Dispersion, Correlation and Index Numbers Questions and Answers AP Inter 2nd Year Economics Chapter 2

Reviewing AP Inter 2nd Year Economics Study Material Chapter 2 Dispersion, Correlation and Index Numbers Questions and Answers can help students prepare confidently for exams.

AP Inter 2nd Year Economics 2nd Lesson Dispersion, Correlation and Index Numbers Questions and Answers

Very Short Answer Questions

Question 1.
Write the formula for coefficient of variation (CV).
Answer:
Formula for the Coefficient of Variation is
CV = \(\frac{\text { SD }}{\text { Mean }}\) × 100

CV is the ratio of the standard deviation(SD) to the mean, expressed in percentage.

Question 2.
Define variance.
Answer:
Variance (σ²) is defined as the average of the squared deviations from the mean.

Variance is a statistical measure of dispersion that shows how far the values in a dataset are spread from the mean.

Question 3.
What is rank correlation?
Answer:
Rank correlation is a method of measuring the relationship between variables by using their ranks instead of actual values.

It is used when data is in the form of qualitative characteristics like beauty, honesty.

Question 4.
The marks obtained by 10 students are: 12, 18, 25, 30, 32, 28, 20, 15, 40, 35. Find the Range and the Coefficient of Range.
Answer:
Given data:12, 18, 25, 30, 32, 28, 20, 15, 40, 35.
Largest value (L) = 40,
Smallest value (S) = 12 ;
(i) Range = L – S
= 40 – 12 = 28

(ii) Coefficient of Range = \(\frac{L-S}{L+S}\)
= \(\frac{40-12}{40+12}\)
= \(\frac{28}{52}\) = 0.538

Question 5.
Find Mean Deviation for dataset: 3, 6, 9
Answer:
Given dataset: 3, 6, 9
Step 1: Mean = \(\frac{3+6+9}{3}\)
= \(\frac{18}{3}\) = 6

Step 2: Deviations from Mean (ignoring signs):
|3 – 6| = 3 |6 – 6|
= 0 |9 – 6| = 3

Step 3: Mean Deviation MD = \(\frac{\sum|\mathrm{D}|}{\mathrm{n} \mid}\)
= \(\frac{3+0+3}{3}=\frac{6}{3}\) = 2.

Question 6.
Base year price of rice is Rs.45 per kg. Current year price is Rs.50. Find simple price index. (Base year = 100)
Answer:
Given Current year price P₁ = 50,
Base year price P₀ = 45
∴ Price Index = \(\frac{P_1}{P_0}\) × 100
= \(\frac{50}{45}\) × 100 = 111.11

Short Answer Questions

Question 1.
Distinguish between absolute measures and relative measures of dispersion with examples.
Answer:

Question 2.
From the following data, calculate the Quartile deviation.

Answer:
The Cumulative frequency table for the given data is as follows:

1) Calculating the First Quartile (Q₁)
Total frequency N = 50,
Position of Q₁ : \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Here Q₁ Class = Class containing 12.5th item
The value in the cf column where it is just above 12.5 is 20.
It belongs to class 50-100.
∴ Q₁ Class is 50-100.
Here L = 50,
i = 100-50 = 50,
f = 5,
cf = 5 (cf of the above class).

Formula: Q₁ = L + \(\left(\frac{\frac{\mathrm{N}}{4}-\mathrm{cf}}{\mathrm{f}}\right)\) × i
= 50 + \(\left(\frac{12.5-5}{15}\right)\) × 50
= 50 + \(\left(\frac{7.5}{15}\right)\) × 50
= 50 + 25 = 75

2) Calculating the Third Quartile (Q₃)
Total frequency N = 50,
Position of Q₃ : \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 50}{4}\) = 37.5
Q₃ Class = Class containing 37.5th item
The value in the cf column where it is just above 37.5 is 40.
It belongs to class 100-150
∴ Q₃ Class is 100-150.
Hence L = 100,
i = 150 – 100 = 50,
f = 20,
cf = 20 (cf of the before class)

Formula:
Q₃ = L + \(\left(\frac{\frac{\mathrm{N}}{4}-\mathrm{cf}}{\mathrm{f}}\right)\) × i
= 100 + \(\left(\frac{37.5-20}{20}\right)\) × 50
= 100 + \(\left(\frac{17.5}{20}\right)\) × 50
= 100 + 43.75 = 143.75

3) Calculating Quartile Deviation (Q.D.)
Formula: QD = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{143.75-75}{2}\)
= \(\frac{68.75}{2}\) = 34.375
The quartile Deviation for the given data is 34.375.

Question 3.
Calculate the standard deviation of the following data.

Answer:
Calculating the Mean (\(\overline{\mathrm{X}}\))
∑X = 10 + 16 + 10 + 16 + 10 + 10 + 16 + 16 = 104,
Number of observations N = 8
Mean (\(\overline{\mathrm{X}}\)) = \(\frac{\Sigma X}{N}\)
= \(\frac{104}{8}\) = 13
Finding Deviations at d = X – \(\overline{\mathrm{X}}\) and their squares d²

Standard Deviation σ = \(\sqrt{\frac{\sum(X-\bar{X})^2}{n}}\)
= \(\sqrt{\frac{\sum d^2}{n}}\)
= \(\sqrt{\frac{72}{8}}\)
= √9 = 3
The standard deviation for the given data is 3.

Question 4.
Define correlation. What are the properties of correlation?
Answer:
Correlation measures the degree and direction of the relationship between two or more variables.

It shows how variables move in relation to each other. It may be positive, negative, or zero.

Properties of Correlation:

1. Range of Values: The value of the correlation coefficient (r) always lies between – 1 and + 1
2. Direction and Strength: A value close to +1 indicates strong positive relationship, close to – 1 indicates strong negative relationship, and 0 indicates no relationship.
3. Association vs. Causation: Correlation measures only association, but not cause-and-effect relationship.
4. Unit-Free Measurement: The coefficient of correlation has not units. Hence, it is unit-free.
5. Invariance: The coefficient remains unaffected by change in origin and scale.

Question 5.
What are the uses of Index numbers?
Answer:
Uses of Index Numbers:
1) An Aid to Framing Policies: Index numbers are used to guide economic policies, such as fixing wages and dearness allowance based on the Consumer Price Index (CPI).
They also help in determining policies related to the volume of trade and price fixation.

2) To Find Trends: They measure changes overtime, helping to study general trends in economic activity and are useful for forecasting future conditions.

3) To Assess the Purchasing Power of Money : Index numbers (especially CPI) help in calculating real wages and show whether the purchasing power of money has increased or decreased.

4) For Adjusting National Income: They are used for deflating national income (converting current prices into constant prices) to get a more accurate picture of economic growth.

Question 6.
Distinguish between Consumer Price Index (CPI) and Wholesale Price Index (WPI).
Answer:

Long Answer Questions

Question 1.
What is correlation? What are the measures of correlation?
Answer:
Correlation is a statistical tool used to measure the degree and direction of the relationship between two or more variables. It explains how one variable changes in relation to another.
Ex: An increase in income leads to an increase in expenditure.

Correlation can be of three types:

• Positive correlation: Both variables move in the same direction (e.g., income and expenditure).
• Negative correlation: Variables move in opposite directions (e.g., price and demand).
• Zero correlation: No relationship exists between variables.

3 Measures of Correlation:

1) Scatter Diagram (or Dot Diagram):
This is a simple and visual method of studying correlation.

In this method, each pair of values is plotted as a point on a graph.

This method is easy to understand but does not give a precise numerical value of correlation.

• lf the points are closely clustered around an upward sloping line, it indicates strong positive correlation.
• lf they slope downward, it indicates negative correlation.
• lf the points are widely scattered without any definite pattern, it shows no correlation.

2) Karl Pearson’s Coefficient of Correlation:
This is the most widely used and accurate numerical method for measuring correlation between two variables that have a linear relationship.

This method provides a precise and reliable measure, mostly suitable for quantitative data.

• lt is the ratio of covariance between variables to the product of their standard deviations.
• The value of the coefficient lies between – 1 and + 1.
• A value of +1 indicates perfect positive correlation.

– 1 indicates perfect negative correlation, and 0 indicates no correlation.

3) Spearman’s Rank Correlation:
It is used when the data is qualitative or cannot be measured numerically, such as intelligence, beauty, or honesty.

It is particularly useful when exact measurements are not available but ranking is possible.

• Instead of actual values, ranks are assigned to-the observations.
• The differences between ranks of paired items are calculated and used to find the correlation coefficient.

Question 2.
Calculate the Range, Coefficient of Range, and Quartile Deviation from the following data.

Answer:
1) Calculation of Range and Coefficient of Range:
In this continuous series,
upper limit of the largest class (50 – 60) is L = 60
lower limit of the lower class (10 – 20) is S = 10

(i) Range = L- S
= 60 – 10 = 50

(ii) Coefficient of Range = \(\frac{\mathrm{L}-\mathrm{S}}{\mathrm{~L}+\mathrm{S}}\)
= \(\frac{60-10}{60+10}\)
= \(\frac{50}{70}\) = 0.714

2) Calculation of Quartile Deviation (Q.D)

Step 1:
Finding the First Quartile (Q₁).
Total frequency N = 42,
Position of Q₁ : \(\frac{N}{4}=\frac{42}{4}\) = 10.5
Q₁ Class = Class containing 10.5th item
The value in the cf column where it is just above 10.5 is 18. It belongs to class 20-30.
∴ Q₁ Class is 20 – 30.
Here L = 20,
i = 30 – 20=10,
f = 10,
cf = 8 (cf of the before Q₁ class).
Formula:
Q₁ = L + \(\left(\frac{\frac{N}{4}-c f}{f}\right)\) × i
= 20 + \(\left(\frac{10.5-8}{10}\right)\) × 10
= 20 + 2.5 = 22.5

Step 2:
Finding the Quartile (Q₃).
Total frequency N = 42,
Position of Q₃ : \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 42}{4}\) = 31.5
Q₃ Class = Class containing 31.5th item
The value in the cf column where it is just above 31.5 is 38.
It belongs to class 40 – 50.
∴ Q₃ Class is 40 – 50.
Here L = 40,
i = 50 – 40 = 10,
f = 8,
cf = 30 (cf of the before Q₃ class).
Q₃ = L + \(\left(\frac{\frac{3 \mathrm{~N}}{4}-\mathrm{cf}}{\mathrm{f}}\right)\) × i
= 40 + \(\left(\frac{31.5-30}{8}\right)\) × 10
= 40 + 1.875 = 41.875

Step 3:
Quartile Deviation QD = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{41.875-22.5}{2}\)
= \(\frac{19.375}{2}\) = 9.6875.

Question 3.
The following data relate to prices and quantities of certain commodities in the base year and current year. Construct the Weighted Aggregative Price Index using Laspeyres’ and Paasche’s methods.

Weighted Aggregative Price Index

Answer:
Table showing the required products (pq) for both formula based on the prices (p) and quantities (q) for the base year (0) and current year (1)

1) Laspeyres Method:
This method uses base period quantities (q₀) as weights.
From the table we have ∑ p₁q₀ = 1669,
∑ p₀q₀ = 1345,

Formula: Weighted aggressive Price Index
P₀₁ = \(\frac{\sum \mathrm{p}_1 \mathrm{q}_0}{\sum \mathrm{p}_0 \mathrm{q}_0}\) × 100
= \(\frac{1669}{1345}\) × 100 = 124.1
Laspeyres’ index: 124.1

2) Paasche’s Method:
This method uses current year quantities (q₁) as weights.
From the table, we have ∑ p₁q₁ = 1133,
∑ p₀q₁ = 910
Formula:
P₀₁ = \(\frac{\sum p_1 q_1}{\sum p_0 q_1}\) × 100
= \(\frac{1133}{910}\) × 100
Paasche’s index: 124.5

An Excellent Cricket Illustration-2

Consider the scores of Virat and Kohli in Five ODI Matches of Worldcup.

A. Measures of Dispersion:
1. Range: Virat’s Range = 169 – 27 = 142;
Rohit’s Range = 71 – 48 = 23
Interpretation : Rohit is more consistent because his range (23) is much smaller than Virat’s range (142).

2. Standard Deviation (or Variance)
Virat has a high standard deviation because of scores like 27 and 169.
Rohit has a low standard deviation because most scores are close to each other.
So, Rohit has low dispersion variation/ and hence his performance is more stable and dependable

One-Line Summary

Dispersion measures tell us how consistently one scores, but not how much a batsman scores.

B. Measures of Correlation: Consider the scores of Virat and Rohit in 3 ODI Series.

Interpretation for Series 1 : Positive Correlation

• When Virat scores more, Rohit also scores more and vice-versa.
• Cricket lnsight: The two batsmen are contributing well combinedly (both together).
• Conclusion: There is a strong positive correlation between their performances.

Interpretation for Series 2: Negative Correlation

• Interpretation: As Virat’s score decreases, Rohit’s score increases.
• When one performs poor, the other performs well.
• Cricket lnsight: The two batsmen are not contributing well together.

Interpretation for Series 3 : Zero Correlation

• Virat’s scores show no clear pattern with Rohit’s scores.
• One batsman’s performance does not help predict the other’s. They score independently.
• Conclusion: There is no correlation between their performances.

Final Comparison:

One-Sentence Summary

• Measures of Central Tendency describe the centre of performance.
• Measures of Dispersion describe the consistency of performance.
• Measures of Correlation describe the relationship between the performances of two batsmen.

Multiple Choice Questions

Question 1.
Measures of dispersion indicate
1) The central value of data
2)The degree of scatter of values around a central value
3) The average of values
4) The relation between two variables
Answer:
2)The degree of scatter of values around a central value

Question 2.
The difference between the largest and the smallest value in a data set is called:
1) Mean deviation
2) Variance
3) Standard deviation
4) Range
Answer:
4) Range

Question 3.
The interquartile range is equal to:
1) Q₃ – Q₁
2) Q₂ – Q₁
3) Q₄ – Q₁
4) Q₃ — Q₂
Answer:
1) Q₃ – Q₁

Question 4.
Which of the following measures of dispersion is on all observations?
1) Range
2) Quartile deviation
3) Standard deviation
4) Co-efficient of range
Answer:
3) Standard deviation

Question 5.
The value of the correlation coefficient lies between
1) 0 and 1
2) – 1 and + 1
3) 0 and ∝
4) 1 and ∝
Answer:
2) – 1 and + 1

Question 6.
If variance is 25, standard deviation is:
1) 625
2) 125
3) 5
4) 15
Answer:
3) 5

Question 7.
Which measure of dispersion is most affected by extreme values?
1) Range
2) Quartile deviation
3) Mean deviation
4) Standard deviation
Answer:
1) Range

Question 8.
If Mean = 30 and Standard Deviation = 6, then CV = ?
1) 10 %
2) 15 %
3) 18 %
4) 20 %
Answer:
4) 20 %

Question 9.
Correlation measures.
1) The average of a data set
2) The degree and direction of relationship between two variables
3) The difference between maximum and minimum values
4) The spread of one set of values
Answer:
2) The degree and direction of relationship between two variables

Question 10.
The price of a commodity was Rs. 50 in the base year and Rs. 60 in the current year. The simple price index number (base year = 100) is: [3
1) 140
2) 130
3) 120
4) 110
Answer:
3) 120

Fill in the Blanks

Question 1.
Averages show the center of data, but dispersion shows of ____________ the data.
Answer:
the spread

Question 2.
____________ measure of dispersion is called semi-inter quartile range.
Answer:
Quartile Deviation (QD)

Question 3.
The numerical value that expresses size or direction or proportion of a relationship is called ____________.
Answer:
Coefficient.

Question 4.
The square of Standard Deviation is equal to ____________.
Answer:
variance

Question 5.
In Laspeyres’ method of Index Numbers, ____________ year quantities are taken as weights.
Answer:
Base year

One Word Answers

Question 1.
Which quartile divides the data set into two equal parts?
Answer:
Second quartile (Q₂) (or) Median divides the data set into two equal parts.

Question 2.
Which measure of dispersion depends on the unit of measurement?
Answer:
Absolute measure of dispersion depends on the unit of measurement.

Question 3.
What is the relative measure of dispersion expressed as a percentage of the standard deviation to the mean?
Answer:
Coefficient of Variation (CV)

Question 4.
What statistical tool measures the relptive change in a variable over time in percentages?
Answer:
Index Number

Question 5.
Which weighted aggregate index is the geometric mean of Laspeyr’s and Paasche’s index numbers?
Answer:
Fisher’s ideal index