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NCERT Solutions for Class 11th: Ch 5 Measures of Central Tendency Statistics for Economics

Page No: 71

Exercises

1. Which average would be suitable in the following cases?

(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 open-ended 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

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

(iii) The algebraic sum of deviation of a set of n values from A.M. is
(a) n
(b) 0
(c) 1
(d) none of the above
► (b) 0

3. Comment whether the following statements are true or false.

(i) The sum of deviation of items from median is zero.
► False

(ii) An average alone is not enough to compare series.
► True

(iii) Arithmetic mean is a positional value.
► False

(iv) Upper quartile is the lowest value of top 25% of items.
► True

(v) Median is unduly affected by extreme observations.
► False

4. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:
Profit per retail shop (in Rs) 0-10 10-20 20-30 30-40 40-50 50-60
Number of retail shops
12
18
27
-
17
6

Answer

(a) Let the missing frequency be x
Arithmetic mean = 28 (given)

Profit per retail shop (in Rs)
Class Interval
 No. of retail shops
Frequency (f)
 Mid Value
(m)
 fm
0-10
12
5
 60
10-20
18
15
 270
20-30
27
25
 675
30-40
x
35
 35x
40-50
17
45
 765
50-60
6
55
 330

 Î£f = 80 + x
Σfx = 2100 + 35x

Mean = Î£fx/Σf
⇒ 28 = 2100 + 35x/80 + x
⇒ 2240 + 28x = 2100 + 35
⇒ 2240 - 2100 = 35x - 25x
⇒ 140 = 7x
⇒  x = 140/7 = 20
Missing frequency = 20

(b)
Class Interval Frequency (f) Cumulative frequency
(CF)
0-10
12
 12
10-20
18
 30
20-30
27
 57
30-40
x
 77
40-50
17
 94
50-60
6
 100
Total
  Î£f = 100

Median = Size of (N/2)th item
             = 100/2 = 50th item
It lies in class 20-30.


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
  Î£X = 2400

N = 10
Arithmetic Mean = Î£X/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
75-85
10
80
 800
85-95
25
90
 2250
95-105
20
100
 2000
105-115
25
110
 2750
115-125
10
120
 1200
125-135
20
130
 2600
135-145 15
140
 2100
145-155 25
150
 3750

150
17450
Arithmetic Mean = Î£fx/Σf
                            = 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
100-200
200-300
300-400
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)
0-100
40
 40
100-200
89
 129
200-300
148
 277
300-400
64
 341
400-500
39
 380
Total
  Î£f = 380
Σf = N = 380
Median = Size of (N/2)th item
             = 380/2 = 190th item
It lies in class 200-300.

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)
10-14
15-19
20-24
25-29
30-34
35-39
Number of workers
5
10
15
20
10
5
(Hint: Compute median, lower quartile and upper quartile.)

Answer

Daily Income (in Rs)
      Class Interval
No of workers
(f)
 Cumulative frequency
(CF)
9.5-14.5
5
 5
14.5-19.5
10
 15
19.5-24.5
15
 30
24.5-29.5
20
 50
29.5-34.5
10
 60
34.5-39.5
5
 65
Total
  Î£f = 65

(a) Î£f = N = 65
Median = Size of (N/2)th item
             = 65/2 = 32.5th item
It lies in class 24.5-29.5.

Highest income of lowest 50% workers = Rs 25.12

(b) First, we need to find Q1
Class interval of Q1 = (N/4)th items
                                 = (65/4)th items = 16.25th item
It lies in class 19.5-24.5.

Minimum income earned by the top 25% workers = Rs 19.92

(c) First, we need to find Q3
Class interval of Q3 = 3 (N/4)th items
                                 = 3 (65/4)th items = 3 × 16.25th item
                                 = 48.75th item
It lies in class 24.5-29.5.

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)
50-53
53-56
56-59
59-62
62-65
65-68
68-71
71-74
74-77
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)
50-53
3
51.5
 154.5
3
53-56
8
54.5
 436
11
56-59
14
57.5
 805
25
59-62
30
60.5
 1815
55
62-65
36
63.5
 2286
91
65-68
28
66.5
 1862
119
68-71
 16
69.5
 1112
135
71-74
 10
72.5
 725
145
74-77
 5
75.5
 377.5
150

Σf = 150
Σfx = 9573


Mean = Î£fx/Σf  = 9573/150 = 63.82 hectare

Modal Class = 62-65


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