0.384

Hence, range of the above series is 50 kg and coefficient of range is 0.384

Page No 10.88:

Question 38:

Calculate standard deviation of the following data:

Age in years (below) 10 20 30 40 50 60 70 80
No. of persons 15 30 53 75 100 110 115 125

Answer:

Age
(X)
Cumulative Frequency
(c.f.)
Frequency
(f)
Mid-Values
(m)
fm m2 fm2
0−10 15 15-0=15 5 75 25 375
10−20 30 30-15=15 15 225 225 3375
20−30 53 53-15=23 25 575 625 14375
30−40 75 75-23=22 35 770 1225 26950
40−50 100 100-75=25 45 1125 2025 50625
50−60 110 110-100=10 55 550 3025 30250
60−70 115 115-110=5 65 325 4225 21125
70−80 125 125-115=10 75 750 5625 56250
    Σf=125   Σfm=4395   Σfm2=203325

Mean,  X¯=ΣfmΣf=4395125=35.16Standard deviation σ=Σfm2Σf-X¯2or,  σ=203325125-35.162or,  σ=1626.6-1236.23or,  σ=390.37  σ=19.76

Hence, standard deviation of the above series is 19.76 years.

Page No 10.88:

Question 39:

Price of a particular item in 10 years in two cities are given below, which city has more stable prices?

City A 55 54 52 53 56 58 52 50 51 49
City B 108 107 105 105 106 107 104 103 104 101

Answer:

City A City B
XA xA=XA-X¯A xA2 XB xB=XB-X¯B xB2
55 2 4 108 3 9
54 1 1 107 2 4
52 −1 1 105 0 0
53 0 0 105 0 0
56 3 9 106 1 1
58 5 25 107 2 4
52 −1 1 104 −1 1
50 −3 9 103 −2 4
51 −2 4 104 −1 1
49 −4 16 101 −4 16
ΣXA=530   ΣxA2=70 ΣXB=1050   ΣxB2=40

X¯A=ΣXAN=53010=53Aσ=ΣxA2N=7010=7=2.64C.V (City A)=σAXA¯×100=2.6453×100=4.99%

X¯B=ΣXBN=105010=105σB=ΣxB2N=4010=2C.V (City B)=σBXB¯×100=2×100105=200105=1.90%

Since C.V. of city B is less. Therefore, prices are more stable in city B.

Page No 10.88:

Question 40:

For a distribution, the coefficient of variation is 22.5% and mean is 7.5. Calculate standard deviation.

Answer:

Given,

Coefficient of variation= 22.5%
Mean, X= 7.5

Coefficient of Variation=σX¯×10022.5=σ7.5×10022.5×7.5100=σσ=168.75100σ=1.687=1.69 approx

Hence, standard deviation is 1.69

Page No 10.88:

Question 41:

Find out the arithmetic mean and standard deviation from the following data:

Variable 5−10 10−15 15−20 20−25 25−30 30−35
Frequency 2 9 29 54 11 5

Answer:

Variable
(X)
Frequency
(f)
Mid-Values
(m)
fm m2 fm2
5−10 2 7.5 15 56.25 112.5
10−15 9 12.5 112.5 156.25 1406.25
15−20 29 17.5 507.5 306.25 8881.25
20−25 54 22.5 1215.0 506.25 27337.5
25−30 11 27.5 302.5 756.25 8318.75
30−35 5 32.5 162.5 1056.25 5281.25
  Σf=110   Σfm=2315   Σfm2=51337.5

Mean,  X¯=ΣfmΣf=2315110=21.05Standard Deviation, σ=Σfm2Σf-X¯2or,  σ=51337.5110-21.052or,  σ=466.70-443.10or,  σ=23.6 σ= 4.86

Page No 10.88:

Question 42:

The mean and standard deviation of a series of 20 items are 20 and 5 respectively. While calculating these measures, an item of 13 was wrongly read as 30. Find out the correct mean and standard deviation.

Answer:

Calculating correct Mean, by using the following observations:

X¯wrong=ΣXwrongNΣXwrong=X¯wrong×NΣXwrong=20×20ΣXwrong=400 ΣXcorrect=ΣXwrong -incorrect item+correct item ΣXcorrect=400-30+13 ΣXcorrect=383X¯correct=ΣXcorrectN=38320=19.15Calculation of Correct S.D.σ=Σx2N-X¯25=Σx220-202

Squaring both sides

25=Σx220-400

425 × 20 = Σx2

 Σx2wrong= 8500

Σx2correct = Σx2wrong − (Incorrect item)2 + (Correct item)2

Σx2correct= 8500 − (30)2 + (13)2
Σx2correct= 8500 − 900 + 169 = 7769

 σCorrect=Σx2Correct N- XCorrect2or,  σCorrect=776920-19.152or,  σCorrect=388.45-366.72or,  σCorrect=21.73 σCorrect=4.66

Hence, correct mean and standard deviation are 19.15 and 4.66 respectively.

Page No 10.88:

Question 43:

Following are the marks obtained by two students: Mollie and Isha, in 10 sets of examinations:

Marks of obtained by Mollie 44 80 76 48 52 72 68 56 60 54
Marks of obtained by Isha 48 75 54 60 63 69 72 51 57 66
Out of Mollie and Isha, who is more consistent?

Answer:

 
Mollie Isha
XM xM=XM-X¯M xM2 XI xI=XI-X¯I xI2
44 −17 289 48 −13.5 182.25
80 19 361 75 13.5 182.25
76 15 225 54 −7.5 56.25
48 −13 169 60 −1.5 2.25
52 −9 81 63 1.5 2.25
72 11 121 69 7.5 56.25
68 7 49 72 10.5 110.25
56 −5 25 51 −10.5 110.25
60 −1 1 57 −4.5 20.25
54 −7 49 66 4.5 20.25
ΣXM=610   ΣxM2=1370 ΣXI=615   ΣxI2=742.5

X¯M=ΣXMN=61010=61 marksσM=ΣxM2N=137010=137=11.70C.V of Mollie's marks=σMXM¯×100=11.7061×100=19.18%


X¯I=ΣXIN=61510=61.5σI=ΣxI2N=742.510=74.25=8.62C.V of Isha's marks=σX¯×100=8.6261.5×100=14.01%

Isha is more consistent in securing makes as her C.V. (14.01)% is less than that of Mollie's C.V. (19.18)%



Page No 10.89:

Question 44:

Calculate coefficient of variation from the following data:

Marks (more than) 0 10 20 30 40 50 60 70
No. of students 100 90 75 50 20 10 5 0

Answer:


Converting more than cumulative frequency distribution into simple frequency distribution:
 

Marks
(X)
No. of Students
(f)
Mid-Values
(m)
fm m2 fm2
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 −70
70 −80
100-90=10
90-75=15
75-50=25
50-20=30
20-10=10
10-5=5
5-0=5
0-0=0
5
15
25
35
45
55
65
75
50
225
625
1050
450
275
325
0
25
225
625
1225
2025
3025
4225
5625
250
3375
15625
36750
20250
15125
21125
0
  Σf =100   Σfm = 3000   Σfm2 =112500


Mean, X=ΣfmΣf=3000100=30Standard deviation, σ=Σfm2Σf-X2                          =112500100-302                          =1125-900                          =225= 15Coefficient of variation=σX×100                                 =1530×100                                 =50%

Page No 10.89:

Question 45:

In two Towns A and B, daily pocket money and the standard deviation are given below:

Town Average Daily Pocket Standard Deviation No. of Teenagers
A 34.5 5.0 476
B 28.5 4.5 524

(i) Which town, A or B, pays out the larger amount of daily pocket money?
(ii) What is the average daily pocket money of all teenagers taken together?
(iii) Calculate coefficient of variation of each town. Which town is more variable in terms of pocket money?

Answer:

(i) Town A
Daily pocket money = Average Daily Pocket × No. of teenagers
                                 = 34.5 × 476
                                 = Rs 16422

Town B
Daily Pocket Money = 28.5 × 524
                                 = Rs 14934

Town 'A' pays larger amount of daily pocket money.

(ii) Solution:

In order to find out average daily pocket money of all teenagers, we need find out the combined mean of their pocket money.

Thus,

XA=34.5, NA=476XB=28.5, NB=524

XA,B=XA NA+XB NBNA+NBor, XA,B =34.5×476+28.5×524476+524or, XA,B =16422+149341000or, XA,B =313561000 XA,B = Rs 31.356=Rs 31.36 approx

(iii)
C.V. of A=σAXA×100=534.5×100  =14.49%          C.V. of B=σBXB×100  =4.528.5×100=15.79%                          

Town B is more variable in terms of pocket money as its C.V. is higher than that of C.V. of Town A.

Page No 10.89:

Question 46:

The prices of share of Company X and Company Y are given below. State, which company is more stable?

Company X 25 50 45 30 70 42 36 48 34 60
Company Y 10 70 50 20 95 55 42 60 48 80

Answer:

 

Company X Company Y
X x=X-X x2 Y y=Y-Y y2
25
50
45
30
70
42
36
48
34
60
−19
6
1
−14
26
−2
−8
4
−10
16
361
36
1
196
676
4
64
16
100
256
10
70
50
20
95
55
42
60
48
80
−43
17
−3
−33
42
2
−11
7
−5
27
1849
289
9
1089
1764
4
121
49
25
729
ΣX = 440   Σx2 = 1710 ΣY = 530   Σy2 = 5928

X=Σx2N=44010=44σx=Σx2N=171010=171=13.08C.V. of prices of Shares of X Co.=σxX×100=13.0844×100 =29.72%        


Y=ΣYN=53010=53σy=Σy2N=592810=24.35C.V. of prices of Share of Y co.=σyX×100=24.3553×100=45.94%

As C.V of prices of shares of Co. X is less than that of the prices of shares of Co. Y therefore,  price of share of  Co. X is more stable.

Page No 10.89:

Question 47:

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50 instead of 40. Find the correct mean and standard deviation.

Answer:

Calculating correct Mean, by using the following observations:

X¯wrong=ΣXwrongNΣXwrong=X¯wrong×NΣXwrong=40×100ΣXwrong=4000 ΣXcorrect=ΣXwrong -incorrect item+correct item ΣXcorrect=4000-50+40 ΣXcorrect=3990X¯correct=ΣXcorrectN=3990100=39.9Calculation of Correct S.D.σ=Σx2N-X¯25.1=Σx220-202

Squaring both sides

26.01=Σx2100-1600

1626.01×100=Σx2Σx2wrong=162601 Σx2correct=Σx2wrong-Incorrect value2+Correct value2            Σx2correct  =162601-502+402             Σx2correct=162601-2500+1600 =161701 σCorrect=Σx2correctN- Xcorrect or, σCorrect = 161701100-39.92 or, σCorrect  =1617.01-1592.01  or, σCorrect =25   σCorrect =5


Hence, correct mean and standard deviation are 39.9 and 5 respectively.

Page No 10.89:

Question 48:

From the following data, calculate standard deviation of the two groups A and B. Which group is more consistent?

Class Interval Group A Group B
5−10 2 9
10−15 9 11
15−20 29 18
20−25 54 32
25−30 11 27
30−35 5 13

Answer:

For group A
 

Class Interval Frequency
(f)
Mid-Values
(m)
fm m2 fm2
5 − 10
10 − 15
15 − 20
20 − 25
25 − 30
30 − 35
2
9
29
54
11
5
7.5
12.5
17.5
22.5
27.5
32.5
15
112.5
507.5
1215
302.5
162.5
56.25
156.25
306.25
506.25
756.25
1056.25
112.5
1406.25
8881.25
27337.5
8318.75
5281.25
  Σf = 110   Σfm = 2315   Σfm2 = 51337.5

XA=ΣfmΣf=2315110=21.05σA=Σfm2Σf-X2 =51337.5110-21.052 =466.70-443.10 =23.6 =4.86C.V. of group A=σAXA×100=4.8621.05×100=23.08%      

For group B
 
Class Interval Frequency
(f)
Mid-Values
(m)
fm m2 fm2
5 − 10
10 − 15
15 − 20
20 − 25
25 − 30
30 − 35
9
11
18
32
27
13
7.5
12.5
17.5
22.5
27.5
32.5
67.5
137.5
315
720
742.5
422.5
56.25
156.25
306.25
506.25
756.25
1056.25
506.25
1718.75
5512.5
16200
20418.75
13731.25
  Σf = 110   Σfm = 2405   Σfm2 = 58087.5

XB=ΣfmΣf=2405110=21.86σB=Σfm2Σf-X2 =58087.5110-21.862 =528.06-477.85 =50.21 =7.08C.V. of group B=σBXB×100=7.0821.86×100=32.4%

Group A is more consistent as C.V. of group A is less than C.V. of group B.

Page No 10.89:

Question 49:

From the following data of marks, calculate standard deviation. What will be the value of standard deviation, if marks obtained by each student is increased by one?

Marks Obtained 1 2 3 4 5 6 7 8 9
No. of students 32 41 57 98 123 83 46 17 3

Answer:

Marks
(X)
Frequency
(f
)
fX x=X-X x2 fx2
1
2
3
4
5
6
7
8
9
32
41
57
98
123
83
46
17
3
32
82
171
392
615
498
322
136
27
−3.55
−2.55
−1.55
−0.55
0.45
1.45
2.45
3.45
4.45
12.60
6.50
2.40
0.30
0.20
2.10
6.00
11.90
19.80
403.2
266.5
136.8
29.4
24.6
174.3
276
202.3
59.4
  Σf = 500 Σfx = 2275     Σfx2 = 1572.5

X=ΣfXΣf=2275500=4.55S.D. σ=Σfx2N  =1572.5500=3.145=1.77 marks                        

As, standard deviation is independent of the change in origin, i.e. it will not get affected if the value of the series is increased or decreased by a constant quantity. Therefore it will remain the same.



Page No 10.90:

Question 50:

From the following data of two workers, identify who is a more consistent worker?

  Worker
A B
Average Time in completing a job 40 42
Standard Deviation 8 6

Answer:

C.V. of worker A=σAXA×100                         =840×100                         =20%C.V. of worker B=σBXB×100                          =642×100                          =14.29%

Worker B is more consistent as his C.V. (14.29)% is less than that of C.V of worker A(20%).

Page No 10.90:

Question 51:

Find the standard deviation and coefficient of standard deviation:

X (less than) 10 20 30 40 50 60 70 80
Frequency 12 30 65 107 157 202 222 230

Answer:

Converting less than frequency distribution into simple frequency distribution:
 

X Frequency
(f)

Mid-Values
(m)
 
fm m2 fm2
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
12-0=12
30-12=18
65-30=35
107-65=42
157-107=50
202-157=45
222-202=20
230-222=8
5
15
25
35
45
55
65
75
60
270
875
1470
2250
2475
1300
600
25
225
625
1225
2025
3025
4225
5625
300
4050
21875
51450
101250
136125
84500
45000
  Σf = 230   Σfm = 9300   Σfm2 = 444550


Mean (X)=ΣfmΣf=9300230=40.43Standard Deviation (σ)=Σfm2Σf-X2 =444550230-40.432 =1932.83-1634.58 =298.25=17.26Coefficient of standard deviation=σX        =17.2640.47        =0.43

Page No 10.90:

Question 52:

Find mean, standard deviation and coefficient of variation.

Class-interval 0−4 4−8 8−12 12−16 16−20 20−24
Frequency 10 15 20 25 20 10

Answer:

Class Interval Frequency
(f)
 Mid-Values
(m)
fm m2 fm2
0 − 4
4 − 8
8 − 12
12 − 16
16 − 18
20 − 24
10
15
20
25
20
10
2
6
10
14
18
22
20
90
200
350
360
220
4
36
100
196
324
484
40
540
2000
4900
6480
4840
  Σf = 100   Σfm = 1240   Σfm2 = 18800

Mean ( X)=ΣfmΣf=1240100=12.4S.D.σ=Σfm2Σf-X2          =18800100-12.42         =188-153.76         =34.24=5.85C.V.=σX×100     =5.8512.4×100     =47.17%

Page No 10.90:

Question 53:

The following table shows the marks obtained by 60 students. Calculate mean and standard deviation.

Marks (more than) 70 60 50 40 30 20
No. of students 7 18 40 40 55 60

Answer:


Converting more than cumulative frequency into ordinary continuos series:
 

Marks
X
Frequency
(f)
Mid-Values
(m)
fm m2 fm2
20 −30
30 −40
40 −50
50 −60
60 −70
70 −80
60-55=5
55-40=15
40-40=0
40-18=22
18-7=11
7-0=7
25
35
45
55
65
75
125
525
0
1210
715
525
625
1225
2025
3025
4225
5625
3125
18375
0
66550
46475
39375
  Σf = 60   Σfm = 3100   Σfm2 = 173900


Mean (X)=ΣfmΣf=310060=51.67Standard deviation (σ)=Σfm2Σf-X2 =17390060-51.672 =2898.33-2669.79 =228.54 =15.11

Hence, mean and standard deviation of the above series are 51.67 marks and 15.11 marks respectively.

Page No 10.90:

Question 54:

Calculate standard deviation and coefficient of dispersion from the data below:

Mid-Points 5 15 25 35 45 55 65 75
Frequency 5 8 7 12 28 20 10 10

Answer:

Mid-Values
(m)
Frequency
(f)
fm m2 fm2
5
15
25
35
45
55
65
75
5
8
7
12
28
20
10
10
25
120
175
420
1260
1100
650
750
25
225
625
1225
2025
3025
4225
5625
125
1800
4375
14700
56700
60500
42250
56250
  Σf = 100 Σfm= 4500   Σfm2 = 236700

Mean (X)=ΣfmΣf=4500100=45Standard Deviation (σ)=Σfm2Σf-X2 =236700100-452 =2367-2025 =342=18.49C.V=σX×100=18.4945×100=41.09%    

Page No 10.90:

Question 55:

The following data shows the expected life of two models of T.V.: A and B:

Life (no. of years) 0−2 2−4 4−6 6−8 8−10 10−12
Model A 5 16 13 7 5 4
Model B 2 7 12 19 9 1
Which model has greater uniformity.

Answer:

 

For A
Life
(No. of years)
Frequency
(f)
Mid Values
(m)
fm m2 fm2
0 − 2
2 − 4
4 − 6
6 − 8
8 −10
10 − 12
5
16
13
7
5
4
1
3
5
7
9
11
5
48
65
49
45
44
1
9
25
49
81
121
5
144
325
343
405
484
  Σf = 50   Σfm = 256   Σfm2 = 1706

XA=ΣfmΣf=25650=5.12S.D. σ=Σfm2Σf-XA2           =170650-5.122           =34.12-26.21           =7.91=2.81C.V. of T.V.=σAXA×100                   =2.8125.12×100                   =54.91%
For B

Life
(No. of years)

 
Frequency
(f)
Mid Values
(m)
fm m2 fm2
0 − 2
2 − 4
4 − 6
6 − 8
8 −10
10 − 12
2
7
12
19
9
1
1
3
5
7
9
11
2
21
60
133
81
11
1
9
25
49
81
121
2
63
300
931
729
121
  Σf = 50   Σfm = 308   Σfm2 = 2146

XB=ΣfmΣf=30850=6.16S.D.σB=Σfm2Σf-XB2            =214650-6.162            =42.92-37.95            =4.97            =2.23C.V. of T.VB=σBXB×100                  =2.236.16×100                  =36.21%

As,  C.V of Model B is less than that of C.V of model A therefore model B has greater uniformity.



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