Tr Jain & Vk Ohri (2017) Solutions for Class 11 Science Economics Chapter 9 Measures Of Central Tendency Arithematic Mean are provided here with simple step-by-step explanations. These solutions for Measures Of Central Tendency Arithematic Mean are extremely popular among Class 11 Science students for Economics Measures Of Central Tendency Arithematic Mean 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 9 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 Class 11 Science Economics are prepared by experts and are 100% accurate.

Page No 148:

Question 1:

Following is the monthly income of eight families in a locality
Monthly Income (₹) : 70, 10, 500, 75, 13, 250, 8, 42
Find out arithmetic mean using Direct Method and Short-cut Method.

Answer:

Families
Monthly income
(X)
Deviation from the Assumed average
(d = X − A) (A = 75)
1
2
3
4
5
6
7
8
70
10
500
75 (A)
13
250
8
42
−5
−65
425
0
−62
175
−67
−33

(N) = 8

(ΣX)=968 Σd = −232 + 600 = 368

where,
N= Number of items
ΣX= Total of monthly income of all the families
Σd= net sum of deviations


Calculation of Mean by Direct Method
X=ΣXN=9688=121

Calculation of mean by short cut method
 X=A+ΣdNor, X=75+3688or, X=75+46X=121

So, mean monthly income of a family is Rs 121.

Page No 148:

Question 2:

Students of Class XI secured following marks in their Statistics paper.
Calculate arithmetic mean using Direct Method and Short-cut Method.
20, 44, 65, 28, 45, 67, 30, 50, 68, 39, 53, 70, 40, 60, 75

Answer:

Sr. No. Marks
(X)
Deviation from the Assumed mean
d = X − A where A = 50
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
20
44
65
28
45
67
30
     50 (A)
68
39
53
70
40
60
75
-30
-06
15
-22
-05
17
-20
0
18
-11
3
20
-10
10
25
N = 15 ΣX = 754 Σd = -104 + 108 = 4

where,
N= Number of items
ΣX= Total of monthly income of all the families
Σd= net sum of deviations

Calculation of Mean by direct method
X=ΣXNor, X¯=75415or, X¯=50.27X¯ =50.27

Calculation of mean by short cut method
X=A+ΣdNor, X=50+415or, X=50+0.27=50.27X=50.27

So, mean marks in statistics is 50.27.

Page No 148:

Question 3:

Find out mean height of girl-students in your school. Compare it with the mean height of boy-students.

Answer:

Sr. No. Height of girls
(in cm) (X)
Height of Boys
(in cm) (Y)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
152
153.2
151.4
152
156
154
152.7
158
152
153.5
170
172.5
168.2
175
172
170.3
176.5
178
180
172
N = 10 ΣX = 1534.8 ΣY = 1734.5

Mean height of girls X=ΣXN=1534.810 X¯=153.48 cmMean height of boyx Y=ΣYN=1734.510Y¯=173.45 cm

Boys' mean height is more than girls's mean height.



Page No 152:

Question 1:

Calculate arithmetic mean of the following series.
Take 15 as assumed average and use Short-cut Method.

Size 20 19 18 17 16 15 14 13 12 11
Frequency 1 2 4 8 11 10 7 4 2 1

Answer:

Size
(X)
Frequency
(f)
Deviation
d = X − A
(A = 15)
Multiple of Deviation and Frequency
(fd)
20
19
18
17
16
15
14
13
12
11
1
2
4
8
11
10
7
4
2
1
5
4
3
2
1
0
−1
−2
−3
−4
5
8
12
16
11
0
−7
−8
−6
−4
  Σf = 50   Σfd = 52 − 25 = 27


Mean size X=A+ΣfdΣfor, X¯=15+2750or, X¯=15+0.54X¯=15.54

So the average size is 15.54.

Page No 152:

Question 2:

Students of your class obtained following marks in Statistics in the weekly test:

Marks 4 5 6 7 8
Number of Students 12 10 18 20 24
Calculate arithmetic mean using Direct and Short-cut Methods.

Answer:

Marks
(X)
Frequency
(f)
fX Deviation
d = X − A
(A = 6)
Multiplication of deviation and frequency
(fd)
4
5
   6(A)
7
8
12
10
18
20
24
48
50
108
140
192
−2
−1
0
1
2
−24
−10
0
20
48
  Σf = 84 ΣfX = 538   Σfd = −34 + 68 = 34

Calculation of mean by Direct method
X=ΣfXΣfor, X¯=53884X¯=6.40

Calculation of mean by short cut method
X=A+ΣfdΣfor, X¯=6+3484or, X¯=6+0.40X¯=6.40

So, average statistics marks is 6.40.

Page No 152:

Question 3:

The following table gives height of certain persons:

Height (inches) 57 59 61 63 65 67 69 71 73
Number of Persons 1 3 7 10 11 15 10 7 2
Calculate arithmetic mean.

Answer:

Heights (inches)
(X)
No. of Persons
(f)
Multiplication of the value of X and frequency (fX)
57
59
61
63
65
67
69
71
73
1
3
7
10
11
15
10
7
2
57
177
427
630
715
1005
690
497
146
  Σf = 66 Σfx = 4344

Arithmetic mean will be:
X=ΣfxΣfor, X¯=434466X¯=65.82

So, the average height is 65.82 inches.

Page No 152:

Question 4:

Following is the number of heads of 8 coins when tossed 205 times:

Number of Heads 0 1 2 3 4 5 6 7 8
Frequency 2 19 46 62 47 20 5 2 2
Calculate arithmetic mean (mean number of heads per toss), using Direct Method and Short-cut Method.

Answer:

No. of Heads
(X)
Frequency
(f)
fX Deviation
d = X − A
(A = 4)
Multiplication of deviation and frequency (fd)
0
1
2
3
      4 (A)
5
6
7
8
2
19
46
62
47
20
5
2
2
0
19
92
186
188
100
30
14
16
−4
−3
−2
−1
0
1
2
3
4
−8
−57
−92
−62
0
20
10
6
8
  Σf = 205 Σfx = 645   Σfd = −219 + 44 = −175

Calculation of mean by direct method.
X=ΣfXΣfor, X¯ =645205or, X¯=3.146X¯=3.15 approx

Calculation mean by short cut method
X=A+ΣfdΣfor, X¯=4+-175205or, X¯=4-0.853or, X¯=3.147X¯=3.15 approx

So, mean number of heads per toss is 3.15.



Page No 155:

Question 1:

Find out the mean of the following distribution:

Items 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:

Items Mid value
m=l1+l22
Frequency
(f)
Multiplication of mid value and frequency (fm)
0 − 5 0+52=2.5 2 5
5 − 10 5+102=7.5 5 37.5
10 − 15 10+152=12.5 7 87.5
15 − 20 15+202=17.5 13 227.5
20 − 25 20+252=22.5 21 472.5
25 − 30 25+302=27.5 16 440
30 − 35 30+353=32.5 8 260
35 − 40 35+402=37.5 3 112.5
    Σf = 75 Σfm = 1642.5

X=ΣfmΣfor, X¯=1642.575X¯=21.9

Hence, the mean of the above distribution is 21.9

Page No 155:

Question 2:

Find arithmetic mean from the following distribution:

Marks 0−4 4−8 8−12 12−16
Number of Students 4 8 2 1

Answer:

Marks Mid value (m)
m=l1+l22
No. of student
(f)
Multiplication of mid-value and frequency
(fm)
0 − 4 0+42=2 4 8
4 − 8 4+82=6 8 48
8 − 12 8+122=10 2 20
12 − 16 12+162=14 1 14
    Σf = 15 Σfm = 90

X=ΣfmΣfor, X¯=9015X¯=6

Hence, the mean marks of the above distribution is 6.



Page No 160:

Question 1:

A polling agency interviewed 200 persons. The age distribution of those persons was recorded as under:

Age (Years) 80−89 70−79 60−69 50−59 40−49 30−39 20−29 10−19
Frequency 2 2 6 20 56 40 42 32
Calculate mean age of the persons by Direct Method.

Answer:

Age Mid Value (m)
m=l1+l22
Frequency
(f)
fm
10 − 19 10+192=14.5 32 464
20 − 29 20+292=24.5 42 1029
30 − 39 30+392=34.5 40 1380
40 − 49 40+492=44.5 56 2492
50 − 59 50+592=54.5 20 1090
60 − 69 60+692=64.5 6 387
70 − 79 70+792=74.5 2 149
80 − 89 80+892=84.5 2 169
    Σf = 200   Σfm = 7160

Calculating the mean by Direct method:

X=ΣfmΣfor, X¯=7160200X¯=35.8

Hence,mean age of the persons is 35.8 years.

Page No 160:

Question 2:

Calculate mean of the following series by Direct Method:

Mid-value 10 12 14 16 18 20
Frequency 5 8 12 20 10 5

Answer:

Mid Value
(m)
Frequency
(f)
fm
10
12
14
16
18
20
5
8
12
20
10
5
50
96
168
320
180
100
  Σf = 60 Σfm = 914

Calculating the mean by Direct method:

X=ΣfmΣfor, X¯=91460 X¯=15.23

Hence, the mean of the above series is 15.23.

Page No 160:

Question 3:

In an examination in Hindi, the students of a class secured following marks. Calculate mean marks by Direct Method.

Marks Number of Students
More than 0

More than 10

More than 20

More than 30

More than 40

More than 50
50

46

40

20

10

5

Answer:

Converting cumulative frequency Distribution into a simple frequency distribution
 

Marks No. of students
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
50 − 46 =  4
46 − 40 =  6
40 − 20 = 20
20 − 10 = 10
10 − 3   =  7
 3 − 0    =  3
 

 
Marks Mid Value
(m)
No. of students
(f)
Multiple of mid value and frequency (fm)
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
5
15
25
35
45
55
4
6
20
10
7
3
20
90
500
350
315
165
    Σf = 50 Σfm = 1440

Calculating the mean by Direct method:

X=ΣfmΣfor, X¯=144050X¯=28.8

Hence, the mean marks in Hindi is 28.8



Page No 161:

Question 1:

The following table shows prices per 100g of tea of different brands. Using quantities as weight, find out weighted arithmetic mean of the prices.

Prices per 100g in ₹ 2.50 3 3.50 4 4.25 5
Quantity 10 8 8 4 4 2

Answer:

Price
(X)
Quantity
(W)
WX
2.50
3
3.50
4
4.25
5
10
8
8
4
4
2
25
24
28
16
17
10
  ∑W = 36 ∑WX=120

Calculating the weighted arithmetic mean of the prices:

XW=ΣWXΣWor, X¯w=12036X¯w=3.33

Hence, the weighted arithmetic mean of the prices is 3.33.

Page No 161:

Question 2:

Calculate weighted mean of the following data:

Items (X) 5 10 15 20 25 30
Weight (W) 8 4 5 10 7 6

Answer:

Item
(X)
Weight
(W)
WX
5
10
25
20
25
30
8
4
5
10
7
6
40
40
125
200
175
180
  ΣW = 40 ΣWX = 760


Calculating the weighted mean:

XW=ΣWXΣWor, X¯w=76040X¯w=19

Hence, the weighted mean of the above data is 19



Page No 164:

Question 1:

The arithmetic mean of the following series is 18, find out the missing frequency.

Class Interval 11−13 13−15 15−17 17−19 19−21 21−23 23−25
Frequency 3 6 9 13 ? 5 4

Answer:

Class interval Mid value
(m)
Frequency
(f)
Multiple of mid value and frequency
(fm)
11 − 13
13 − 15
15 − 17
17 − 19
19 − 21
21 − 23
23 − 25
12
14
16
18
20
22
24
3
6
9
13
f
5
4
36
84
144
234
20f
110
96
    Σf = 40 + f Σfm = 704 + 20f

X=ΣfmΣfor, 18=704+20f40+for, 18 (40 +f) = 704 + 20for, 720 + 18f = 704 + 20for, 720 − 704 = 20f − 18for, 16 = 2for, 2f = 16or, f=162 f= 8

Hence, the missing frequency is 8.

Page No 164:

Question 2:

The following series gives the weekly wages. The mean wages are ₹ 50.75. Find out the missing value.

Weekly Wages 40−43 43−46 46−49 49−52 52−55
Frequency 3 6 9 13 ?

Answer:

Weekly wages Mid value
(m)
Frequency
(f)
Multiple of mid value and frequency (fm)
40 − 43
43 − 46
46 − 49
49 − 52
52 − 55
41.5
44.5
47.5
50.5
53.5
3
6
9
13
f
124.5
267
427.5
656.5
53.5f
    Σf = 31 + f Σfm = 1475.5 + 53.5f

X=ΣfmΣfor, 50.75=147.5+53.5f31+for, 50.75(31 +f)= 1475.5 + 53.5for, 1573.25 + 50.75f= 1475.5 + 53.5for, 1573.25 − 1475.5 =53.5f− 50.75for, 97.75=2.75for, 2.75f=97.75or, f=97.752.75 f=35.54 or 36 approx.

Hence, the missing frequency or value is 36.

Page No 164:

Question 3:

Mark the missing value of the following data. Mean marks are 11.66.

Weekly Wages 5 6 7 8 12 ? 15 17 18
Frequency 3 10 5 10 12 15 11 13 1

Answer:

Marks
(x)
Frequency
(f)
fx
5
6
7
8
12
x
15
17
18
3
10
5
10
12
15
11
13
1
15
60
35
80
144
15x
165
221
18
  Σf = 80 Σfx = 738 + 15x

X=ΣfxΣfor, 11.66=738+15x80or, 932.8=738+15xor, 932.8-738=15xor, 194.8=15xor, x=194.815x=12.99 or 13 approximately

So, the missing weekly wage is Rs 13.



Page No 176:

Question 1:

Eight workers earn the following income:
30, 36, 34, 40, 42, 46, 54, 62
Find out arithmetic mean.

Answer:

Arithmetic Mean = X=ΣXNor, X =30 + 36 +34 + 40 + 42 + 46 + 54 + 62 8or, X =3448=43Thus, the mean income of the workers is Rs 43.

Page No 176:

Question 2:

Pocket allowance of 5 students respectively are:
125, 75, 150, 175, 200
Find out arithmetic mean.

Answer:

Arithmetic Mean = X=ΣXNor, X = 125 + 75 + 150 + 175 + 2005 or, X =7255=145Thus, the mean pocket allowance is Rs 145. 

Page No 176:

Question 3:

Following is the height of 10 students:

Students A B C D E F G H I J
Height (cm) 155 153 168 160 162 166 164 180 157 165
Calculate arithmetic mean using DIrect and Short-cut Methods.

Answer:

Students Height
(X)
 X − A
(d)
A
B
C
D
E
F
G
H
I
J
155
153
168
160
      162 = A
166
164
180
157
165
−7
−9
6
−2
0
4
2
18
−5
3
  ∑ X = 1630 d = 10

Direct Method:
A.M. = X=ΣXN       =163010=163

Shortcut Method:
A.M.=X=A+ΣdNHere,A represents assumed meand represents the deviations of the values from the assumed mean

Here, we take 162 as the assumed mean. So, we take deviations of each item in the series from 162.
X=162+1010   =162+1=163

Thus, the mean height is 163 cm. 

Page No 176:

Question 4:

Weight of 15 persons is as follows:

Weight (kg) 20 28 34 39 42 50 53 54 59 64 72 74 74 78 79
Find out mean weight, using Direct Method as well as Short-cut Method.

Answer:

S. No. Weight
(X)
d = X − A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
20
28
34
39
42
       50 = A
53
54
59
64
72
74
74
78
79
−30
−22
−16
−11
−8
0
3
4
9
14
22
24
24
28
29
  X = 820 d = 70

Direct Method:
A.M. = X=ΣXN  =82015=54.67 

Shortcut Method:
A.M.= X=A +ΣdNHere, A represents the assumed mean d represents the deviations of the values from the assumed mean. 

Here, we take 50 as assumed mean. Thus, we take deviations of the values from 50.
X = 50+7015     =  50+4.67=54.67 

Thus, the mean weight is 54.67 kg.



Page No 177:

Question 5:

Calculate average of the following discrete series. Use Short-cut Method by taking 25 as assumed average.

Size 30 29 28 27 26 25 24 23 22 21
Frequency (f) 2 4 5 3 2 7 1 4 5 7

Answer:

Size
(X)
Frequency
(f)
Deviation
d = XA
fd
30
29
28
27
26
     25=A
24
23
22
21
2
4
5
3
2
7
1
4
5
7
5
4
3
2
1
0
−1
−2
−3
−4
10
16
15
6
2
0
−1
−8
−15
−28
  f = 40   fd = −3
A.M. = X=ΣXN       =163010=163
Shortcut Method:-
Here, we take 25 as the assumed mean, so we tale the deviation of each item in the series from 25. 
X=A+ΣfdΣf
=25 +-340= 25-0.075=24.925
Thus, the mean of the series is 24.925

Page No 177:

Question 6:

Marks secured by 42 students in Economics are:

Marks 15 20 22 23 27 35 18
Number of Students 8 4 7 3 8 7 5
Find average marks.

Answer:

Marks
(X)
No. of students
(f)
fX
15
20
22
23
27
35
18
8
4
7
3
8
7
5
120
80
154
69
216
245
90
  f = 42 fX = 974

Here, we use the direct method to find the mean of the given series.
A.M. = X=ΣfXΣfor, X =97442=23.20
Hence, the mean marks of the students is 23.20

Page No 177:

Question 7:

Average age of the people of a country is shown in the following table:

Age (Years) 10−20 20−30 30−40 40−50 50−60
People ('000) 30 32 15 12 9
Find out mean age by Direct Method.

Answer:

Age
(X)
Mid-Values
(m
People
(f)
fm
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
15
25
35
45
55
30
32
15
12
9
450
800
525
540
495
    f = 98 fm = 2810

Here, we use the direct method to compute the value of mean.
X=ΣfmΣfor, X=281098=28.7

Hence, the average age of the people in the country is 28.7 years.

Page No 177:

Question 8:

Calculate the arithmetic mean of the following frequency distribution by Direct Method:

Class Interval 10−20 20−40 40−70 70−120 120−200
Frequency 4 10 26 8 2

Answer:

Class Interval Mid-Value
(m)
Frequency
(f)
fm
10 − 20
20 − 40
40 − 70
70 − 120
120 − 200
15
30
55
95
160
4
10
26
8
2
60
300
1430
760
320
    f = 50 fm = 2870

Using the direct method, arithmetic mean is calculated using the following formula.
X=ΣfmΣfor, X=287050=57.4
Hence, the arithmetic mean of the series is 57.4

Page No 177:

Question 9:

Calculate arithmetic mean from the following data by Short-cut Method:

Class Interval 20−25 25−30 30−35 35−40 40−45 45−50 50−55
Frequency 10 12 8 20 11 4 5

Answer:

Class Interval Mid-Value
(m
Frequency
(f)
Deviation
 mA
(d)
fd
20 − 25
25 − 30
30 − 35
35 − 40
40 − 45
45 − 50
50 − 55
22.5
27.5
32.5
       37.5 = A
42.5
47.5
52.5
10
12
8
20
11
4
5
−15
−10
−5
0
5
10
15
−150
−120
−40
0
55
40
75
    f = 70   fd = −140

Here, the assumed mean is decided from the mid-points. Let us take 37.5 as the assumed mean. So, we calculate the deviations of the mid-values from 37.5
X=A+ΣfdΣfHere,A represents the assumed meand represents the deviation of the mid-values from the assumed mean X =37.5+-14070or, X  =37.5 -2 = 35.5

Hence, the mean of the series is 35.5

Page No 177:

Question 10:

Find out arithmetic mean from the following distribution by Short-cut Method:

Items 10−8 8−6 6−4 4−2 2−0
Frequency 10 8 6 4 2

Answer:

Items
(X)
Mid-Values
(m)
Frequency
(f)
Deviation
 mA
(d)
fd
0 − 2
2 − 4
4 − 6
6 − 8
8 − 10
1
3
5 = A
7
9
2
4
6
8
10
−4
−2
0
2
4
−8
−8
0
16
40
    f = 30   fd = 40

Here, the assumed mean is decided from the mid-points. Let us take 5 as the assumed mean. So, we calculate the deviations of the mid-values from 5
X=A+ΣfdΣfHere,A represents the assumed meand represents the deviations of the mid-values from the assumed mean=5+4030= 5+1.33 = 6.33

Hence, the mean of the given distribution is 6.33

Page No 177:

Question 11:

Sachin made the following runs in different matches:

Runs 5−15 15−25 25−35 35−45 45−55
Frequency 10 12 17 19 22
Calculate the average mean of the runs by Step-deviation Method.

Answer:

Run
(X)
Mid-Values
(m
Frequency
(f)
Deviation
 m − A
(d)
Step deviation
d'=m-Ai
fd'
5 − 15
15 − 25
25 − 35
35 − 45
45 − 55
10
20
30 = A
40
50
10
12
17
19
22
−20
−10
0
10
20
−2
−1
0
1
2
−20
−12
0
19
44
    f = 80     fd' = 31

Using the step-deviation method, arithmetic mean is calcualted using the following formula.X=A+Σfd'Σfd×iwhere,d'= m-AiSubstituting the calculated values in the formulaX=30+3180×10=30+3.875=33.875

Hence, the average runs scored by Sachin is 33.875

Page No 177:

Question 12:

Calculate arithmetic mean of the following frequency distribution:

Class less than 10 10−20 20−30 30−40 40−50 50−60 more than 60
Frequency 5 12 18 22 6 4 3

Answer:

It must be noted that the given distribution is an open ended distribution, that is, the first and the last class interval are not explicitly defined. As mean is based on all the items in the series, so for such distributions mean cannot be calculated accurately. ​  
However, going by the symmetry of the distribution we assume the first and the last class interval to be (0- 10) and (60 - 70), respectively

Class Interval Mid-Values
(m
Frequency
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
5
15
25
35
45
55
65
5
12
18
22
6
4
3
25
180
450
770
270
220
195
    f = 70 fm = 2110
Using the direct method to find the mean, we get
X=ΣfmΣf  =211070=30.14
Hence, the mean of the given distribution is 30.14

Page No 177:

Question 13:

Mean marks obtained by a student in his five subjects are 15. In English he secures 8 marks, in Economics 12, in Mathematics 18, and in Commerce 9. Find out the marks he secured in Statistics.

Answer:

Given:
Mean marks of five subjects = X = 28
Marks in English = 8
Marks in Economics = 12
Marks in Mathematics = 18
Marks in Commerce = 9

To find: Marks in Statistics
Let marks scored in statistics be S

Substituting the given values in the formula for mean
X=ΣXN15=S+8+12+18+95
15 × 5 = S + 47
75 − 47 = S
S = 28
Hence, marks scored in statistics is 28

Page No 177:

Question 14:

Mean value of the weekly income of 40 families is 265. But in the calculation, income of one family was read as 150 instead of 115. Find the "Corrected" mean.

Answer:

Given:
Mean ,X = 265
Number of observations, N=40
Incorrect observation = 150
Correct observation = 115
 
XWrong=ΣXNor, ΣX = XWrong ×N or, ΣX = 265 ×40 = 10,600

Thus, wrong summation X, i.e., ΣXwrong is 10,600. 
Now, correct mean is calculated using the following formula.
Correct X=ΣXwrong+correct value - incorrect valueNSubstituting the given values in the formula. Correct X=10600+115-15040               =1056540=264.125
Hence, the correct mean is 264.125



Page No 178:

Question 15:

Average pocket allowance of 6 students is ₹45. Of these, pocket allowance of 5 students is 20, 30, 22, 24 and 32 respectively. What is the pocket allowance of the sixth student?

Answer:

Given:
Average, X = 45
Number of students, N = 6

Let sixth student's allowance be Z

Substituting the given values in the formula for mean
X=ΣXN45 = 20+30+22+24+32+Z6
or, 45 × 6 = 128 + Z
or, 270 − 128 = Z
or, Z = 142
Hence, sixth student's allowance is Rs. 142

Page No 178:

Question 16:

The following table shows wages of the workers. Calculate the average wage of the workers.

Wages (₹) 10−19 20−29 30−39 40−49 50−59
Number of Workers 8 9 12 11 6

Answer:

Wages
(X)
Mid Value
(m
Frequency
(f)
Deviation
d = mA
d'=di fd'
10 − 19
20 − 29
30 − 39
40 − 49
50 − 59
14.5
24.5
     34.5 =A
44.5
54.5
8
9
12
11
6
−20
−10
0
10
20
−2
−1
0
1
2
−16
−9
0
11
12
    f = 46     fd = −2

We use the step-deviation method to find the value of mean.
X=A+Σfd'Σf×i=34.5+-246×10=34.07

Hence, the mean of the given distribution is 34.07

Page No 178:

Question 17:

Ten players of the Australian team made an average of 63 runs and ten players of the Indian team made an average of 77 runs. Calculate the average run made by both the teams.

Answer:

The information given in the question can be summarised as follows.

  Average Runs  Number of Players
Australian Team X1 = 63 N1 = 10
Indian Team X2 = 77 N1 = 10

The combined average is calculated using the following formula.
X12=N1X1+N2X2N1+N2Substituting the given values in the formula.X12=10×63+10×7710+10       =140020=70
Hence, the average scored by both the teams is 70.

Page No 178:

Question 18:

Average income of 50 families is ₹3,000. Of these average income of 12 families is ₹1,800. Find out the average income of the remaining families.

Answer:

The information given in the question can be summarised as follows.
Average income of all the 50 families = X12 = 3000

  Average income  Number of families
12 families   X1 = 1800 N1 = 12
38 families X2 = ? N1 = 38
 

Substituting the given values in the formula for combined mean
X12=N1X1+N2X2N1+N23000 =12×1800+38×X250or, 3000 × 50 = 21600 + 38 X2or, 150000 − 21600 = 38X2or, X2 = 3378.95

Hence, the  average income of remaining 38 families is Rs 3,378.95

Page No 178:

Question 19:

In the following frequency distribution, ifthe arithmetic mean is 45.6, find out missing frequency.

Wages (₹) 10−20 20−30 30−40 40−50 50−60 60−70 70−80
Number of Workers 5 6 7 X 4 3 9

Answer:

Wages
(X)
Mid Value
(m
No. of worker
(f)
fm
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
70 − 80
15
25
35
45
55
65
75
5
6
7
X
4
3
9
75
150
245
45
220
195
675
    f = 34 + X fm = 1560 + 45X

Substituting the values in the formula for mean.
X=ΣfmΣf45.6=1560 + 45X34 + Xor, 45.6 (34 + X) = 1560 + 45 Xor, 1550.4 + 45.6 X = 1560 + 45 Xor, 45.6 X − 45 X = 1560  1550.4or, 0.6X = 9.6or, X = 16

X=9.6.6
X = 16
Hence, the missing frequency is 16.

Page No 178:

Question 20:

Calculate the weighted mean of the following data:

Items 96 102 104 124 148 164
Weight 5 6 3 7 12 9

Answer:

Items
(X)
Weight
(W)
WX
96
102
104
124
148
164
5
6
3
7
12
9
480
612
312
868
1776
1476
  W = 42 WX = 5524

XW=ΣWXΣX     =552442=131.52
Hence, the weighted mean is 131.52

Page No 178:

Question 21:

A student obtained 60 marks in English, 75 in Hindi, 63 in Mathematics, 59 in Economics and 55 in Statistics. Calculate weighted mean of the marks if weights are respectively 2, 1, 5, 5 and 3.

Answer:

The information given in the question can be presented as follows.

Subject Marks
(X)
Weight
(W)
WX
English
Hindi
Mathematics
Economics
Statistics
60
75
63
59
55
2
1
5
5
3
120
75
315
295
165
    W = 16 WX = 970

XW=ΣWXΣW     =97016     =60.625
Hence, the weighted mean marks is 60.625

Page No 178:

Question 22:

A housewife uses 10 kg of wheat, 20 kg of Fuel, 5 kg of Sugar, and 2 kg of oil. Prices (per kg) of these items are ₹1.50, 50 paise, ₹ 2.80 and ₹10 respectively. Taking quantities used as weights find out the weighted arithmetic average of the prices.

Answer:

The information given in the question can be summarised as follows. 

Price
(X)
Quantity
(W)
WX
1.5
0.5
2.8
10
10
20
5
2
15
10
14
20
  W = 37 WX = 59

XW=ΣWXΣW     =5937=1.59
Hence, the weighted average of the prices is Rs 1.59

Page No 178:

Question 23:

Calculate weighted mean of the following data by using Direct and Short-cut Methods:

Items 81 76 74 58 70 73
Weight 2 3 6 7 3 7

Answer:

Items
(X)
Weight
(W)
WX Deviation
d = X − A
Wd
81
76
     74 = A
58
70
73
2
3
6
7
3
7
162
228
444
406
210
511
7
2
0
−16
−4
−1
14
6
0
−112
−12
−7
  W = 28 WX = 1961   Wd = −111

Direct Method
Xw=ΣWXΣW    =196128=70.04

Shortcut Method
Xw=A+ΣWdΣW    =74+-11128    = 74  3.964      = 70.04
Hence, the weighted mean is 70.04



View NCERT Solutions for all chapters of Class 11