Question

NCERT Solutions Class 11, Economics, Statistics for Economics, Chapter- 5, Measures of Central Tendency

For a comprehensive understanding of Class 11 Economics and to achieve excellence in board and competitive examinations, NCERT Solutions are crucial. These resources, developed by experts, focus on important chapter concepts and are tailored to the CBSE curriculum, providing vital assistance in your studies.

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Concepts covered in Class 11 Economics, Statistics for Economics, Chapter- 5 Measures of Central Tendency, are-

• Introduction
• Arithmetic Mean
• How Arithmetic Mean is Calculated
• Calculation of Arithmetic Mean for Grouped Data
• Discrete Series
• Continuous Series
• Weighted Arithmetic Mean
• Median
• Quartiles
• Percentiles
• Mode
• Computation of Mode

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Answer 1

NCERT Solutions Class 11, Economics, Statistics for Economics, Chapter- 5, Measures of Central Tendency

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

(i) Average size of readymade garments.
(ii) Average intelligence of students in a class.
(iii) Average production in a factory per shift.
(iv) Average wage in an industrial concern.
(v) When the sum of absolute deviations from average is least.
(vi) When quantities of the variable are in ratios.
(vii) In case of open-ended frequency distribution.

Solution:

(i) Mode: Average size of readymade garments should be the size for which demand is the maximum. As the modal value represents the value with the highest frequency should be taken as the average size to be produced.

(ii) Median: It is the middlemost term of the series. Therefore, the Median will be the best measure to calculate the average intelligence of students in a class as it will give the average intelligence such that there is an equal number of students above and below this average. It will not be affected by extreme values.

(iii) Arithmetic Mean: The average production in a factory per shift can best be calculated by arithmetic mean as it will acquires all types of fluctuations in production during the shifts.

(iv) Arithmetic Mean: Arithmetic mean will be the most suitable measure to calculate average wages. It is calculated by dividing the sum of the wages of all the workers by the total number of workers in the industrial concern. It gives a fair idea of the average wage bill taking into account all the workers.

(v) Arithmetic Mean: The algebraic sum of the deviations of values about the arithmetic mean is zero. Hence, when the sum of absolute deviations from the average is the least, then mean could be used to calculate the average.

(vi) Median: Median will be the most suitable measure when the variables are in ratios as it is least affected by the extreme values.

(vii) Median: Median is the most suitable measure because it can be easily computed even in an open-ended frequency distribution and will not get affected by extreme values.

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

Solution:

(b) median

Median is the most relevant average for qualitative measurement because it divides a series into two equal parts representing the average qualitative measure without being affected by extreme values.

(ii) Which average is affected most by the presence of extreme items?

(a) median
(b) mode
(c) arithmetic mean
(d) none of the above

Solution:

(c) arithmetic mean

Arithmetic mean is mostly affected by the presence of extreme items. It is easily affected by extreme values. The value of the arithmetic mean may not figure out at all in the series.

(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 these

Solution:

(b) 0

The algebraic sum of deviation of a set of n values from A.M. is zero. This is one of the mathematical properties of the arithmetic mean.

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

(i) The sum of deviation of items from median is zero.
(ii) An average alone is not enough to compare series.
(iii) Arithmetic mean is a positional value.
(iv) Upper quartile is the low est value of top 25% of items.
(v) Median is unduly affected by extreme observations.

Solution:

(i) False

The mathematical property stated here applies to the arithmetic mean and not to the median.

(ii) True

The average is not enough to compare the series as it does not explain the scope of deviation of different items from the central tendency and the difference in the frequency of values. These are measured by the measures of dispersion.

(iii) False

The given statement is false as arithmetic mean is not a positional value, but median and mode are because the calculation of them is based on the position of the items.

(iv) True

The upper quartile also called the third quartile has 75% of the items below it and 25% of the items above it.

(v) False

Median is a positional average as it is calculated by observation of a series and not by extreme values. Hence, arithmetic mean is unduly affected by extreme observations.

Answer 2

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)	Number of retail shops
0 – 10	12
10 – 20	18
20 – 30	27
30 – 40	-
40 – 50	17
50 – 60	6

Solution:

(a) The missing frequency

Profit per Retail Shop (in ₹)	No. of Shops (f)	Mid-value (m)	fm
0 – 10	12	5	60
10 – 20	18	15	270
20 – 30	27	25	675
30 – 40	f1	35	35f1
40 – 50	17	45	765
50 – 60	6	55	330
Σf = 80 + f1			2100 + 35f1

Arithmetic mean(\(\bar X\))  = 28 (given)

\((\bar X) = \frac{\sum fm}{\sum f}\)

\(28 = \frac{2100+35f_1}{80 + f_1}\)

28(80 + f₁) = 2100 + 35f₁

2240 + 28f₁ = 2100 + 35f₁

2240 – 2100 = 35f₁ – 28f₁

140 = 7f₁

20 = f₁ 

Hence, the missing frequency is 20.

(b) The median of the series

CI	Frequency (f)	Cumulative Frequency (cf)
0 – 10	12	12
10 – 20	18	30
20 – 30	27	57
30 – 40	20	77
40 – 50	17	94
50 – 60	6	100
	Σf = 100

 \(M= L+\frac{\frac N2 - cf}{f}\times i\)

\(= 20+\frac{\frac{100}2 - 30}{27}\times10\)

\(= 20+\frac{50-30}{27} \times10\)

\(M = 20 + \frac{20}{27} \times 10\)

\(M = 20 + 7.40\)

\(M =27.40\)

5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	Daily Income (in ₹)
A	120
B	150
C	180
D	200
E	250
F	300
G	220
H	350
I	370
J	260

Solution:

Workers	Daily Income (in ₹)
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

\(\bar X = \frac{\sum X}N\)

\(\bar X = \frac{2400}{10} = 240\)

Arithmetic mean = ₹240

Answer 3

6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Income (in ₹)	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

Solution:

C.I.	cf	f	m	fm
75 – 85	150	10	80	800
85 – 95	140	25	90	2250
95 – 105	115	20	100	2000
105 – 115	95	25	110	2750
115 – 125	70	10	120	1200
125 – 135	60	20	130	2600
135 – 145	40	15	140	2100
145 – 155	25	25	150	3750
		150		17450

\(\bar X = \frac{\sum fm}{\sum f}\) 

\(\bar X = \frac{17450}{150}\)

\(\bar X = 116.33\)

Arithmetic mean = ₹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)	Number of families
Less than 100	40
100 – 200	89
200 – 300	148
300 – 400	64
400 and above	39

Solution:

Size of Land Holdings C.I.	Number of families (f)	Cumulative Frequency (c.f.)
0 – 100	40	40
100 – 200	89	129
200 – 300	148	277
300 – 400	64	341
400 – 500	39	380
	Σf = 380

Σf = N = 380

M class = Size of (N/2)^th item

Median class = 190^th item

Hence, median class interval is 200 – 300.

\(M= L_1 + \frac{\frac N2 + cf}f\times i\)

\(= 200+\frac{\frac{380}2-129}{148}\times 100\)

\(M = 200+\left(\frac{190-129}{148}\right)\times 100\)

\(M = 200 + 41.22\)

\(M = 2 41.22\) 

Answer 4

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.

[Hint: Compute median, low er quartile and uapper quartile.]

Daily Income (in ₹)	10-14	15-19	20-24	25-29	30-34	35-39
Number of workers	5	10	15	20	10	5

Solution:

Daily Income (in `)	Exclusive Group	Number of workers (f)	Cumulative frequency (cf)
10 – 14	9.5 – 14.5	5	5
15 – 19	14.5 – 19.5	10	15
20 – 24	19.5 – 24.5	15	30
25 – 29	24.5 – 29.5	20	50
30 – 34	29.5 – 34.5	10	60
35 – 39	34.5 – 39.5	5	65
	n = Σf = 65

(a) Highest income of lowest 50% workers will be given by the median.

Σf = N = 65

Median class = \(\frac n2\) th item = \(\frac{65}2 = 32.5\)

\(M = L_1 + \left(\frac{\frac N2 - cf}f \right) \times i\)

\( =24.5 + \left(\frac{32.5 - 30}{20} \right) \times 5\)

\( =24.5 + 0.625\)

\( =25.125\)

(b) Minimum income earned by the lowest 25% workers will be given by the Quartile (Q₁).

Class interval of Q₁ ​= \(\frac n2\) th item

Q₁ ​= \(\frac{65}4\) th item = 16.25^th item

Quartile class = 19.5 – 24.5

\(Q_1 = L_1 + \left(\frac{\frac N2 - cf}f \right) \times i\)

\( =19.5 + \left(\frac{16.25 - 15}{15} \right) \times 5\)

\( =19.5 + \frac{1.25}{15} \times 5\)

\( =19.5 +0.42\)

\( =19.92\)

(c) Maximum income earned by the top 25% workers will be given by the upper quartiler Q₃.

Class interval of Q₃ = \(3(\frac n2)\) th item

\(= 3(\frac{65}4)\) th itema

= 3 x 16.25^th item

= 48.75^th item

It lies in 24.5 – 29.5

Quartile class = 19.5 – 24.5

\(Q_3= L_1 + \left(\frac{\frac N2 - cf}f \right) \times i\)

\( =24.5 + \left(\frac{48.75 - 30}{20} \right) \times 5\)

\( =24.5 +\frac{18.75}{20} \times 5\)

\(= 24.5 +4.69\)

\(= 29.19\)

Answer 5

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)	Number of farms
50 – 53	3
53 – 56	8
56 – 59	14
59 – 62	30
62 – 65	36
65 – 68	28
68 – 71	16
71 – 74	10
74 – 77	5

Solution:

(i) Mean

(Production Yield) C.I.	Number of Farms (f)	Mid-value (m)	fm
50 – 53	3	51.5	154.5
53 – 56	8	54.5	436.0
56 – 59	14	57.5	805.0
59 – 62	30	60.5	1815.0
62 – 65	36	63.5	2286.0
65 – 68	28	66.5	1862.0
68 – 71	16	69.5	1112.0
71 – 74	10	72.5	725.0
74 – 77	5	75.5	377.5
	Σf = 150	Σfm = 9573.0

(ii) Median 

CI	f	cf
50 – 53	3	3
53 – 56	8	11
56 – 59	14	25
59 – 62	30	55
62 – 65	36	91
65 – 68	28	119
68 – 71	16	135
71 – 74	10	145
74 – 77	5	150

(iii) Mode

Mode = 3 Median – 2 Mean

= 3 × 63.67 – 2 × 63.82

= 191.01 – 127.64

= 63.37.