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NCERT Class 11 Economics Chapter 6 Measures of Dispersion
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Measures of Dispersion
Chapter: 6
PART – (A) STATISTICS FOR ECONOMICS
TEXTUAL QUESTION ANSWERS
1. A measure of dispersion is a good supplement to the central value in understanding a frequency distribution. Comment.
Ans: In order to understand the frequency distribution fully, it is essential to study the variability of the observations. Measures of dispersion enhance the understanding of a distribution considerably by providing information about how much the actual value of items in a series deviate from the central value, e.g., per capita income gives only the average income but a measure of dispersion can tell you about income inequalities, thereby improving the understanding of the relative living standards of different sections of the society. The average measures the centre of the data whereas the quantum of the variation is measured by the measures of dispersion like range, quartile deviation, mean deviation, and Standard Deviation.
2. Which measure of dispersion is the best and how?
Ans: Standard deviation is considered to be the best measure of dispersion and is therefore, the most widely used measure of dispersion:
(i) It is based on all values and thus provides information about the complete series. Because of this reason, a change in even one value affects the value of the standard deviation.
(ii) It is independent of origin but not of scale.
(iii) It is useful in advanced statistical calculation like the comparison of variability in two data sets.
(iv) It can be used in testing hypotheses.
(v) It is capable of further algebraic treatment.
3. Some measures of dispersion depend upon the spread of values whereas some are estimated on the basis of the variation of values from a central value. Do you agree?
Ans: Yes, it is true that some measures of dispersion depend upon the spread of values, whereas some calculate the variation of values from the central value. Range and Quartile Deviation measure the dispersion by calculating the spread within which the value lies. Mean Deviation and Standard Deviation calculate the extent to which the values differ from the average or the central value.
4. In a town, 25% of the persons earned more than ₹45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion.
Ans: 25% of the persons earned more than ₹45,000, which implies that upper quartile (Q₃) = 45,000
75% earned more than 18,000, which implies that lower quartile (Q₁) = 18,000
Absolute Measure of Dispersion
= Q₃- Q₁
= 45,000-18,000
= 27,000
Relative Measure of Dispersion, i.e., Coefficient of Quartile Deviation
5. The yield of wheat and rice per acre for 10 districts of a state is as under:
| District | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Wheat | 12 | 10 | 15 | 19 | 21 | 16 | 18 | 9 | 25 | 10 |
| Rice | 22 | 29 | 12 | 23 | 18 | 15 | 12 | 34 | 18 | 12 |
Calculate for each crop:
(a) Range
(b) Q.D.
(c) Mean deviation about Mean
(d) Mean deviation about Median
(e) Standard deviation
(f) Which crop has greater variation?
(g) Compare the values of different measures for each crop.
Ans: WHEAT:
(a) Range = L – S
= 25-9
= 16
(b) Quartile Deviation (Q.D.)
Data in ascending order
9, 10, 10, 12, 15, 16, 18, 19, 21, 25
= 10 + 0.75(10 – 10)
= 10.75
= 19 + 0.25(21 – 19)
= 19.50
(c) Mean Deviation about mean
(d) Mean Deviation about median
| X | | dₘ | = | X – M | |
| 9 | 6.5 |
| 10 | 5.5 |
| 10 | 5.5 |
| 12 | 3.5 |
| 15 | 0.5 |
| 16 | 0.5 |
| 18 | 2.5 |
| 19 | 3.5 |
| 21 | 5.5 |
| 25 | 9.5 |
| ∑ X = 155 N = 10 | ∑ | dₘ | = 43 |
= 5.5ᵗʰ item
(e) Standard Deviation
| X | X² |
| 9 | 81 |
| 10 | 100 |
| 10 | 100 |
| 12 | 144 |
| 15 | 255 |
| 16 | 256 |
| 18 | 324 |
| 19 | 361 |
| 21 | 441 |
| 25 | 625 |
| ∑ X = 155 N = 10 | ∑ X² = 2657 |
RICE:
(a) Range = L – S = 34 – 12 = 22
(b) Quartile Deviation Q.D
Data in ascending order
12, 12, 12, 15, 18, 18, 22, 23, 29, 34
= 12 + 0. 75 (12 – 12) = 12
= 23 + 0.25 ( 29 – 23 ) = 24.5
(c) Mean Deviation about mean
(d) Mean Deviation about median
| X | | dₘ | = | X – M | |
| 12 | 6 |
| 12 | 6 |
| 12 | 6 |
| 15 | 3 |
| 18 | 0 |
| 18 | 0 |
| 22 | 4 |
| 23 | 5 |
| 29 | 11 |
| 34 | 16 |
| ∑X = 195 N = 10 | ∑ | dₘ | = 57 |
(e) Standard Deviation
| X | X² |
| 12 | 144 |
| 12 | 144 |
| 12 | 144 |
| 15 | 225 |
| 18 | 324 |
| 18 | 324 |
| 22 | 484 |
| 23 | 529 |
| 29 | 841 |
| 34 | 1156 |
| ∑X = 195 N = 10 | ∑X² = 4315 |
(f) Coefficient of Variation
Since the coefficient of variation of Rice is more, therefore, Rice crop has greater variation.
(g) Comparison of the values of different measures:
| Measures of dispersion | Wheat | Rice |
| 1. Range | 16 | 22 |
| 2. Q.D. | 4.37 | 6.25 |
| 3. Mean Deviation about mean | 4.3 | 6 |
| 4. Mean Deviation about median | 4.3 | 5.7 |
| 5. Standard Deviation | 5.044 | 7.158 |
| 6. Coefficient of Variation | 32.54 | 36.70 |
6. In the previous question, calculate the relative measures of variation and indicate the value which, in your opinion, is more reliable.
Ans: Following are the relative measures of variations:
(i) Coefficient of Range for Wheat
Here L = 25,
S = 9
∴ Coefficient of Range
Coefficient of Range for Rice
Here, L = 34,
S = 12
(ii) Coefficient of Quartile Deviation
(iii) Coefficient of Mean Deviation from mean
(iv) Coefficient of Mean Deviation from median
(v) Coefficient of Standard Deviation
Thus, the coefficient of standard deviation is more reliable among all relative measures of variation.
7. A batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their scores in five previous tests which are:
| X | 25 | 85 | 40 | 80 | 120 |
| Y | 50 | 70 | 65 | 45 | 80 |
Which batsman should be selected if we want
(i) a higher run getter, or (ii) a more reliable batsman in the team?
Ans: Batsman-X
Coefficient of S.D.
= 0.484
= 48.44%
Batsman-Y
Average score = 62 runs
Coefficient of S.D.
= 0.207
Coefficient of Variance
= 20.77%
(i) Average score of Batsman X is more than that of Y. Thus we can conclude that X is a higher run getter.
(ii) Bating of Bastman X showed a greater CV than Y.
Hence, Y is a more reliable batsman in the team.
8. To check the quality of two brands of lightbulbs, their life in burning hours was estimated as under 100 bulbs of each brand.
(a) Which brand gives a higher life?
(b) Which brand is more dependable?
Ans: For Brand A:
For Brand B:
(a) The average life of a bulb of Brand B is comparatively higher than that of Brand A.
(b) The bulbs of Brand B are more dependable as CV of Brand B is lesser than CV of Brand A.
9. Average daily wage of 50 workers of a factory was ₹200 with a standard deviation of ₹40. Each worker is given a raise of ₹20. What is the new average daily wage and standard deviation? Have the wages become more or less uniform?
Ans: Increase in each worker wages = ₹20 Total increase in wages = 50 x 20 = ₹1,000 Total wages before increase in worker’s wages = 50 x 200 = ₹10,000
Total wages after increase in wages = 10,000 + 1,000 = ₹11,000
Hence, mean wages will be affected but standard deviation will not be affected as the standard deviation is independent of origin.
In order to calculate uniform wages, we will have to calculate coefficients of variation.
Afterwards
Now, more uniformity in wages has taken place.
10. If in the previous question, each worker is given a hike of 10% in wages, how are the mean and standard deviation values affected?
Ans: When each worker is given a hike of 10% in wages both the mean and standard deviation values will be affected.
New Mean = ₹ 200 + 10% of ₹ 200
= ₹ 200
New Standard Deviation
= ₹ 40 + 10% of ₹ 40
= ₹ 44
11. Calculate the mean deviation using mean and Standard Deviation for the following distribution.
| Classes | Frequencies |
| 20-40 | 3 |
| 40-80 | 6 |
| 80-100 | 20 |
| 100-120 | 12 |
| 120-140 | 9 |
| Total | 50 |
Ans: Mean Deviation:
Standard Deviation:
12. The sum of 10 values is 100 and the sun of their squares is 1090. Find out the coefficient of variation.
Ans: ∑X = 100
N = 10
∑X² = 1090
= √9 = 3
