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TR Jain & Vk Ohri (2017) Solutions for Class 11 Science Economics Chapter 11 - Measures Of Dispersion

TR Jain & Vk Ohri (2017) Solutions for Class 11 Science Economics Chapter 11 Measures Of Dispersion are provided here with simple step-by-step explanations. These solutions for Measures Of Dispersion are extremely popular among class 11 Science students for Economics Measures Of Dispersion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the TR Jain & Vk Ohri (2017) Book of class 11 Science Economics Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s TR Jain & Vk Ohri (2017) Solutions. All TR Jain & Vk Ohri (2017) Solutions for class 11 Science Economics are prepared by experts and are 100% accurate.

Page No 248:

Question 1:

5 students obtained following marks in statistics:
20, 35, 25, 30, 15
Find out range and coefficient of range.
 

Answer:

Here,

Highest value (H)= 35
Lowest value (L) = 15

Range= Highest value − Lowest value

i.e. R= H− L

Substituting the given values in the formula

R= 35 − 15= 20

Coefficient of Range is as follows:

 CR =H-LH+Lor, CR =35-1535+15=2050 CR =0.4

Hence, the range (R) of the above data is 20 and coefficient of Range (CR) is 0.4

Page No 248:

Question 2:

Prices of shares of a company were note as under from Monday through Saturday. Find out range and the coefficient of range.

Day Mon. Tues. Wed. Thu. Fri. Sat.
Price (₹) 200 210 208 160 220 250

Answer:

Here,
Highest value among the prices of shares= 250
Lowest Value among the prices of shares= 160

Range (R) = Highest value (H)− Lowest Value (L)
or,   R  = 250 − 160
    R  = 90
Coefficient of Range (CR)=H-LH+L or, CR=250-160250+160=90410 CR =0.219 or 0.22 ( approx.)

Hence, the Range (R) of the above data is 90 and Coefficient of Range (CR) is 0.22

Page No 248:

Question 3:

You know share market is going bullish during the last several months. Collect weekly data on the share price of any two important industries during the past six months. Calculate the range of share prices. Comment on how volatile are the share prices.

Answer:

Month Price of shares
Tata Motors
Price of shares Reliance
Oct.
Nov.
Dec.
Jan.
Feb.
Mar.
325
397
405
415
420
388
913.35
900.25
750.90
780.70
799.25
850.35

For Tata Motors
Highest Value=420
Lowest Value=325

Range (R) = Highest Value (H)− Lowest Value (L)
or, R   = 420 − 325
  R   = 95
Coefficient of Range (CR1)=H-LH+L=420-325420+325=95745=0.127

For Reliance
Highest Value= 913.35
Lowest Value= 750.90

Range (R)= Highest value (H)− Lowest value (L)
or, R   = 913.35 − 750.90
  R   = 162.45

Coefficient of Range (CR2)=H-LH+L=913.35-750.90913.35+750.90=162.451664.25=0.097

From the above results we can observe that the prices of the Tata Motors shares are less volatile as compared to the prices of Reliance shares.

Page No 248:

Question 5:

Find out the range and the coefficient of range from the following data:

Daily Wage (₹) 6 7 8 9 10 11 12 15
Number of Workers 10 15 12 18 25 20 10 4

Answer:

Daily Wage 6 7 8 9 10 11 12 15
No. of workers 10 15 12 18 25 20 10 4

Here,
Highest value =15
Lowest value  = 6

Range (R) = Highest value (H) − Lowest value (L)
                = 15 − 6
                = 9

Coefficient of Range (CR)=H-LH+L=15-615+6=921=0.429

Hence, the Range (R) of the above series is 9 and Coefficient of Range (CR) is 0.429



Page No 250:

Question 1:

Marks obtained by 100 students of a class are given below. Find out range and coefficient of range of the marks.

Marks 10−20 20−30 30−40 40−50 50−60 60−70 70−80 80−90 90−100
Number of Students 4 10 16 22 20 18 8 2 5

Answer:

Marks No. of Student
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
80 − 90
90 − 100
4
10
16
22
20
18
8
2
5

Range (R)= Upper limit of last class interval − Lower limit of the first class interval
          (R)= 100 − 10
          (R) = 90

Coefficient of Range (CR)=H-LH+Lor, CR=100-10100+10or, CR=90110 CR=0.81 

Hence, the Range (R) of the marks is 90 and Coefficient of Range (CR) is 0.8

Page No 250:

Question 2:

In an examination, 25 students obtained the following marks. Find out coefficient of range of the marks.

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

Answer:

Converting the inclusive series into exclusive series,
 

Marks No. of Students
4.5 − 9.5
9.5 − 14.5
14.5 −19.5
19.5 −24.5
24.5 −29.5
29.5 −34.5
34.5 −39.5
1
3
8
5
4
2
2

Here,
Highest Marks value (H)=39.5
Lowest Marks value  (L)=4.5

Coefficient of Range (CR)=H-LH+Lor, CR=39.5-4.539.5+4.5or, CR=3544 CR=0.79 

Hence, the Coefficient of range of marks in the above series is 0.79



Page No 253:

Question 1:

Estimate quartile deviation and the coefficient of quartile deviation of the following data:
8, 9, 11, 12, 13, 17, 20, 21, 23, 25, 27
Show that QD is the average of the difference between two quartiles.

Answer:

Sr. No. 1 2 3 4 5 6 7 8 9 10 11
Data 8 9 11 12 13 17 20 21 23 25 27


In order to find the quartile deviation in case of individual series, find out the values of  first quartile and third quartile using the following equations:

Q1 = Size of N+14th item
or,  Q1     = Size of 11+14th item
or,  Q1     = Size of 3rd item
  Q1     = 11

Q3 = Size of 3N+14th item

or,  Q3  = Size of 311+14th item
or,   Q3 = Size of 9th item
   Q3 = 23

Calculating Quartile Deviation and Coefficient of Quartile Deviation

Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=23-112or, Q.D.=122 Q.D.=6Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=23-1123+11=1234=0.353                              

We know that the difference between the third quartile (Q3) and first Quartile (Q1) of a series is called the inter Quartile range i.e.
                             Inter Quartile Range= Q3-Q1
Where as Quartile deviation is  half of the Inter Quartile Range. i.e. 
                             Quartile Deviation (Q.D.)=
Q3-Q12

Hence, from the above formula it is clear that quartile deviation is the average of the difference between the two quartiles.

 

Page No 253:

Question 2:

Find out quartile deviation and coefficient of quartile deviation of the following series:
28, 18, 20, 24, 30, 15, 47, 27

Answer:

Arranging the data in ascending order.
 

Sr. No. 1 2 3 4 5 6 7 8
Observation 15 18 20 24 27 28 30 47


In order to find the quartile deviation in case of individual series, we need to find out the values of third quartile and first quartile using the following equations:

Q1 = Size of N+14th item

or, Q1= Size of 8+14th item

or, Q1= Size of 2.25th item

or, Q1= Size of 2nd item + 14Size of 3rd item-Size of 2nd item
or, Q1=18+1420-18

Q1 = 18.5

Q3 = Size of 3N+14th

or, Q3 = Size of 6.75th item
or, Q3 = Size of 6th item + 34Size of 7th item-Size of 6th item
or, Q3=28+3430-28or, Q3=27+342or, Q3=28+1.5 Q3=29.5Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=29.5-18.52or, Q.D.=112 Q.D.=5.5Coefficient of Quartile deviation=Q3-Q1Q3+Q1 or, Coefficient of Q.D. =29.5-18.529.5+18.5 or, Coefficient of Q.D. =1148Coefficient of Q.D. =0.229=0.23

Hence the Q.D. of the series is 5.5 and Coefficient of the Q.D. is 0.23



Page No 254:

Question 1:

Find out quartile deviation and the coefficient of quartile deviation of the following data:

Age 20 30 40 50 60 70 80
Members 3 61 132 153 140 51 3

Answer:

Age Numbers
(f)
Cumulative Frequency (c.f.)
20
30
(Q₁)40
50
(Q₃)60
70
80
3
61
132
153
140
51
3
3
3 + 61 = 64
64 + 132 = 196
196 + 153 = 349
349 + 140 = 489
489 + 51 = 540
540 + 3 = 543
  Σf=N=543  

Q1 = Size of N + 14th item
      = Size of 543+14th item
      = Size of 136th item

136th item lies in 196th cumulative frequency of the series. Age corresponding to 196th (c.f.) is 40.

Hence, Q1 = 40

Q3 = Size of 3N+14th item
     = Size of 3543+14th item
     = Size of 408th item

408th item lies in the 489th c.f. of the series. Age corresponding to 489th c.f. is 60.

Hence,  Q3 = 60.

Calculating Quartile Deviation and Coefficient of Quartile Deviation:

Quartile deviation (Q.D.)=Q3-Q12 or, Q.D. =60-402 or, Q.D. =202  Q.D. =10Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=60-4060+40or, Coefficient of Q.D.=20100Coefficient of Q.D. =0.2

Hence, Q.D. of the series is 10 and Coefficient of Q.D. is 0.2

Page No 254:

Question 2:

Estimate quartile deviation and the coefficient of quartile deviation of the following series:

Height (inches) 58 59 60 61 62 63 64 65 66
Number of Students 15 20 32 35 33 22 20 10 8

Answer:


Height
(inches)

 
No. of student (f) Cumulative frequency (c.f.)
58
59
60
61
62
63
64
65
66
15
20
32
35
33
22
20
10
8
15
15 + 20 = 35
35 + 32 = 67
67 + 35 = 102
102 + 33 = 135
135 + 22 = 157
157 + 20 = 177
177 + 10 = 187
187 + 8 = 195
  Σf=N=195  

Q1 = Size of N+14th item
     = Size of 195+14th item
     = Size of 49th item

49th item lies in 67th c.f. of the series. Height corresponding to 67th (c.f.) is 60.

Hence, Q1 = 60 inches

Q3 = Size of 3N+14th item
      = Size of 3195+14th item
      = Size of 147th item

147th item lies in 157th c.f. of the series. Height corresponding to 157th c.f. is 63.

Hence, Q3 = 63 inches

Calculating Quartile Deviation and Coefficient of Quartile Deviation


Quartile deviation (Q.D.)=Q3-Q12or, Q.D.=63-602or, Q.D.=32 Q.D.=1.5Coefficient of quartile deviation=Q3-Q1Q3+Q1or, Coefficient of Q.D.=63-6063+60or, Coefficient of Q.D.=3123 Coefficient of Q.D.=0.024

Hence, Q.D. of the series is 1.5 and Coefficient of Q.D. is 0.024



Page No 256:

Question 1:

Given the following data, find out quartile deviation and the coefficient of quartile deviation:

Wages (₹) 0−5 5−10 10−15 15−20 20−25 25−30
Number of Workers 4 6 3 8 12 7

Answer:

Wage No. of worker (f) Cumulative Frequency (c.f.)
0 − 5
5 − 10
10 − 15
15 − 20
20 −25
25 − 30
4
6
3
8
12
7
4
4 + 6 = 10
10 + 3 = 13
13 + 8 = 21
21 + 12 = 33
33 + 7 = 40
  Σf=N=40  

Q1 = Size of N4th item
     = Size of 404th item
     = Size of 10th item

10th item lies in group 5 − 10 and falls within 10th c.f. of the series.

Q1=l1+N4-c.f.f×i  

Here,
l=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=5+10-46×5Q1 =5+66×5Q1 =10

Like wise,
Q3 = Size of 3N4th item
      = Size of 3404th item
      = Size of 30th item

30th item lies in the group 20-25 within the 33rd c.f. of the series.

Thus, Q3=l1+3N4-c.ff×ior,  Q3 =20+30-2112×5or,  Q3 =20+9×512or,  Q3 =20+3.75  Q3 =23.75Quartile deviation (Q.D.)=Q3-Q12=23.75-102=13.752=6.87Coefficient of quartile deviation=Q3-Q1Q3+Q1=23.75-1023.75+10=13.7533.75=0.4

Page No 256:

Question 2:

Find out quartile deviation and coefficient of quartile deviation from the following data:

Class Interval 0−10 10−20 20−30 30−40 40−50 50−60
Frequency 4 8 5 4 9 10

Answer:

Class interval Frequency
(f)
Cumulative Frequency (c.f.)
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
4
8
5
4
9
10
4
4 + 8 = 12
12 + 5 = 17
17 + 4 = 21
21 + 9 = 30
30 + 10 = 40
  Σf=N = 40  

Q1 = Size of N4th item
     = Size of 404th item
     = Size of 10th item

10th item lies in group 10 − 20 and fall within 12th c.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 =10+10-48×10or, Q1 =10+608or, Q1 =10+7.5 Q1 =17.5

Likewise,
Q3 = Size of 3N4th item
     = Size of 3404th item
     = Size of 30th item

30th item lies in group 40-50 and fall within the 30th c.f. of the series

Thus, Q3=l1+3N4-c.f.f×i or, Q3 =40+30-219×10 or, Q3 =40+99×10  or, Q3=40+10  Q3=50Quartile deviation (QD)=Q3-Q12=50-17.52=32.52=16.25Coefficient of Q.D=Q3-Q1Q3+Q1=50-17.550+17.5=32.567.5=0.48



Page No 259:

Question 1:

Find out mean deviation of the monthly income of the five families given below, using arithmetic mean of the data:
852, 635, 792, 836, 750

Answer:

Sr. No. Monthly Income (X) Deviation from Arithmetic mean
dX=X-X
 X¯=773

 
1
2
3
4
5
852
635
792
836
750
852-773=79
138
19
63
23
N=5 ΣX = 3865 ΣdX¯=322

Mean X=ΣXN=38655=773Mean deviation MDx¯=ΣdXN=3225=64.4

Hence, mean deviation of the monthly income is 64.4.

Page No 259:

Question 2:

Weight of nine students of a class is given below. Calculate mean deviation, using median and arithmetic mean of the series. Also calculate coefficient of mean deviation:
Weight (kg) : 47, 50, 58, 45, 53, 59, 47, 60, 49

Answer:

Sr. No Weight
(X)
Deviation from Median
dm=X-M
(M = 50)
1
2
3
4
5
6
7
8
9
45
47
47
49
50
53
58
59
60
5
3
3
1
0
3
8
9
10
N=9   Σdm=42
 
Sr. No Weight
(X)
Deviation from Mean
dX=X-X¯X=52
1
2
3
4
5
6
7
8
9
45
47
47
49
50
53
58
59
60
7
5
5
3
2
1
6
7
8
N=9 X=468 Σdx¯=44

From Median
(a) M = Size of N+12th item
          = Size of 9+12th item
          = Size of 5th item
          = 50 kgs

(b) MDm=ΣdmN=429=4.666=4.67 kgs approx.

(c) Coefficient of MDm=MDmM=4.6750=0.0934

From Mean

(a) X=ΣXN=4689=52 kgs

(b) MDX=ΣdXN=449=4.89 kgs

(c) Coefficient of MDX=MDXX=4.8952=0.094



Page No 260:

Question 1:

Find out mean deviation and its coefficient of the following data:

Items 5 10 15 20 25 30 35 40
Frequency 8 16 18 22 14 9 6 7

Answer:

Items (X) Frequency (f) fX Deviation from Mean
X-X= dX
fdX
5
10
15
20
25
30
35
40
8
16
18
22
14
9
6
7
40
160
270
440
350
270
210
280
15.2
10.2
5.2
0.2
4.8
9.8
14.8
19.8
121.6
163.2
93.6
4.4
67.2
88.2
88.8
138.6
  Σf = 100 Σfx = 2020   ΣfdX=765.6

(a) Mean (X¯)=ΣfxΣf=2020100=20.2

(b) Mean deviation from Arithmetic Mean

MDX=ΣfdXΣf=765.6100=7.656

(c) Coefficient of MDX¯ =MDXX=7.65620.2=0.379=0.38 approx.

Page No 260:

Question 2:

Calculate mean deviation from the following data, using mean and median, respectively.

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

Answer:

Calculation of mean deviation from median

Size (X) Frequency (f) Cumulative frequency (c.f.) Deviation from median
|dM| = |X − M|
M = 8
f|dM|
4
6
8
10
12
14
16
2
4
5
3
2
1
4
2
2 + 4 = 6
6 + 5 = 11
11 + 3 = 14
14 + 2 = 16
16 + 1 = 17
17 = 4 = 21
4
2
0
2
4
6
8
8
8
0
6
8
6
32
  Σf=N= 21     Σf|dM| = 68

(a) Median or (M) = Size of N+12th item
                             = Size of 21+12th item
                             = Size of 11th item
                             = 8

(b) Mean Deviation from median

MDM=ΣfdMN=6821=3.24

Calculation of mean deviation from mean
Size (X) Frequency (f) fX From mean deviation from mean
dX=X-XX=9.71
fdX¯
4
6
8
10
12
14
16
2
4
5
3
2
1
4
8
24
40
30
24
14
64
5.71
3.71
1.71
0.29
2.29
4.29
6.29
11.42
14.84
8.55
0.87
4.58
4.29
25.16
  Σf = 21 ΣfX = 204   ΣfdX¯= 69.71

(a) Mean (X¯)=ΣfXΣf=20421=9.71

(b) Mean deviation from A.M.

MDX=ΣfdXΣf=69.7121=3.32



Page No 262:

Question 1:

The following table gives distribution of marks for 50 students of a class. Calculate mean deviation from the mean and median respectively from the data:

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

Answer:

Calculation of M.D from Median
 

Marks Mid value m Frequency
(f)
Cumulative frequency |dM| = |m − M|
M = 172.78
f|dM|
140 − 150
150 − 160
160 − 170
170 − 180
180 − 190
190 − 200
145
155
165
175
185
195
4
6
10
   18(f)
9
3
4
4 + 6 = 10
10 + 10 = 20(c.f)
20 + 18 = 38
38 + 9 = 47
47 + 3 = 50
27.78
17.78
7.78
2.22
12.22
22.22
111.12
106.68
77.8
39.96
109.98
66.66
    Σf = N = 50     Σf|dM| = 512.2

Median = Size of N2th item
             = Size of 502th item
             = Size of 25th item

25th item lies under the 38th cumulative frequency therefore 170 − 180 is the median class interval.

Thus, median value will be as follows:

M=l1+N2-c.ff×ior, M =170+25-2018×10or, M  =170+5018or, M  =170+2.78 M =172.78Hence, Mean deviation MDM=ΣfdMΣf=512.250=10.24

Calculation of M.D from Mean
 
Marks Mid value
m
Frequency
(f)
fm dX=m-XX=171.2 fdX
140 − 150
150 − 160
160 − 170
170 − 180
180 − 190
190 − 200
145
155
165
175
185
195
4
6
10
18(f)
9
3
580
930
1650
3150
1665
585
26.2
16.2
6.2
3.8
13.8
23.8
104.8
97.2
62
68.4
124.2
71.4
    Σf = 50 Σfm = 8560   ΣfdX=528

Mean (X)=ΣfmΣf=856050=171.2

Hence, Mean deviation from the arithmetic mean is:

MDX=ΣfdXΣf=52850=10.56

Page No 262:

Question 2:

Estimate the coefficient of mean deviation from the median from the following data:

Age Group 20−30 30−40 40−50 50−60 60−70
Number of Workers 8 12 20 16 4

Answer:

Calculation of MD. from Median
 

Age
Mid value
(m)

 
No. of workers (f) Cumulative Frequency |dm| = |m − M|

M = 45
f|dm|
20 − 30
30 − 40
40 − 50
50 − 60
60 −70
25
35
45
55
65
8
12
20
16
4
8
8 + 12 = 20
20 + 20 =40
40 + 16 =56
56 + 4 = 60
20
10
0
10
20
160
120
0
160
80
    Nf = 60     Σf|dm| = 520

Median = Size of N2th item
             = Size of 602th item
             = Size of 39th item

39th item lies under the 40th cumulative frequency of the series. So, 40 − 50 will be the median class interval.

Hence, the median value of the series is:

M=l1+N2-c.f.f×ior, M  =40+30-2020×10or, M  =40+10×1020 M =45Thus,Mean Deviation from median MDM=ΣfdmΣf=52060=8.67Coefficient of MDM=MDMMedian=8.6745=0.19



Page No 266:

Question 1:

The following table gives marks obtained by 7 students of a class. Find out standard deviation of the marks.
40, 42, 38, 44, 46, 48, 50

Answer:

S. No. Marks
(X)
Deviation
x=X-XX=44
Square of the deviation
x2=X-X2
1
2
3
4
5
6
7
40
42
38
44
46
48
50
−4
−2
−6
0
2
4
6
16
4
36
0
4
16
36
N = 7 ΣX = 308   Σx2 = 112

Mean (X)=ΣXN=3087=44Standard deviation (σ )=x2N=1127=16=4
Hence, standard deviation of the marks is 4.

Page No 266:

Question 2:

Weight of some students is given below in kilograms. Find out standard deviation.
41, 44, 45, 49, 50, 53, 55, 55, 58, 60

Answer:

S. No. Weight
(X)
Deviation
x=X-XX=51
Square of deviation
x2=X-X2
1
2
3
4
5
6
7
8
9
10
41
44
45
49
50
53
55
55
58
60
−10
−7
−6
−2
−1
2
4
4
7
9
100
49
36
4
1
4
16
16
49
81
N = 10 ΣX = 510   Σx2 = 356

Mean (X)=ΣXN=51010=51Standard deviation (σ )=x2N=35610=35.6=5.97

Hence, the Standard deviation is 5.97.

Page No 266:

Question 3:

Using step-deviation method, calculate standard deviation of the following series:

Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Number of Students 5 10 20 40 30 20 10 4

Answer:

Marks
(X)
Mid value
(m)
Frequency
(f)
fm Deviation from mean value
x=X-X

X¯=39.39
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
5
10
20
40
30
20
10
4
25
150
500
1400
1350
1100
650
300
−34.39
−24.39
−14.39
−4.39
5.61
15.61
25.61
35.61
1182.67
594.87
207.07
19.27
31.47
243.67
655.87
1268.07
5913.35
5948.7
4141.4
770.8
944.1
4873.4
6558.7
5072.28
    Σf = 139 Σfm = 5475     Σfx2 = 34222.73

Mean (X)=ΣfmN=5475139=39.39Standard deviation (σ )=Σfx2N=34222.73139=246.21=15.7

Hence, the standard deviation of the above series is 15.7



Page No 269:

Question 1:

Find out standard deviation of the savings of the following 10 persons:

Persons 1 2 3 4 5 6 7 8 9 10
Savings (₹) 114 108 100 98 101 109 117 119 121 126

Answer:

Persons Savings
(X)

Deviation
x=X-X

X¯=111.3
 
Square of the deviation
x2
1
2
3
4
5
6
7
8
9
10
114
108
100
98
101
109
117
119
121
126
2.7
−3.3
−11.3
−13.3
−10.3
−2.3
5.7
7.7
9.7
14.7
7.29
10.89
127.69
176.89
106.09
5.29
32.49
59.29
94.09
216.09
N = 10 ΣX = 1113   Σx2 = 836.1

Mean (X)=ΣXN=111310=111.3Standard deviation (σ )=Σx2N=836.110=83.61=9.14

Hence, standard deviation of the savings is 9.14

Page No 269:

Question 2:

Find out standard deviation and its coefficient of the following series:

Size 10 20 30 40 50 60 70
Frequency 6 8 16 15 33 11 12

Answer:

Size
(X)
Frequency
(f)
fX Deviation from mean
x=X-X
Square of deviation x2 fx2
10
20
30
40
50
60
70
6
8
16
15
33
11
12
60
160
480
600
1650
660
840
−34.06
−24.06
−14.06
−4.06
5.94
15.94
25.94
1160.08
578.88
197.68
16.48
35.28
254.08
672.88
6960.48
4631.04
3162.88
247.2
1164.24
2794.88
8074.60
  Σf = 101 Σfx = 4450     Σfx2 = 27035.32

Mean (X)=ΣfXΣf=4450101=44.06Standard deviation (σ )=Σfx2Σf=27035.32101=267.68=16.36Coefficient of S.D.=σX=16.3644.06=0.37

Hence, the S.D. is 16.36 and Coefficient of S.D. is 0.37



Page No 273:

Question 3:

Find out standard deviation of the distribution of population in 104 villages of a Tehsil, as given below by step-deviation method.

Distribution of Population
Population No. of Villages
0−200

200−400

400−1,000

1,000−2,000

2,000−5,000
10

28

42

18

6

Answer:

 

Population Mid Value
(m)
No. of villages
(f)


Deviation assumed from average
dx=m-A

 

dx'=dxi=dx40

fdx'=f×dx'

fdx'2=f×dx'2

0-200
200-400
400-1000
1000-2000
2000-5000
100
300
700=A
1500
3500
10
28
42
18
6
-600
-400
0
800
2800
-15
-10
0
20
70
-150
-280
0
360
420
2250
2800
0
7200
29400
     N=104     fdx'=-350 fdx'2=41650



Calculating Standard deviation through Step deviation method:

Standard Deviation (σ)=Σfdx'2N-Σfdx'N2×ior, σ =41650104--3501042×40or, σ  =400.480-11.296×40or, σ  =389.184×40or, σ  =19.72775×40 σ =789.11

Hence, standard deviation of the above series is 789.11

Page No 273:

Question 1:

Calculate standard deviation of the following series:

Size 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

 
Frequency Deviation from assumed Average
(dx = mA)
dx'=dxC   C=5
 
fdx' fdx'2
0 − 5
5 − 10
10 − 15
15 − 20
20 − 25
25 − 30
30 − 35
35 − 40
2.5
7.5
12.5
    17.5(A)
22.5
27.5
32.5
37.5
2
5
7
13
21
16
8
3
−15
−10
−5
0
5
10
15
20
−3
−2
−1
0
1
2
3
4

−6
−10
−7
0
21
32
24
12
 
18
20
7
0
21
64
72
48
    N = 75     Σfdx' = 66 Σfdx'2 = 250

Calculating Standard deviation through Step deviation method:

Standard deviation (σ )=Σfdx'2N-Σfdx'N2×Cor, σ=Σfdx'2N-Σfdx'N2×5or, σ  =25075-66752×5or, σ  =3.33-0.7744 ×5or, σ  =2.5556×5or, σ  =1.598×5 σ  =7.99

Hence, standard deviation of the above series is 7.99.

Page No 273:

Question 2:

Calculate standard deviation of the following data, using step-deviation method.

Age 20−30 30−40 40−50 50−60 60−70 70−80 80−90
Frequency 3 61 132 153 140 51 2

Answer:

Age
Mid value
m

 
Frequency
(f)

 
Deviation assumed from average
(dx = m − A)
dx'=dxc=dx10 fdx'=f×dx' fdx'2=f×dx'2
20−30
30 −40
40 −50
50 −60
60 −70
70 −80
80 −90
25
35
45
  55=A
65
75
85
3
61
132
153
140
51
2
−30
−20
−10
0
10
20
30
−3
−2
−1
0
1
2
3
−9
−122
−132
0
140
102
6
27
244
132
0
140
204
18
    N = 542     Σfdx' = -15 Σfdx'2 = 765

Standard deviation (σ)=Σfdx'2N-Σfdx'N2×cor, σ  =765542--155422×10or, σ   =1.41--0.032×10or, σ   =1.41-0.0009×10or, σ   =1.19×10 σ   =11.9

Hence, standard deviation of the above data is 11.9

Page No 273:

Question 4:

Calculate mean and standard deviation of the following data by short-cut method:

Class Interval 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Frequency 5 10 15 20 25 18 7

Answer:

C.I. Mid Value
(m)
Frequency
(f)
Deviation from Assumed mean
dx = m − A
(A = 45)
fdx f×dx2=fdx2
10 − 20
20 − 30 30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
15
25
35
45=A
55
65
75
5
10
15
20
25
18
7
−30
−20
−10
0
10
20
30
−150
−200
−150
0
250
360
210
4500
4000
1500
0
2500
7200
6300
    Σf = N = 100   Σfdx = 320 Σfdx2 = 26000

Calculating Mean and standard deviation through Short-cut method:Mean (X)=A+ΣfdxΣf or, X=45+320100 or, X=45+3.2 X=48.2Standard deviation (σ)=Σfdx2N-ΣfdxN2or, σ =26000100-3201002or, σ =260-3.22or, σ =260-10.24or, σ =249.76 σ =15.80



Page No 280:

Question 1:

Following are the marks obtained by 20 students in statistics. Find out coefficient of variation of the marks.

62 85 73 81 74 58 66 72 54 84
65 50 83 62 85 52 80 86 71 75

Answer:

S. No. Marks
(X)
Deviation from mean
x=X-X
Square deviation x2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
62
85
73
81
74
58
66
72
54
84
65
50
83
62
85
52
80
86
71
75
−8.9
+14.1
+2.1
10.1
3.1
−12.9
−4.9
1.1
−16.9
13.1
−5.9
−20.9
12.1
−8.9
14.1
−18.9
9.1
15.1
.1
4.1
79.21
198.81
4.41
102.01
9.61
166.41
24.01
1.21
285.61
171.61
34.81
436.81
146.81
79.21
198.81
357.21
82.81
228.01
.01
16.81
  ΣX = 1418   Σx2 = 2623.8


Mean (X)=ΣXN=141820=70.9Standard deviation (σ)=x2Nor, σ =2623.820or, σ =131.19 σ =  11.45Coefficient of variation (CV)=σX×100or,  CV=11.4570.9×100  CV=16.14

Hence, coefficient of variation (CV) of the marks is 16.14

Page No 280:

Question 2:

Calculate coefficient of variation of the following data:

S. No. 1 2 3 4 5 6 7 8
Value 25 42 33 48 45 29 43 39

Answer:

S. No. Value (X) Deviation from mean
x=X-X
x2
1
2
3
4
5
6
7
8
25
42
33
48
45
29
43
39
−13
4
−5
10
7
−9
5
1
169
16
25
100
49
81
25
1
N = 8 ΣX = 304   Σx2 = 466

Mean (X)=ΣXN=3048=38Standard deviation (σ )=Σx2N=4668=58.25=7.63Coefficient of variation (CV)=σX×100=7.6338×100=76338=20.08

Hence, coefficient of variation is 20.08

Page No 280:

Question 3:

Given the following data, calculate coefficient of variation:

Age 20−30 30−40 40−50 50−60 60−70 70−80 80−90
Number of Students 3 61 132 153 140 51 2

Answer:

Age Mid Value
(m)
Frequency
(f)

 
Deviation from Assumed mean
(dx = X − A)
(A = 55)
dx'=dxi=dx10 f×dx'=fdx' f×dx'2=fdx'2
20−30
30 −40
40 −50
50 −60
60 −70
70 −80
80 −90
25
35
45
55=A
65
75
85
3
61
132
153
140
51
2
−30
−20
−10
0
10
20
30
−3
−2
−1
0
1
2
3
−9
−122
−132
0
140
102
6
 
27
244
132
0
140
204
18
    N = 542     Σfdx'= -15 Σfdx'2 =765

Mean (X)=A+Σfdx'N×ior,  X=55+-15542×10or,  X =55-.28  X =54.72Standard Deviation (σ)=Σfdx'2N-Σfdx'N2×ior, σ =765542--155422×10or, σ  =1.41--0.032×10or, σ  =1.41-0.0009×10or, σ  =1.187×10 σ =11.87Coefficient of Variation (CV)=σX×100or, CV=11.8754.72×100or, CV=119054.72CV =21.69

Hence, Coefficient of variation is 21.69



Page No 283:

Question 1:

The following table shows number of firms in two different areas 'A' and 'B' according to their annual profits. Present the data by way of Lorenz curve.

Profit ('000' ₹) 6 25 60 84 105 150 170 400
Firms in Area A 6 11 13 14 15 17 10 14
Firms in Area B 2 38 52 28 38 26 12 4

Answer:

  Firm in Area A Firms in Area B
Profit
(x)
Cumulative sum Cumulative
%
No. of firm c.f Cumulative
%
No. of firms c.f. Cumulative
%
6
25
60
84
105
150
170
400
6
31
91
175
280
430
600
1000
.6
3.1
9.1
17.5
28
43
60
100
6
11
13
14
15
17
10
14
6
17
30
44
59
76
86
100
6
17
30
44
59
76
86
100
2
38
52
28
38
26
12
4
20
40
92
120
158
184
196
200
10
20
46
60
79
82
98
100
Σx = 1000                


Page No 283:

Question 2:

Make a Lorenz curve of the following data:

Income 500 1,000 2,000 3,000 3,500
Number in Class 'A' (000) 4 6 8 12 10
Number in Class 'B' (000) 8 7 5 3 2

Answer:

  CLASS A CLASS B
Income
(x)
Cumulative sum Cumulative
%
No. of class c.f. Cumulative
%
No. of class c.f. cumulative
500
1000
2000
3000
3500
500
1500
3500
6500
10,000
5
15
35
65
100
4
6
8
12
10
4
10
18
30
40
10
25
45
75
100
8
7
5
3
2
8
15
20
23
25
32
60
80
92
100




Page No 296:

Question 1:

Calculate range and coefficient of range from the following data:
4, 7, 8, 46, 53, 77, 8, 1, 5, 13

Answer:

Given:
Highest value (H) = 77
Lowest value (L) = 1
Range = Highest value − Lowest value
i.e. R = HL
Substituting the given values in the formula.
R = 77 − 1 = 76
Coefficient of Range=H-LH+L=77-177+1=7678=0.97
Thus, range is 76 and coefficient of range is 0.97

Page No 296:

Question 2:

Given the following data-set, calculate range and the coefficient of range:

Size 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5
Frequency 4 5 6 3 2 1 3 5

Answer:

Highest value (H) = 11.5
Lowest value (L) = 4.5
Range = Highest value − Lowest value
i.e. R = HL
or, R = 11.5 − 4.5 = 7
Coefficient of Range=H-LH+L=11.5-4.511.5+4.5=716=0.437
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 297:

Question 3:

Find out quartile deviation and the coefficient of range, given the following data-set:

Class Interval 1−5 6−10 11−15 16−20 21−25 26−30 31−35
Frequency 2 8 15 35 20 10 14

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 Mid Value= L1+L22 Frequency
1 − 5
6 − 10
11 − 15
16 − 20
21 − 25
26 − 30
31 − 35
0.5 − 5.5
5.5 − 10.5
10.5 − 15.5
15.5 − 20.5
20.5 − 25.5.
25.5. − 30.5
30.5 − 35.5
3
8
13
18
23
28
33
2
8
15
35
20
10
14
Range = Upper Limit of the Highest Class Interval - Lower Limit of the Lowest Class Intervalor, Range = 35.5 - 0.5 = 35

Coefficient of Range=H-LH+L=35.5-0.535.5+0.5=3536=0.972

Page No 297:

Question 4:

Find out quartile deviation and the coefficient of quartile deviation of the following series.
Wages of  9 workers in Rupees:
170, 82, 110, 100, 150, 150, 200, 116, 250

Answer:

Arranging the data in the ascending order as presented below.
82, 100, 110, 116, 150, 150, 170, 200, 250
Here, N = 9
Q1=size of 9+14th=size of 2.5th item    =size of 2nd item+0.5 size of 3rd item-size of 2nd item    =100+0.5110-100=105Q3=size of 39+14thitem=size of 7.5th item    =size of 7th item+0.5 size of 8th item-size of 7th item    =170+0.5200-170=185Q.D.=Q3-Q12=185-1052=40

Coefficient of Q.D.=Q3 - Q1Q3 + Q1=185-105185+105=0.276                      

Page No 297:

Question 5:

Given the following data, estimate the coefficient of QD:
15, 20, 23, 23, 25, 25, 27, 40

Answer:

Given data series:
15, 20, 23, 23, 25, 25, 27, 40
Here, N = 8
Q1=size of 8+14th=size of 2.25th item    =size of 2nd item+0.25 size of 3rd item-size of 2nd item    =20+0.2523-20=20.75Q3=size of 38+14thitem=size of 6.75th item    =size of 6th item+0.75 size of 7th item-size of 6th item    =25+0.7527-25=26.5
Coefficient of Q.D.=Q3 - Q1Q3 + Q1=26.5-20.7526.5+20.75=0.121

Page No 297:

Question 6:

Find out mean deviation of the following series from mean and median:

Size 4 6 8 10 12 14 16
Frequency 2 4 5 31 2 1 4

Answer:

x f fx dX=X-X fdX
4
6
8
10
12
14
16
2
4
5
31
2
1
4
8
24
40
310
24
14
64
5.87
3.87
1.87
0.13
2.13
4.13
6.13
11.74
15.48
9.35
4.03
4.26
4.13
24.52
  f = 49 fx = 484    fdX = 73.51

Mean X=fxf=48449=9.87Thus calculate the deviation of the values from 9.87.Mean deviation from mean MDX=fdXf=73.5149=1.50
x f c.f. dM = |xM| f|dM|
4
6
8
10
12
14
16
2
4
5
31
2
1
4
2
6
11
42
44
45
49
6
4
2
0
2
4
6
12
16
10
0
4
4
24
  f = 49     f|dM| = 70

N = 49
Median=size of N+12thor, Median=size of  49+12thitem= size of 25th item
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th is 42 (in the c.f. column), which is corresponding to 10.
Hence, median is 10.
Thus, we calculate the deviation of the values from 10. 

Mean Deviation from median=MDM=fdMf=7049=1.428 (or 1.43 approximately) 

Page No 297:

Question 7:

Calculate mean deviation and coefficient of mean deviation with the help of median:

Class Interval 0−10 10−20 20−30 30−40 40−50 50−60
Frequency 15 19 14 20 18 14

Answer:

x Mid Value
(m)
f c.f. |mM| = |dM| f|dM|
0 − 10
10 − 20
20 − 30
5
15
25
15
19
14
15
34
48
26
16
6
390
304
80
30-40 35 20 68 4 252
40 − 50
50 − 60
45
55
18
14
86
100
4
14
252
336
    fx = 100     f|dM| = 1446
Here, N or ∑fx = 100
Median class is given by the size of N2th item, i.e.1002th item, which is 50th item.
This corresponds to the class interval of 30 40, so this is the median class.
Median=l1+N2-c.f.f×iso, Median=30+1002-4820×10or, Median=30+220×10or, Median=30+1Therefore, Median=31

Thus, we calculate the deviation of values from 31.
Mean Deviation from median=MDM=fdMf=1446100=14.46Coefficient of Mean Deviation = Mean DeviationMedian=14.4631= 0.466

Page No 297:

Question 8:

Calculate mean deviation from mean of the following series:

Size of Items 3−4 4−5 5−6 6−7 7−8 8−9 9−10
Frequency 3 7 22 60 85 32 9

Answer:

Class Interval Mid Value
(m
)
f fm m-X=dX¯ fdX
3 − 4
4 − 5
5 − 6
6 − 7
7 − 8
8 − 9
9 − 10
3.5
4.5
5.5
6.5
7.5
8.5
9.5
3
7
22
60
85
32
9
10.5
31.5
121
390
637.5
272
85.5
3.6
2.6
1.6
0.6
0.4
1.4
2.4
10.8
18.2
35.2
36
34
44.8
21.6
    f = 218 fm  = 1548   fdX = 200.6

Mean X=fmf=1548218=7.10Thus, we calcualte the value of deviations from 7.10Mean Deviation from Mean MDX=fdXf=200.6218=0.920

Page No 297:

Question 9:

Given below are the marks obtained by the students of a class. Calculate mean deviation, and its coefficient, using median of data.

Marks 17 35 38 16 42 27 19 11 40 25

Answer:

The given data is an individual series, therefore, to calculate median, we first need to arrange the data in ascending order. This is done as below.

11, 16, 17, 19, 25, 27, 35, 38, 40, 42Here, N =10 (even)Median=size of N2thitem+size of N2+1th item2or, Median=size of 5thitem+size of 6th item2=25+272=26so, Median is 26.
Thus, we calculate the deviation of values from 26

X |x - M| = |dM|
11
16
17
19
25
27
35
38
40
42
15
10
9
7
1
1
9
12
14
16
  |dM| = 94

Mean Deviation from Median MDM=dMn=9410=9.4Coefficient of MDM=MDMMedian=9.426=0.36

Page No 297:

Question 10:

Nine students of a class obtained following marks. Calculate mean deviation from median.

S. No. 1 2 3 4 5 6 7 8 9
Marks 68 49 32 21 54 38 59 66 41

Answer:

The given data series is discrete in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.
21, 32, 38, 41, 49, 54, 59, 66, 68Here, N =9 (odd)Median=size of N+12=9+12=size of 5th item
Thus, we calculate the deviation of values from 49.

Marks Ascending Order |X- M| = |dM|
68
49
32
21
54
38
59
66
41
21
32
38
41
49
54
59
66
68
28
17
11
8
0
5
10
17
19
    |dM| = 115

Mean Deviation from Median MDM=dMN=1159=12.77

Page No 297:

Question 11:

Following data relate to the age-difference of husbands and wives of a particular community. Find out mean deviation from mean.

Age-difference 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40
Frequency 449 705 507 281 109 52 16 4

Answer:

X Mid Value
m
f fm m-X=dX fdX
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
449
705
507
281
109
52
16
4
1122.5
5287.5
6337.5
4917.5
2452.5
1430
520
150
7.96
2.96
2.04
7.04
12.02
17.04
22.04
27.04
3574.04
2086.8
1034.28
1978.24
1312.36
886.08
352.64
108.16
    f = 2123 fm = 22217.5   fdX = 11332.6

Mean X=fmf=22217.52123=10.46Mean Deviation from Mean MDX=fdXX=11332.62123=5.34



Page No 298 :

Question 13:

Calculate standard deviation, given the following data:
10, 12, 14, 16, 18, 22, 24, 26, 28

Answer:

X x=X-X x2
10
12
14
16
18
22
24
26
28
−8.88
−6.88
−4.88
−2.88
−.88
3.12
5.12
7.12
9.12
78.85
47.33
23.81
8.29
0.77
9.73
26.21
50.69
83.17
X = 170   x2 = 328.85
Mean X=XN=1709=18.88Standard Deviation σ=x2N=328.859=6.045



Page No 298:

Question 12:

Find out the mean deviation and its coefficient using median of the following data:

S. No. 1 2 3 4 5 6 7 8 9 10 11 12
Number of Victims of Accidents 16 21 10 17 8 4 2 1 2 2 2 2

Answer:

The given data series is individual in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.

1, 2, 2, 2, 2, 2, 4, 8, 10, 16, 17, 21Here, N =12 (even)Median=size of N2thitem+size of N2+1th item2or, Median=size of 6thitem+size of 7th item2=2+42=3so, Median is 3.

No. of Victims Ascending Order |x - M| = |dM|
16
21
10
17
8
4
2
1
2
2
2
2
1
2
2
2
2
2
4
8
10
16
17
21
2
1
1
1
1
1
1
5
7
13
14
18
    |dM| = 65
Mean Deviation from Median MDM=dMn=6512=5.42Coefficient of MDM=MDMMedian=5.423=1.81

Page No 298:

Question 14:

Calculate standard deviation and the coefficient of standard deviation, given the following data:

Income (₹) 5 10 15 20 25 30 35 40
Number of Workers 26 29 40 35 26 18 14 12

Answer:

Income
(X)
No. of workers
(​f)
fX x=X-X x2 fx2
5
10
15
20
25
30
35
40
26
29
40
35
26
18
14
12
130
290
600
700
650
540
490
280
−14.4
−9.4
−4.4
0.6
5.6
10.6
15.6
20.6
207.36
88.36
19.36
0.36
31.36
112.36
243.36
424.36
5391.36
2562.44
774.4
12.6
815.36
2022.48
3407.04
5092.32
  f = 200 fX = 3880     fx2 = 20078

X=fXf=3880200=19.4Standard Deviation σ=fx2f=20078200=10.02Coefficient of Standard Deviation=σX=10.0219.4=0.516

Page No 298:

Question 15:

Of the two sets of income distribution of five and seven persons respectively, as given below calculate standard deviation:

(i) Income (₹) 4,000 4,200 4,400 4,600 4,800
(ii) Income (₹) 3,000 4,000 4,200 4,400 4,600 4,800 5,800

Answer:

For Income Group I

X1 x1=X1-X1 xI2
4000
4200
4400
4600
4800
−400
−200
0
200
400
160000
40000
0
40000
160000
X1 = 22,000   xI2 = 4,00,000

X1=X1n1=220005=4400Standard Deviation σ1=x12n1=4,00,0005=282.84Thus, Standard Deviation for the first group of income is Rs 282.84

For Income Group II
X2 x2=X2-X2 x22
3000
4000
4200
4400
4600
4800
5800
−1400
−400
−200
0
200
400
1400
1960000
160000
40000
0
40000
160000
1960000
30,800   x22 = 4320000

X2=X2n2=30,8007=4,400Standard Deviation σ2=x22n2=4,32,0007=785.58Thus, Standard Deviation for the second group of income is Rs 785.58

Page No 298:

Question 16:

Find out the standard deviation of the marks secured by 10 students:

S. No. 1 2 3 4 5 6 7 8 9 10
Marks 43 48 65 57 31 60 37 48 78 59

Answer:

S.No. Marks
(X)
x=X-X x2
1
2
3
4
5
6
7
8
9
10
43
48
65
57
31
60
37
48
78
59
−9.6
−4.6
12.4
4.4
−21.6
7.4
−15.6
−4.6
25.4
6.4
92.16
21.16
153.76
19.36
466.56
54.76
243.36
21.16
645.16
40.96
  X = 526   x2 = 1758.4
Mean X=XN=52610=52.6Standard Deviation σ=x2N=1758.410=13.26

Page No 298:

Question 17:

Data of daily sale proceeds of a shop are as below. Calculate mean deviation and standard deviation.

Daily Sales 102 100 110 114 118 122 126
Days 3 9 25 35 17 10 1

Answer:

Daily Sales
(X)
Days
(​f)
fX X-X=dX fdX or fX-X fX-X2 or fdX2
102
100
110
114
118
122
126
3
9
25
35
17
10
1
306
900
2750
3990
2006
1220
126
10.98
12.98
2.98
1.02
5.02
9.02
13.02
32.94
116.82
745
35.7
85.34
90.2
13.02
361.68
1516.32
222
36.
428.4
813.6
169.52
  f = 100 fX = 11298   fdX = 1119.02 ∑ = 3547.92
Mean X=fXf=11298100=112.98Mean Deviation from mean MDX=fdXf=1119.02100=11.19Standard Deviation (σ) =fX-X2f=3547.92100= 5.96

Page No 298:

Question 18:

Calculate range, standard deviation and coefficient of variation of marks secured by students.

50 55 57 49 54 61 64 59 58 56

Answer:

X x=X-X x2
50
55
57
49
54
61
64
59
58
56
−6.3
−1.3
.7
−7.3
−2.3
4.7
7.7
2.7
1.7
−.3
39.69
1.69
.49
53.29
5.29
22.09
59.29
7.29
2.89
.09
X=563   x2=192.1
Here,
N = 10
Highest value (H) = 64
Lowest value (Z) = 49
Range (R) = H − L = 64 − 49 = 15
So, range is 15.
Mean X=XN=56310=56.3Standard Deviation σ=x2N=192.110=4.38Coefficient of Variation=σX¯×100=4.3856.3×100=7.78

Page No 298:

Question 19:

Following data show the number of runs made by Sachin and Sourabh in different innings. Find out who is a good scorer and who is a consistent player?

Sachin 92 17 83 56 72 76 64 45 40 32
Sourabh 28 70 31 00 59 108 82 14 3 95

Answer:

Sachin Saurav
X1 x1 x12 X2 x2 x22
92
17
83
56
72
76
64
45
40
30
34.3
−40.7
25.3
−1.7
14.3
18.3
6.3
−12.7
−17.7
−25.7
1176.49
1656.49
640.09
2.89
204.49
334.89
39.69
161.29
313.29
660.49
28
70
31
00
59
108
82
14
3
95
−27
25
−24
−55
4
53
27
−41
−52
40
729
625
576
3025
16
2809
729
1681
2704
1600
X1= 577   x12 = 5190.1 X2 = 550   x22 = 14494

XSachin=X1n1=57710=57.7σSachin=x12n1=5190.110=22.78Coefficient of Variation (Sachin)=σSachinXSachin×100=22.7857.7×100=39.48XSaurav=X2n2==55010=55σSaurav=x22n2=1449410=38.07Coefficient of Variation (Saurav)=σSauravXSaurav×100=38.0755×100=69.21
Observation and Conclusion
  • In order to regard anyone as a better batsman, we have to analyse the mean runs of both the players. Analysing the mean runs, we can say that Sachin is a better scorer, since his mean runs (average of 57.7) is higher than that of Saurav (average of 55).
  • In order to regard one as more consistent, we have to take the help of coefficient of variation. The one who has lower coefficient of variation will be regarded as more consistent than the other. In this way, we can say that Sachin is more consistent than Saurav, as the coefficient of variation of Sachin (39.48) is lower than that of Saurav's (69.21).
Thus, on the basis of the above two observations, we can conclude that Sachin is not only a better scorer, but also a more consistent player than Saurav.



Page No 299:

Question 20:

Calculate standard deviation of marks secured by 100 examinees in the examination:

Marks 10−20 20−30 30−40 40−50 50−60 60−70 70−80 80−90
Number of Examinees 19 3 2 49 24 2 0 1

Answer:

X Mid Values
m
f fX x=X- X x2 fx2
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
80 − 90
15
25
35
45
55
65
75
85
19
3
2
49
24
2
0
1
285
75
70
2205
1320
130
0
85
−26.70
−16.70
−6.7
3.3
13.3
23.3
33.3
43.3
712.89
278.89
44.89
10.89
176.89
542.89
1108.89
1874.89
13544.91
836.67
89.78
533.61
4245.36
1085.78
0
1874.89
    f = 100 fX = 4170     fx2 = 22211

X=fXf=4170100=41.70Standard Deviation σ=fx2f=22211100=14.90

Page No 299:

Question 21:

Calculate coefficient of variation from the following data:

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

Answer:

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

X=fXf=4400100=44Standard Deviation σ=fx2f=27000100=16.43Coefficient of Variation =σX×100=27000100×100 = 37.34

Page No 299:

Question 22:

Estimate coefficient of variation of the following data:

Weight (kg) 0−20 20−40 40−60 60−80 80−100
Number of Persons 81 40 66 49 14

Answer:

Class Interval Mid Value
(m)
f fm x=X- X x2 fx2
0 − 20
20 − 40
40 − 60
60 − 80
80 − 100
10
30
50
70
90
81
40
66
49
14
810
1200
3300
3430
1260
−30
−10
10
30
50
900
100
100
900
2500
72900
4000
6600
44100
35000
    f = 250 fm = 10000     fx2 = 162600

X=fmf=10000250=40Standard Deviation σ=fx2f=162600250=25.50Coefficient of Variation =σX×100=25.5040×100 = 63.75



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