Measures of Central Tendency - Class 11 Economics - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Notes - Measures of Central Tendency | Class 11 Statistics For Economics | Economics
In-Depth Notes on Measures of Central Tendency for Class 11 Economics
Understanding Measures of Central Tendency
Measures of central tendency are essential statistical tools that help summarise a set of data into a single representative value. By doing so, they provide a straightforward way to understand large volumes of data quickly. This approach is commonly used in various fields such as economics, social sciences, and business.
Arithmetic Mean: The Average
Definition and Formula
The arithmetic mean, often simply referred to as the "mean," is the sum of all values divided by the number of values. It is given by the formula:
$$ \text{Arithmetic Mean} = \frac{\sum X}{N} $$
where (\sum X) represents the sum of all observations, and (N) is the number of observations.
Calculation for Ungrouped Data
To calculate the arithmetic mean for ungrouped data, sum all the observations and divide by the total number of observations.
Example:
Given data: 40, 50, 55, 78, 58
$$ \text{Mean} = \frac{40 + 50 + 55 + 78 + 58}{5} = 56.2 $$
Thus, the average mark of students in an economics test is 56.2.
Calculation for Grouped Data
For grouped data, calculations may differ based on whether the data is discrete or continuous.
Direct Method:
For discrete series, multiply each value by its frequency, sum these products, and then divide by the total number of frequencies.
Assumed Mean Method:
If the data is extensive, an assumed mean simplifies the calculation:
$$ \text{Mean} = A + \frac{\sum d}{N} $$
where (A) is the assumed mean, (d) is the deviation from the assumed mean.
Step Deviation Method:
Further simplifies using a common factor (c):
$$ \text{Mean} = A + c \times \frac{\sum d'}{N} $$
where $d' = \frac{d}{c}$.
Properties of Arithmetic Mean
- Affected by Extreme Values: Large or small values can significantly influence the mean.
- Sum of Deviations: The sum of deviations of items about the mean is always zero.
Understanding the Median
Definition and Significance
The median is the middle value that separates the higher half from the lower half of the data set when it is ordered. It is less affected by extreme values compared to the mean.
Computation for Ungrouped Data
Arrange the data in ascending order and find the middle value.
Example:
Given data: 5, 7, 6, 1, 8, 10, 12, 4, 3
Ordered: 1, 3, 4, 5, 6, 7, 8, 10, 12
Median = 6
Computation for Grouped Data
For grouped data, the median is found within a class interval using cumulative frequency.
Example:
Income ranges: 10–20, 20–30, 30–40, 40–50 Frequencies: 2, 4, 10, 4
Cumulative Frequencies: [2, 6, 16, 20] Median position: (20+1)/2 = 10.5th
Median = 30 (in the 20–30 range).
Quartiles and Percentiles
Quartiles divide the data into four equal parts, while percentiles divide it into one hundred equal parts.
Determining the Mode
Definition and Importance
Mode is the value that occurs most frequently in a data set. It is helpful in identifying the most common value in a distribution.
Computation for Discrete Data
Identify the value with the highest frequency.
Example:
Frequency distribution of the variable: [10, 20, 30, 40, 50] Frequencies: [2, 8, 20, 10, 5]
Mode = 30.
Continuous Data
For continuous data, mode is found using the class interval with the highest frequency.
$$ \text{Mode} = L + \frac{D1}{D1 + D2} \times h $$
where (L) is the lower limit of the modal class, (D1) and (D2) are the differences between frequencies of the modal class and its adjacent classes, and (h) is the class interval.
Comparing Arithmetic Mean, Median, and Mode
When to Use Each Measure
Each measure has its use case based on data distribution and the analytical need.
Sensitivity to Extreme Values
- Mean: Highly sensitive
- Median: Less affected
- Mode: Not affected
Applying Measures of Central Tendency in Real Life
These measures are pivotal in various scenarios such as average income calculation, evaluating academic performance, and market analysis.
Choosing the Right Measure of Central Tendency
Selecting the appropriate measure depends on the data characteristics and the specific analytical requirement.
Mermaid.js Diagrams to Illustrate Concepts
flowchart LR
A[Data Set] -->|Summarises Data| B[Measures of Central Tendency]
B --> C[Arithmetic Mean]
B --> D[Median]
B --> E[Mode]
graph TD
A[Arithmetic Mean] --> B[Properties]
A --> C[Ungrouped Data]
A --> D[Grouped Data]
Conclusion
Understanding and correctly applying measures of central tendency is crucial for accurate data analysis. Whether you use the mean, median, or mode depends on the specific context and nature of your data. Equipped with these notes, you can confidently summarise and interpret data sets, an essential skill in economics and beyond.
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Extra Questions - Measures of Central Tendency | Statistics For Economics | Economics | Class 11
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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.
Average size of readymade garments:
Median: It is less affected by extreme values and provides a central tendency for size distribution, which is useful for clothing where extreme sizes might skew the average.
Average intelligence of students in a class:
Mean: Given that intelligence scores are generally measured on a continuous scale with a normal distribution, the mean would provide a good measure of central tendency.
Average production in a factory per shift:
Mean: Since production data is typically quantitative and regularly recorded, the mean would appropriately represent overall production output.
Average wage in an industrial concern:
Mean: Wages often have a broad range, but for general purposes and comparisons, the mean wage would be helpful. However, in cases of significant skewness due to a few high earners, the median might also be reported.
When the sum of absolute deviations from average is least:
Median: By definition, the median minimizes the sum of absolute deviations.
When quantities of the variable are in ratios:
Geometric Mean: The geometric mean is most appropriate when dealing with ratios and percentages, as it gives a more accurate central tendency in multiplicative processes.
In case of open-ended frequency distribution:
Median: The median is less sensitive to the open-ended nature of the distribution and can be calculated without knowing the extreme values.
The most suitable average for qualitative measurement is
(a) arithmetic mean
(b) median
(c) mode
(d) geometric mean
(e) none of the above
The most suitable average for qualitative measurement is:
(c) mode
The mode is the most frequently occurring value in a dataset and is particularly useful for qualitative data, where we observe categories or names rather than numerical values.
Which average is affected most by the presence of extreme items?
(a) median
(b) mode
(c) arithmetic mean
(d) none of the above
The average most affected by the presence of extreme items is:
(c) arithmetic mean
Extreme values, or outliers, can significantly influence the arithmetic mean because it takes into account all data points in its calculation.
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
The algebraic sum of deviations of a set of ( n ) values from its Arithmetic Mean (A.M.) is always zero. This is a fundamental property of the arithmetic mean.
So, the correct answer is:
$\text{(b) 0}$
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.
The sum of deviations of items from the median is zero:
False. The sum of deviations of items from the median is not zero. It is the mean for which the sum of deviations is zero.
An average alone is not enough to compare series:
True. An average provides a measure of central tendency but does not give information about the variability or distribution. To compare series, measures like standard deviation, range, etc., are also needed.
Arithmetic mean is a positional value:
False. The arithmetic mean is a measure of central tendency calculated by summing all values and dividing by the number of values. Positional values, like the median and mode, are based on the position of values within an ordered dataset.
Upper quartile is the lowest value of the top 25% of items:
True. The upper quartile (also known as the third quartile or Q3) is the value below which 75% of the data lies and above which 25% lies. It represents the lowest value within the top 25%.
Median is unduly affected by extreme observations:
False. The median is a positional measure and is not unduly affected by extreme observations. Extreme values have more effect on the arithmetic mean than on the median.
So the answers are:
False
True
False
True
False
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 |
To find the missing frequency and the median of the series, we need to follow these steps:
Given Data:
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, ( f ) (missing frequency), 17, 6
The arithmetic mean ($\overline{X}$) of the data is given as 28.
Step 1: Finding the Missing Frequency ($ f $)
The arithmetic mean is given by: $$ \overline{X} = \frac{\sum fX}{\sum f} $$
Let’s denote the class midpoints ($ m $) for each class interval:
0-10: midpoint = 5
10-20: midpoint = 15
20-30: midpoint = 25
30-40: midpoint = 35
40-50: midpoint = 45
50-60: midpoint = 55
Now, let's calculate the total sum of frequencies and the sum of the products of frequencies and midpoints.
$$ \sum f = 12 + 18 + 27 + f + 17 + 6 = 80 + f $$
$$ \sum fm = 12 \cdot 5 + 18 \cdot 15 + 27 \cdot 25 + f \cdot 35 + 17 \cdot 45 + 6 \cdot 55 $$
$$ = 60 + 270 + 675 + 35f + 765 + 330 $$
$$ = 2100 + 35f $$
Given the arithmetic mean is 28:
$$ 28 = \frac{2100 + 35f}{80 + f} $$
Solving for $ f $:
$$ 28(80 + f) = 2100 + 35f $$
$$ 2240 + 28f = 2100 + 35f $$
$$ 2240 - 2100 = 35f - 28f $$
$$ 140 = 7f $$
$$ f = 20 $$
The missing frequency ( f ) is 20.
Step 2: Finding the Median
To find the median, we first need to calculate the cumulative frequencies.
Profit per retail shop (in Rs) | No. of Retail Shops (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 |
The total number of observations (N) is 100. The median will be at the $ \left(\frac{N}{2}\right) $th position:
$$ \text{Median Position} = \frac{100}{2} = 50 $$
From the cumulative frequency table, the 50th position falls in the 20-30 class interval.
Using the median formula for continuous series:
$$ \text{Median} = L + \left(\frac{\frac{N}{2} - \text{cf}}{f}\right) \cdot h $$
Where:
$ L $ = Lower boundary of the median class = 20
$ \text{cf} $ = Cumulative frequency of the class preceding the median class = 30
$ f $ = Frequency of the median class = 27
$ h $ = Class width = 10
$$ \text{Median} = 20 + \left(\frac{50 - 30}{27}\right) \cdot 10 $$
$$ = 20 + \left(\frac{20}{27}\right) \cdot 10 $$
$$ = 20 + 7.41 $$
$$ = 27.41 $$
The median of the series is 27.41.
The following table gives the daily income of ten workers in a factory. Find the arithmetic mean. Explain with steps
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 |
To find the arithmetic mean of the daily income of ten workers, follow these steps:
List all the daily incomes:(120, 150, 180, 200, 250, 300, 220, 350, 370, 260)
Sum all the daily incomes:
$$ 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 $$
Divide the total sum by the number of workers (which is 10):
$$ \text{Arithmetic Mean} = \frac{\sum \text{Daily Incomes}}{\text{Number of Workers}} $$
Let's calculate the sum and then divide it by 10 to find the arithmetic mean.
First, calculate the sum:
$$ 120 + 150 + 180 + 200 + 250 + 300 + 220 + 350 + 370 + 260 = 2400 $$
Now, divide this sum by the number of workers:
$$ \text{Arithmetic Mean} = \frac{2400}{10} = 240 $$
Therefore, the arithmetic mean of the daily incomes of the ten workers is (240) Rs.
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 115 | 115 |
More than 95 | 95 |
More than 70 | 70 |
More than 60 | 60 |
More than 40 | 40 |
More than 25 | 25 |
The sum of the products of mid-point and frequency $( \Sigma f x )) is ( 11862.5) $.
The total number of families $( \Sigma f )$ is 150.
Now, we can calculate the arithmetic mean:
$$ \overline{X} = \frac{\Sigma f x}{\Sigma f} = \frac{11862.5}{150} = 79.0833 $$
So, the arithmetic mean of the daily income of the families is approximately Rs. 79.08.
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 |
To find the median of a grouped data set, we need to follow these steps:
Calculate the cumulative frequency for each class interval.
Identify the median class interval, which is the class interval where the middle value (N/2) lies.
Apply the median formula for grouped data:
$$ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h $$
where:
$ L $ is the lower boundary of the median class.
$ N $ is the total number of families.
$ CF$ is the cumulative frequency of the class before the median class.
$f $ is the frequency of the median class.
$ h $ is the class width (assuming all class intervals are of equal width).
Let's break this down step-by-step with the given data:
Calculate the cumulative frequency (CF):
Less than 100: $ 40$
100-200: $40 + 89 = 129$
200-300: $129 + 148 = 277$
300-400: $277 + 64 = 341$
400 and above: $ 341 + 39 = 380$
The total number of families, ( N ), is $ 380$.
Identify the median class:
Calculate $ N/2 = 380/2 = 190 $.
The cumulative frequency just greater than 190 is 277, which lies in the 200-300 class interval. Therefore, the median class is 200-300.
Apply the median formula:
Given:
$ L = 200$
$ N = 380$
$ CF = 129$ (cumulative frequency before the median class)
$f = 148 $ (frequency of the median class)
$h = 100 $ (width of the class interval)
$$ \text{Median} = 200 + \left(\frac{190 - 129}{148}\right) \times 100 $$
$$ \text{Median} = 200 + \left(\frac{61}{148}\right) \times 100 $$
$$ \text{Median} = 200 + 41.22 \approx 241.22 $$
Therefore, the median size of land holdings is approximately 241.22 acres.
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.
Daily Income (in Rs) | Number of workers |
10–14 | 5 |
15–19 | 10 |
20–24 | 15 |
25–29 | 20 |
30–34 | 10 |
35–39 | 5 |
(Hint: compute median, lower quartile and upper quartile.)
To solve this problem, we need to compute the median, lower quartile (Q1), and upper quartile (Q3) using the given data, and then derive the required values based on these quartiles. Let's break it down step-by-step:
Step 1: Frequency Distribution Calculation
First, we need to determine the cumulative frequency distribution:
10–14 (5 workers): Cumulative Frequency = 5
15–19 (10 workers): Cumulative Frequency = 15
20–24 (15 workers): Cumulative Frequency = 30
25–29 (20 workers): Cumulative Frequency = 50
30–34 (10 workers): Cumulative Frequency = 60
35–39 (5 workers): Cumulative Frequency = 65
Step 2: Median Calculation
The total number of workers is 65, so the median worker is at position: $ n/2 $ = 65/2 = 32.5
So, the median:
Lies in the 25-29 range since the cumulative frequency before this class is 30 and adding 20 reaches 50.
Step 3: Lower Quartile (Q1) Calculation
The lower quartile (Q1) corresponds to the 25% position, so it's the 16.25th worker.
Since 16.25 lies in the 20-24 range (cumulative frequency before this class is 15 and adding 15 reaches 30).
Step 4: Upper Quartile (Q3) Calculation
The upper quartile (Q3) corresponds to the 75% position, so it's the 48.75th worker.
Since 48.75 lies in the 25-29 range (cumulative freq. is 50).
Step 5: Derive Answers
(a) Highest income of lowest 50% workers (Median):
Highest income in the class preceding the median class (20–24) = Rs 24.
(b) Minimum income of top 25% workers (Upper Quartile):
Minimum income in the class where the Q3 lies (25–29) = Rs 25.
(c) Maximum income of lowest 25% workers (Lower Quartile):
Maximum income in the class where Q1 lies (20–24) = Rs 24.
Thus, we have:
( \text{Highest income of lowest 50% workers} = Rs 24 )
( \text{Minimum income of top 25% workers} = Rs 25 )
( \text{Maximum income of lowest 25% workers} = Rs 24 )
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 |
Calculation of Mean
From the computations, we have:
$$ \overline{X} = 63.57 \text{ kg per hectare} $$
Calculation of Median
Let's proceed by calculating the median step-by-step.
Since there are 150 farms, the median class is at the position: $$ \frac{N}{2} = \frac{150}{2} = 75 $$
We need to generate the cumulative frequency distribution to locate the median class.
Cumulative Frequency Distribution
Class Interval | Frequency (f ) | Cumulative Frequency (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 |
The median class is 62-65 since the 75th observation falls here.
We can use the formula for median: $$ \text{Median} = L + \left( \frac{\frac{N}{2} - CF_p}{f_m} \right) \times h $$
Where:
$ L = 62$(lower limit of the median class)
$CF_p = 55$ (cumulative frequency of the class preceding the median class)
$f_m = 36$ (frequency of the median class)
$h = 3$ (class interval width)
Now let's calculate:
Median Calculation:$$ \text{Median} = 62 + \left( \frac{75 - 55}{36} \right) \times 3 = 62 + \left( \frac{20}{36} \right) \times 3 = 62 + 1.67 = \text{63.67 kg per hectare} $$
Calculation of Mode
The modal class is the class with the highest frequency, which is 62-65 (frequency = 36).
Using the mode formula: $$ \text{Mode} = L + \left( \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \right) \times h $$
Where:
$ L = 62 $(lower limit of the modal class)
$ f_1 = 36 $ (frequency of the modal class)
$f_0 = 30 $ (frequency of the class preceding the modal class)
$ f_2 = 28 $ (frequency of the class succeeding the modal class)
$ h = 3$ (class interval width)
Let's calculate:
Mode Calculation:$$ \text{Mode} = 62 + \left( \frac{(36 - 30)}{(2 \times 36 - 30 - 28)} \right) \times 3 = 62 + \left( \frac{6}{14} \right) \times 3 = 62 + 1.29 = \text{63.29 kg per hectare} $$
Summary
Mean: 63.57 kg/ha
Median: 63.67 kg/ha
Mode: 63.29 kg/ha
These calculations give a summary of the production yield in kg per hectare for the wheat farms in the village.
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