NCERT Solutions for Class 11th: Ch 5 Measures of Central Tendency Statistics for Economics
Page No: 71
Exercises
(i) Average size of readymade garments.
► Mode
(ii) Average intelligence of students in a class.
► Median
(iii) Average production in a factory per shift.
► Arithmetic average
(iv) Average wages in an industrial concern.
► Arithmetic average
(v) When the sum of absolute deviations from average is least.
► Median
(vi) When quantities of the variable are in ratios.
► Arithmetic average
(vii) In case of openended frequency distribution.
► Median
2. Indicate the most appropriate alternative from the multiple choices provided against each question.
(i) The most suitable average for qualitative measurement is
(a) arithmetic mean
(b) median
(c) mode
(d) geometric mean
(e) none of the above
(b) median
(c) mode
(d) geometric mean
(e) none of the above
► (b) median
Page No: 72
(ii) Which average is affected most by the presence of extreme items?
(a) median(b) mode
(c) arithmetic mean
(d) none of the above
► (c) arithmetic mean
(a) n
(b) 0
(c) 1
(d) none of the above
► (b) 0
(i) The sum of deviation of items from median is zero.
► False
(ii) An average alone is not enough to compare series.
(ii) An average alone is not enough to compare series.
► True
(iii) Arithmetic mean is a positional value.
(iii) Arithmetic mean is a positional value.
► False
(iv) Upper quartile is the lowest value of top 25% of items.
(iv) Upper quartile is the lowest value of top 25% of items.
► True
(v) Median is unduly affected by extreme observations.
(v) Median is unduly affected by extreme observations.
► False
Profit per retail shop (in Rs)  010  1020  2030  3040  4050  5060 
Number of retail shops 
12

18

27



17

6

Answer
(a) Let the missing frequency be x
Arithmetic mean = 28 (given)
Mean = Σfx/Σf
⇒ 28 = 2100 + 35x/80 + x
⇒ 2240 + 28x = 2100 + 35Arithmetic mean = 28 (given)
Profit per retail shop (in Rs)
Class Interval

No. of retail shops
Frequency (f)

Mid Value
(m)

fm 
010

12

5

60 
1020

18

15

270 
2030

27

25

675 
3040

x

35

35x 
4050

17

45

765 
5060

6

55

330 
Σf = 80 + x  Σfx = 2100 + 35x 
Mean = Σfx/Σf
⇒ 28 = 2100 + 35x/80 + x
⇒ 2240  2100 = 35x  25x
⇒ 140 = 7x
⇒ x = 140/7 = 20
Missing frequency = 20
(b)
Median = Size of (N/2)th item
= 100/2 = 50th item
It lies in class 2030.
5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
Class Interval  Frequency (f)  Cumulative frequency
(CF)

010

12

12 
1020

18

30 
2030

27

57 
3040

x

77 
4050

17

94 
5060

6

100 
Total

Σf = 100 
Median = Size of (N/2)th item
= 100/2 = 50th item
It lies in class 2030.
5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
Workers 
A

B

C

D

E

F

G

H

I

J

Daily Income (in Rs) 
120

150

180

200

250

300

220

350

370

260

Answer
Workers 
Daily Income (in Rs)
X

A

120

B

150

C

180

D

200

E

250

F

300

G

220 
H

350 
I

370 
J

260 
Total

ΣX = 2400 
N = 10
Arithmetic Mean = ΣX/N
= 2400/10
= 240
Arithmetic Mean = 240
6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
Income (in Rs) 
Number of families

More than 75

150

More than 85

140

More than 95

115

More than 105

95

More than 115

70

More than 125

60

More than 135

40 
More than 145

25 
Answer
Income  No. of families
Frequency (f)

Mid Class
(x)

fx 
7585

10

80

800 
8595

25

90

2250 
95105

20

100

2000 
105115

25

110

2750 
115125

10

120

1200 
125135

20

130

2600 
135145  15 
140

2100 
145155  25 
150

3750 
150  17450 
= 17450/150
= 116.33
7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
Size of Land Holdings (in acres) 
Less than 100

100200

200300

300400

400 and above

Number of families 
40

89

148

64

39

Answer
Size of Land Holdings
Class Interval 
Number of families
(f)

Cumulative frequency
(CF)

0100

40

40 
100200

89

129 
200300

148

277 
300400

64

341 
400500

39

380 
Total

Σf = 380 
Median = Size of (N/2)th item
= 380/2 = 190th item
It lies in class 200300.
Median size of land holdings = 241.22 acres
8. The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.
Daily Income (in Rs) 
1014

1519

2024

2529

3034

3539

Number of workers 
5

10

15

20

10

5

Answer
Daily Income (in Rs) Class Interval 
No of workers
(f)

Cumulative frequency
(CF)

9.514.5

5

5 
14.519.5

10

15 
19.524.5

15

30 
24.529.5

20

50 
29.534.5

10

60 
34.539.5

5

65 
Total

Σf = 65 
(a) Σf = N = 65
Median = Size of (N/2)th item
= 65/2 = 32.5th item
It lies in class 24.529.5.
Highest income of lowest 50% workers = Rs 25.12
(b) First, we need to find Q_{1}
Class interval of Q_{1} = (N/4)th items
= (65/4)th items = 16.25th item
It lies in class 19.524.5.
Minimum income earned by the top 25% workers = Rs 19.92
(c) First, we need to find Q_{3}
Class interval of Q_{3 }= 3 (N/4)th items
= 3 (65/4)th items = 3 × 16.25th item
= 48.75th item
It lies in class 24.529.5.
Maximum income earned by lowest 25% workers = Rs 29.19
9. The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.
Production yield (kg. per hectare) 
5053

5356

5659

5962

6265

6568

6871

7174

7477

Number of workers 
3

8

14

30

36

28

16

10

5

Answer
Production Yield (kg. per hectare) 
No. of farms
Frequency (f)

Mid Class
(x)

fx

Cumulative frequency
(CF)

5053

3

51.5

154.5 
3

5356

8

54.5

436 
11

5659

14

57.5

805 
25

5962

30

60.5

1815 
55

6265

36

63.5

2286 
91

6568

28

66.5

1862 
119

6871

16 
69.5

1112 
135

7174

10 
72.5

725 
145

7477

5 
75.5

377.5 
150

Σf = 150  Σfx = 9573 
Mean = Σfx/Σf = 9573/150 = 63.82 hectare
Modal Class = 6265
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