NCERT Solutions Class 11 Economics Chapter 5 Measures of Central Tendency have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 11 Economics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 11 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 11 Economics are an important part of exams for Class 11 Economics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 11 Economics and also download more latest study material for all subjects. Chapter 5 Measures of Central Tendency is an important topic in Class 11, please refer to answers provided below to help you score better in exams
Chapter 5 Measures of Central Tendency Class 11 Economics NCERT Solutions
Class 11 Economics students should refer to the following NCERT questions with answers for Chapter 5 Measures of Central Tendency in Class 11. These NCERT Solutions with answers for Class 11 Economics will come in exams and help you to score good marks
Chapter 5 Measures of Central Tendency NCERT Solutions Class 11 Economics
NCERT Solution for Class 11 Statistics for chapter 5 Measures of Central Tendency
Exercises
Q1. Which average would be suitable in 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.
Answer.
(i) Average size of readymade garments. Mode
Explanation: Mode is suitable average for average size of readymade garments because it gives the most frequent occurring value.
(ii) Average intelligence of students in a class. Median
Explanation: Median is a suitable average in case of a qualitative nature of the data.
(iii) Average production in a factory per shift. Mean
Explanation: Production can be measured on a quantitative scale so Arithmetic mean is suitable in this case.
(iv) Average wage in an industrial concern. Mean
Explanation: Wage can be measured on a quantitative scale so arithmetic mean is suitable in this case.
(v) When the sum of absolute deviations from average is least. Mean
Explanation: Mean shall be used because sum of deviations from mean is always zero or least than the other averages.
(vi) When quantities of the variable are in ratios. Mean
Explanation: Ratios are quantitative, so it is suitable to use arithmetic mean.
(vii) In case of open-ended frequency distribution. Median
Explanation: Median is used because there is no need to adjust class size or magnitude for using median.
Q2. 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
(ii) Which average is affected most by the presence of extreme items?
(a) Median
(b) Mode
(c) Arithmetic mean
(d) None of the above
(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
Answer.
(i) The most suitable average for qualitative measurement is Median.
(ii) Arithmetic mean is the average affected by the presence of the extreme values.
(iii) 0 is the sum of deviations of a set of n values from AM.
Q3. 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 lowest value of top 25% of items.
(v) Median is unduly affected by extreme observations.
Answer.
(i) The sum of deviation of items from median is zero. False
Explanation: Generally, sum of deviations from mean is zero; but only in the case of symmetric distribution (mean=median=mode) above statement is true.
(ii) An average alone is not enough to compare series. True
Explanation: Averages are very rigid values, they don’t say anything about the variability of the series, and thus they are not enough to compare series.
(iii) Arithmetic mean is a positional value. False
Explanation: Arithmetic mean is not a positional value because it is calculated on the basis of all the observations.
(iv) Upper quartile is the lowest value of top 25% of items. True
Explanation: Quartile refers to a quarter, so when the frequency is arranged in a ascending order the upper quartile refers to the first 25% of the items.
(v) Median is unduly affected by extreme observations. False
Explanation: Median doesn’t get affected by extreme observations because it only takes the median class to calculate it. It is mean which gets affected by extreme observations.
Q4. 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 us take the missing frequency as x
Profit per retail shop (in Rs .) (X) |
Number of retail shops (f) |
Mid values (m) |
fm |
0-10 10-20 20-30 30-40 40-50 50-60 |
12 18 27 x 6 |
5 15 25 35 45 55 |
60 270 675 35x 765 330 |
∑f = 80 + x |
∑Fm = 2100 + 35x |
Mean = ∑fm/∑f Mean = 28
Substituting the values in the formula we get,
28 = 2100 + 35x
80 + x
28 x (80 + x) = 2100 + 35x
2240 + 28x = 2100 + 35x
7x = 140 x = 20
Thus, the missing value frequency is 20.
(b)
|
Formula of median is as follow: Median
= L + (N/2 – c.f) x h
f
By substituting the value in the formula we get,
Median = 20 + (50 – 30) x (30-20)
27
Median = 20 + (20) x (10) = 20 + 200 = 20 + 7.41 = 27.41
27 27
Thus the median value of the series is 27.41.
Q5. 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.
Formula of mean is as follow:
Mean = Sum of all the observations
No. of observations
= 120+150+180+200+250+300+220+350+370+260
10
= 2400 = 240
10
Thus, average income of the workers is Rs 240.
Q6. 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 (in Rs.) |
Number of families f) |
Mid values (x) |
fx |
75-85 85-95 95-105 105-115 115-125 125-135 135-145 145-155 |
10 25 20 25 10 20 15 25 |
80 90 100 110 120 130 140 150 |
800 2250 2000 2750 1200 2600 2100 3750 |
∑f = 150 |
∑fx = 17450 |
Formula of mean is as follow: |
||
Mean = Sum of all the observations |
= |
∑fx |
= 17450/150 = 116.33
Thus, the average mean income for 150 families is Rs . 116.33.
|
Q7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
Answer.
To calculate the mean size of holding, first calculate the cumulative frequency.
Size of land holdings (in acres) (X) |
Number of families (f) |
Cumulative frequency (cf) |
0-100 100-200 200-300 300-400 400-500 |
40 89 148 64 39 |
40 129 277 341 380 |
∑f =380 |
Then, find the median frequency
Median frequency = N/2 = 380/2 = 190
Formula of median is as follow: Median = L + (N/2 – c.f) x h
f
By substituting the value in the formula we get,
= 200 + (190 – 129) x 100 = 241.21
148
Thus, the median size of land holding is 241.21 acres.
Q8. The following series relates to the daily income of workers employed in affirm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by 25% workers.
Daily Income (in Rs ) |
10-14 |
15-19 |
20-24 |
25-29 |
30-34 |
35-39 |
Numbers of workers |
5 |
10 |
15 |
20 |
10 |
5 |
Answer.
Daily income(in Rs ) |
Class interval (X) |
Number of workers (f) |
Cumulative frequency (cf) |
10-14 15-19 20-24 25-29 30-34 35-39 |
9.5-14.5 14.5-19.5 19.5-24.5 24.5-29.5 29.5-34.5 34.5-39.5 |
5 10 15 20 10 5 |
5 15 30 50 60 65 |
∑f = 65 |
a) To compute highest income of lowest 50% we need to calculate median
Median frequency = N/2 = 65/2 = 32.5
Formula for median is s follows
Median = L + (N/2 – c.f) x h
f
By substituting the values in the formula, we get
Median = 24.5 + (32.5 –30) x 5
20
= 24.5 + 0.625 = 25.125
Highest income of lowest 50% workers is Rs 25.125.
b) For minimum income of 25% of workers we need to calculate Q1.
First Quartile frequency = D/4 = 65/4 = 16.25
Q1 = L + (N/4 – c.f) x h
f
By substituting the values in the formula, we get
Q1 = 19.5 + (16.25 – 15) x 5 = 19.9166
15
Minimum income earned by the top 25% workers is Rs 19.92.
c) For maximum income of 25% of workers we need to calculate Q3.
Third quartile frequency = 3(D/4) = 3 (65/4)= 48.75
Q3 = L + (3(N/4) – c.f) x h
f
By substituting the values in the formula, we get
Q3 = 24.5 + (48.75 – 30) x 5
20
= 24.5 + 4.6875 = 29.1875
Maximum income earned by 25% workers is Rs 29.19.
Q9. 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 farms |
3 |
8 |
14 |
30 |
36 |
28 |
16 |
10 |
5 |
Answer.
To calculate mean, median and mode values
Production yield (kg per hectare) (X) |
Number of farms (f) |
Mid values (m) |
Cumulative Frequency (cf) |
fm |
50-53 53-56 56-59 59-62 62-65 65-68 68-71 71-74 74-77 |
3 8 14 30 36 28 16 10 5 |
51.5 54.5 57.5 60.5 63.5 66.5 69.5 72.5 75.5 |
3 11 25 55 91 119 135 145 150 |
154.5 436 805 1815 2286 1862 1112 725 377.5 |
∑f =150 |
∑fm= 9573 |
Formula of mean is as follows:
Mean = ∑fm = 9573/150 = 63.82 kg/hectare
∑f
Formula of median is as follows: Median = L + (N/2 – cf) x h
f
By substituting the value in the formula we get,
= 62 + (75 – 55) x 3
36
= 62 + 1.67 = 63.67 kg/hectare
Formula of mode is as follows: Mode = L + d1 x h
d1 + d2
By substituting the value in the formula we get,
= 62 + 6/6+8 x 3 = 62 + 18/14 = 62 + 1.28
= 63.28 kg/hectare
Mean, median and mode values are 63.82 kg/hectare, 63.67 kg/hectare and 63.28 kg/hectare.
NCERT Solutions Class 11 Economics Chapter 1 Indian Economy on the Eve of Independence |
NCERT Solutions Class 11 Economics Chapter 2 Indian Economy 1950 1990 |
NCERT Solutions Class 11 Economics Chapter 3 Liberalisation Privatisation And Globalisation An Appraisal |
NCERT Solutions Class 11 Economics Chapter 4 Poverty |
NCERT Solutions Class 11 Economics Chapter 5 Human Capital Formation in India |
NCERT Solutions Class 11 Economics Chapter 6 Rural Development |
NCERT Solutions Class 11 Economics Chapter 7 Employment Growth Informalisation and other Issues |
NCERT Solutions Class 11 Economics Chapter 8 Infrastructure |
NCERT Solutions Class 11 Economics Chapter 9 Environment and Sustainable Development |
NCERT Solutions Class 11 Economics Chapter 10 Comparative Development Experiences of India and Its Neighbors |
NCERT Solutions Class 11 Economics Chapter 1 Introduction |
NCERT Solutions Class 11 Economics Chapter 2 Collection of Data |
NCERT Solutions Class 11 Economics Chapter 3 Organisation of Data |
NCERT Solutions Class 11 Economics Chapter 4 Presentation of Data |
NCERT Solutions Class 11 Economics Chapter 5 Measures of Central Tendency |
NCERT Solutions Class 11 Economics Chapter 6 Measures of Dispersion |
NCERT Solutions Class 11 Economics Chapter 7 Correlation |
NCERT Solutions Class 11 Economics Chapter 8 Index Numbers |
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