ECONOMICS XI

Solutions Chapter 6 – Measures of Dispersion

Page No 89:

Question 1:

A measure of dispersion is a good supplement to the central value in understanding a frequency distribution. Comment.

Answer:

The study of the averages is only one sided distribution story. In order to understand the frequency distribution fully, it is essential to study the variability of the observations. The average measures center 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. For example, if a country has very high income group people and very low income group people, then we can say that the country has large income disparity.

Question 2:

Which measure of dispersion is the best and how?

Answer:

Standard Deviation is the best measure of dispersion as it satisfies the most essentials of the good measure of dispersion. The following points make Standard Deviation the best measure of dispersion:

  1. Most of the statistical theory is based on Standard Deviation. It helps to make comparison between variability of two or more sets of data. Also, Standard Deviation helps in testing the significance of random samples and in regression and correlation analysis.
  2. It is based on the values of all the observations. In other words, Standard Deviation makes use of every item in a particular distribution.
  3. Standard Deviation has a precise value and is a well-defined and definite measure of dispersion. That is, it is rigidly defined.
  4. It is independent of the origin.
  5. It is widely used measure of dispersion as all data distribution is nearer to the normal distribution.
  6. It enables algebraic treatment. It has correct mathematical processes in comparison to range, quartile deviation and mean deviation.

Question 3:

Some measures of dispersion depend upon the spread of values whereas some calculate the variation of values from a central value. Do you agree?

Answer:

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. The spread of values is determined by the absolute measures of dispersion like Range, Quartile Mean Deviation, and Standard Deviation. These measures express dispersion in terms of original unit of the series and it cannot be used for the comparison of statistical data having different units. On the other hand, the relative measures of the dispersion calculate the variability of the values from a central value. The relative measure includes coefficient of Range, Mean Deviation and Variation. It is used when the comparison has to be made between two statistical sets. These measures are free from any units.

Question 4:

In a town, 25% of the persons earned more than Rs 45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion.

Answer:

Absolute Value of Dispersion

Relative Value of Dispersion

Question 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,

(i) Range

(ii) Q.D.

(iii) Mean Deviation about Mean

(iv) Mean Deviation about Median

(v) Standard Deviation

(vi) Which crop has greater variation?

(vii) Compare the value of different measures for each crop.

Answer:

(i) Range

  1. Wheat

Highest value of distribution (H) = 25

Lowest value of distribution (L) = 9

Range = H – L

          = 25 – 9

          =16

  1. Rice

Highest value of distribution (H) = 34

Lowest value of distribution (L) = 12

Range = H – L

           =34 – 12

= 22

(ii) Quartile Deviation

  1. Wheat

Arranging the production of wheat in increasing order

9, 10, 10, 12, 15, 16, 18, 19, 21, 25

= 2.75th item

=Size of 2th item + 0.75 (size of 3rd item – size of 2nd item)

= 10 + 0.75 (10 – 10)

= 10 + 0.75 × 0

=10

= 8.25th

=Size of 8th item + 0.25 (size of 9th item – size of 8th item)

= 19 + 0.25 (21 – 19)

= 19 + 0.25 × 2

= 19 + 0.50 = 19.50

= 4.75

  1. Rice

Arranging the data of production of rice

12, 12, 12, 15, 18, 18, 22, 23, 29, 34

= 2.75th item

= size of 2nd item + 0.75 (size of 3rd item – size of 2nd item)

= 12 + 0 .75 (12 – 12)

= 12 + 0.75 × 0

= 12

= 8.25th item

= Size of 8th item + 0.25 (size of 9th item – size of 8th item)

= 23 + 0.25 (29 – 23)

= 23 + 0.25 × 6

= 23 + 1.5

= 24.5

= 6.25

(iii) Mean Deviation about Mean

  1. Wheat
Wheat Production

x

9 6
10 5
10 5
12 3
15 0
16 1
18 3
19 4
21 6
25 10
  1. Rice
Rice Production

 x

12 6
12 6
12 6
15 3
18 0
18 0
22 4
23 5
29 11
34 16

(iv) Mean Deviation about Median

  1. Wheat
Production of Wheat

x

09 6
10 5
10 5
12 3
15 0
16 1
18 3
19 4
21 6
25 10
  1. Rice
Production of rice d = X–18
12 6
12 6
12 6
15 3
0
18 0
22 4
23 5
29 11
34 16

Since n is even

(v) Standard Deviation

  1. Wheat
Reduction of Wheat

x

Ax = 15

d= x – Ax

d2
9 –6 36
10 –5 25
10 –5 25
12 –3 9
15 0 0
16 1 1
18 3 9
19 4 16
21 6 36
25 10 100
  1. Rice
Production of rice

 x

Ax = 18

d = x – A x

d2
12 –6 36
12 –6 36
12 –6 36
15 –3 9
18 0 0
22 4 16
23 5 25
29 11 121
34 16 256

(vi) Coefficient of Variation

  1. Wheat
  2. Rice

(vi) Rice crop has greater variation as the coefficient of variation is higher for rice as compared to that of wheat.

(vii) Rice crop has higher Range, Quartile Deviation, Mean Deviation about mean, Mean Deviation about median, Standard Deviation and Coefficient of Variation.

Question 6:

In the previous question, calculate the relative measures of variation and indicate the value which, in your opinion, is more reliable.

Answer:

(i) Coefficient of Range

  1. a) Wheat
  2. b) Rice

(ii) Coefficient of Quartile Deviation

  1. a) Wheat
  2. b) Rice

Q1 = 12, Q3 = 24.5

(iii) Coefficient of Mean Deviation from mean

  1. a) Wheat
  2. b) Rice

(iv) Coefficient of Variation

  1. a) Wheat
  2. b) Rice

The coefficient of variation is more reliable than all other measures.

Question 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?

Answer:

Batsman X
X X –  = x

X – 70

x2
25 – 45 2025
85 + 15 225
– 30 900
80 10 100
120 50 2500
Batsman Y
Y y2
50 – 12 144
70 8 64
65 3 9
45 – 17 289
80 18 324

(i) Average of Batsman X is higher than that of Batsman Y, so he should be selected if we want to score  higher run.

(ii) The Batsman X is more reliable than Batsman Y. This is because the coefficient of variation of Batsman X is higher than that of Batsman Y.

Page No 90:

Question 8:

To check the quality of two brands of light bulbs, their life in burning hours was estimated as under for 100 bulbs of each brand.

Life  No. of bulbs
(in hrs) Brand A Brand B
0 – 50 15 2
50 – 100 20 8
100 – 150 18 60
150 – 200 25 25
200 – 250 22 5
100 100

(i) Which brand gives higher life?

(ii) Which brand is more dependable?

Answer:

For Brand A
Life

(in hours)

No. of bulbs M A = 125 d2 fd fd2
f x X – A = d
0 – 50 15 25 – 100 – 2 4 – 30 60
50 – 100 20 75 – 50 – 1 1 – 20 20
100 – 150 18 125 0 0 0 0 0
150 – 200 25 175 50 1 1 25 25
200 – 250 22 225 100 2 4 44 88
For Brand B
Life (in hrs) No. of bulbs (f) Mid value m A = 125 d’2 fd’ fd’2
0 – 50 2 25 – 100 – 2 4 – 4 8
50 – 100 8 75 – 50 – 1 1 – 8 8
100 – 150 60 125 0 0 0 0 0
150 – 200 25 175 50 1 1 25 25
200 – 250 5 225 100 2 4 10 20

(i) The average life of bulb of Brand B has comparatively  higher life than the bulb of Brand A.

(ii) The  bulbs  of Brand B is more dependable as CV of Brand B is lesser than CV of Brand A.

Question 9:

Average daily wage of 50 workers of a factory was Rs 200 with a Standard Deviation of Rs 40. Each worker is given a raise of Rs 20. What is the new average daily wage and Standard Deviation? Have the wages become more or less uniform?

Answer:

N = 50

 = 200

s = 40

So, Total Wages = 200 × 50

= Rs 10,000

Now, increased wage rate = Rs 20

Total raise = 50 × 20 = Rs 1,000

Total Wage after raise = 10,000 + 1,000

=Rs 11,000

= Rs 220

Initial Standard Deviation = Rs 40

So, New Standard Deviation = Rs 40 + Rs 20

= Rs 60

Note: New Standard Deviation will rise by the same amount as the wage of each worker has increased.

Question 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?

Answer:

Average wage = Rs 200

Hike in wages = 10% of Rs 200

= Rs 20

Individual raise given to each worker = Rs 20

Total raise in wage = 50 × 20 = Rs 1,000

New Total Wage = Rs 10,000 + Rs 1000

= Rs 11,000

Initial Standard Deviation = Rs 40

So, New Standard Deviation = Rs 40 + 20

= Rs 60

Question 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

Answer:

Classes Frequency m A = 90

d = X – A

fd’ d’2 fd’2
20 – 40 3 30 – 60 60 180 – 6 – 18 36 108
40 – 80 6 60 – 30 30 180 – 3 – 18 9 54
80 – 100 20 90 0 0 0 0 0 0 0
100 – 120 12 110 20 20 240 2 24 4 48
120 – 140 9 130 40 40 360 4 36 16 144

Mean Deviation from mean

Question 12:

The sum of 10 values is 100 and the sum of their squares is 1090. Find out the Coefficient of Variation.

Answer:

Solutions Chapter 5 – Measures of Central Tendency (Prev Lesson)
(Next Lesson) Solutions Chapter 7 – Correlation
Back to ECONOMICS XI

No Comments

Post a Reply

error: Content is protected !!