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Chapter 5 Measures Of Central Tendency
Introduction
Building upon the previous chapter's discussion of data presentation, this chapter introduces measures of central tendency. These are numerical methods used to summarize a set of data with a single value that represents the typical or central point of the dataset. Examples from everyday life include average marks, average rainfall, or average income. Summarizing data using central tendency helps in understanding the data in brief and comparing different datasets.
For instance, to understand the economic condition of a farmer like Baiju in a village with 50 small farmers, you might compare his land holding size to the average land holding size in the village, the size owned by half the farmers (median), or the size owned by most farmers (mode). Measures of central tendency provide this typical or representative value.
There are several statistical measures of central tendency, commonly referred to as "averages". The three most frequently used are the Arithmetic Mean, Median, and Mode. While Geometric Mean and Harmonic Mean also exist for specific situations, this chapter focuses on the three primary measures.
Arithmetic Mean
The **Arithmetic Mean** (A.M.), also known as the average, is the most commonly used measure of central tendency. It is calculated by summing up all the values in a dataset and dividing the sum by the number of observations.
Mathematically, for N observations $X_1, X_2, ..., X_N$, the Arithmetic Mean ($\bar{X}$) is given by:
$\bar{X} = \frac{X_1 + X_2 + ... + X_N}{N} = \frac{\sum X}{N}$
Where $\sum X$ is the sum of all observations and $N$ is the total number of observations.
How Arithmetic Mean Is Calculated
The calculation methods differ for ungrouped and grouped data.
Arithmetic Mean For Series Of Ungrouped Data
For data that is not grouped into classes, A.M. can be calculated using direct method, assumed mean method, or step deviation method.
Direct Method
Sum of all observations divided by the number of observations.
Example 1. Calculate Arithmetic Mean from the data showing marks of students in a class in an economics test: 40, 50, 55, 78, 58.
Answer:
Using the direct method:
$\bar{X} = \frac{\sum X}{N} = \frac{40 + 50 + 55 + 78 + 58}{5} = \frac{281}{5} = 56.2$
The average mark is 56.2.
Assumed Mean Method
This method simplifies calculations for large datasets. Choose an assumed mean (A) from the data (often a central value). Calculate deviations (d) of each observation from the assumed mean ($d = X - A$). Sum the deviations ($\sum d$). The A.M. is $A + \frac{\sum d}{N}$.
Example 2. The following data shows the weekly income of 10 families. Compute mean family income.
Weekly Income (in Rs): 850, 700, 100, 750, 5000, 80, 420, 2500, 400, 360.
Answer:
Let A = 850 (Assumed Mean).
| Family | Income (X) | d = X – 850 |
|---|---|---|
| A | 850 | 0 |
| B | 700 | –150 |
| C | 100 | –750 |
| D | 750 | –100 |
| E | 5000 | +4150 |
| F | 80 | –770 |
| G | 420 | –430 |
| H | 2500 | +1650 |
| I | 400 | –450 |
| J | 360 | –490 |
| Total | 11160 | +2660 |
$\bar{X} = A + \frac{\sum d}{N} = 850 + \frac{2660}{10} = 850 + 266 = 1116$
The average weekly income is Rs 1,116.
Step Deviation Method
This method further simplifies calculations if deviations are divisible by a common factor (c). Calculate $d' = \frac{d}{c} = \frac{X - A}{c}$. Sum $d'$ ($\sum d'$). The A.M. is $A + \frac{\sum d'}{N} \times c$.
In Example 2, using A=850, we can find a common factor. Let's choose c=10 for illustration, although deviations like -750, -150 are not easily divisible by 10. Let's assume a common factor of 10 for all deviations for simplicity as in the text's Table 5.1 (this may not be strictly mathematically correct for this data, but follows the text's representation). $d' = d/10$.
From Example 2 table, $\sum d = 2660$. If we assume a common factor (like 10, which is not exactly correct for all values here, but follows the text's simplified example), $\sum d' = 266$.
$\bar{X} = A + \frac{\sum d'}{N} \times c = 850 + \frac{266}{10} \times 10 = 850 + 266 = 1116$
Average weekly income is Rs 1,116.
Calculation Of Arithmetic Mean For Grouped Data
For grouped data (discrete or continuous series), frequency (f) is given for each value or class. Calculation uses frequencies.
Discrete Series
Multiply each variable value (X) by its frequency (f) to get fX. Sum fX ($\sum fX$). Sum frequencies ($\sum f$). A.M. = $\frac{\sum fX}{\sum f}$. Assumed Mean and Step Deviation methods are adapted similarly, multiplying deviations or step deviations by frequencies (f d or f d').
Example 3. Plots in a housing colony come in only three sizes: 100 sq. metre, 200 sq. meters and 300 sq. metre and the number of plots are respectively 200, 50 and 10. Compute mean plot size.
Answer:
Using the direct method:
| Plot size in Sq. metre (X) | No. of Plots (f) | fX |
|---|---|---|
| 100 | 200 | 20000 |
| 200 | 50 | 10000 |
| 300 | 10 | 3000 |
| Total | 260 ($\sum f$) | 33000 ($\sum fX$) |
$\bar{X} = \frac{\sum fX}{\sum f} = \frac{33000}{260} \approx 126.92$
The mean plot size is 126.92 Sq. metre.
Continuous Series
For continuous series (data grouped into class intervals), use the mid-point (m) of each class as the representative value. Multiply mid-points by frequencies (fm). Sum fm ($\sum fm$). Sum frequencies ($\sum f$). A.M. = $\frac{\sum fm}{\sum f}$. Assumed Mean and Step Deviation methods are adapted using mid-points instead of X values, calculating deviations ($d = m - A$) or step deviations ($d' = (m - A)/c$), and multiplying by frequencies (f d or f d').
Example 4. Calculate average marks of the following students using (a) Direct method (b) Step deviation method.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
No. of Students: 5, 12, 15, 25, 8, 3, 2
Answer:
(a) Direct Method:
| Marks | No. of Students (f) | Mid value (m) | fm |
|---|---|---|---|
| 0–10 | 5 | 5 | 25 |
| 10–20 | 12 | 15 | 180 |
| 20–30 | 15 | 25 | 375 |
| 30–40 | 25 | 35 | 875 |
| 40–50 | 8 | 45 | 360 |
| 50–60 | 3 | 55 | 165 |
| 60–70 | 2 | 65 | 130 |
| Total | 70 ($\sum f$) | 2110 ($\sum fm$) |
$\bar{X} = \frac{\sum fm}{\sum f} = \frac{2110}{70} \approx 30.14$
The average marks are approximately 30.14.
(b) Step Deviation Method:
Let A = 35 (Midpoint of the class 30-40). Let c = 10 (Class interval width).
| Marks | f | m | d = m – 35 | d' = d/10 | fd' |
|---|---|---|---|---|---|
| 0–10 | 5 | 5 | –30 | –3 | –15 |
| 10–20 | 12 | 15 | –20 | –2 | –24 |
| 20–30 | 15 | 25 | –10 | –1 | –15 |
| 30–40 | 25 | 35 | 0 | 0 | 0 |
| 40–50 | 8 | 45 | 10 | 1 | 8 |
| 50–60 | 3 | 55 | 20 | 2 | 6 |
| 60–70 | 2 | 65 | 30 | 3 | 6 |
| Total | 70 ($\sum f$) | –34 ($\sum fd'$) |
$\bar{X} = A + \frac{\sum fd'}{\sum f} \times c = 35 + \frac{-34}{70} \times 10 = 35 - \frac{340}{70} = 35 - 4.857 \approx 30.143$
The average marks are approximately 30.14.
Two Interesting Properties Of A.m.
The sum of deviations of observations from the arithmetic mean is always zero ($\sum (X - \bar{X}) = 0$). A.M. is sensitive to extreme values, which can significantly affect its value.
Weighted Arithmetic Mean
Used when assigning different importance (weights) to observations. The weighted A.M. is calculated by multiplying each observation by its weight, summing the products, and dividing by the sum of the weights: $\frac{\sum WX}{\sum W}$.
Median
The **Median** is a positional average that divides a dataset into two equal halves when the data is arranged in order of magnitude. Half of the values are below the median, and half are above it. Unlike the mean, the median is not affected by extreme values because it depends only on the position of the central value(s).
Computation Of Median
For ungrouped data, sort the data and find the middle value. If N is odd, the median is the value at the $(\frac{N+1}{2})^{th}$ position. If N is even, the median is the average of the values at the $\frac{N}{2}^{th}$ and $(\frac{N}{2}+1)^{th}$ positions. The formula $\frac{N+1}{2}$ item gives the position of the median in an ordered array.
Example 5. Suppose we have the following observation in a data set: 5, 7, 6, 1, 8, 10, 12, 4, and 3. Calculate the median.
Answer:
Arranging in ascending order: 1, 3, 4, 5, 6, 7, 8, 10, 12. N=9.
Position of median = $\frac{9+1}{2} = 5^{th}$ item. The 5th item is 6. Median = 6.
Example 6. The following data provides marks of 20 students. Calculate the median marks.
25, 72, 28, 65, 29, 60, 30, 54, 32, 53, 33, 52, 35, 51, 42, 48, 45, 47, 46, 33.
Answer:
Arranging in ascending order:
25, 28, 29, 30, 32, 33, 33, 35, 42, 45, 46, 47, 48, 51, 52, 53, 54, 60, 65, 72. N=20.
Position of median = $\frac{20+1}{2} = 10.5^{th}$ item. The 10th item is 45, and the 11th item is 46.
Median = Average of 10th and 11th items = $\frac{45 + 46}{2} = 45.5$ marks.
Discrete Series
For discrete series, calculate cumulative frequency (cf). Find the position of the median using $\frac{N+1}{2}$ item. Locate the cf that contains this position. The corresponding variable value is the median.
Example 7. The frequency distributsion of the number of persons and their respective incomes (in Rs) are given below. Calculate the median income.
Income (in Rs): 10, 20, 30, 40
Number of persons: 2, 4, 10, 4
Answer:
| Income (in Rs) | No. of persons (f) | Cumulative frequency (cf) |
|---|---|---|
| 10 | 2 | 2 |
| 20 | 4 | 6 |
| 30 | 10 | 16 |
| 40 | 4 | 20 |
| Total | 20 (N) |
Position of median = $\frac{20+1}{2} = 10.5^{th}$ item.
The cf of 16 contains the 10.5th item. The corresponding income is Rs 30. Median income = Rs 30.
Continuous Series
For continuous series, find the cumulative frequency. Locate the median class where the $\frac{N}{2}^{th}$ item (not $\frac{N+1}{2}$) lies. Use the formula: $Median = L + \frac{\frac{N}{2} - c.f.}{f} \times h$. Where L=lower limit of median class, c.f.=cf of preceding class, f=frequency of median class, h=class interval width.
Example 8. Following data relates to daily wages of persons working in a factory. Compute the median daily wage.
Daily wages (in Rs): 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55, 55–60
Number of workers: 14, 28, 33, 30, 20, 15, 13, 7
Answer:
| Daily wages (in Rs) | No. of Workers (f) | Cumulative Frequency (cf) |
|---|---|---|
| 20–25 | 14 | 14 |
| 25–30 | 28 | 42 |
| 30–35 | 33 | 75 |
| 35–40 | 30 | 105 |
| 40–45 | 20 | 125 |
| 45–50 | 15 | 140 |
| 50–55 | 13 | 153 |
| 55–60 | 7 | 160 |
| Total | 160 (N) |
N=160. N/2 = 160/2 = 80. The 80th item lies in the cf of 105, which corresponds to the class 35–40. This is the median class.
L = 35, c.f. of preceding class = 75, f of median class = 30, h = 5.
$Median = L + \frac{\frac{N}{2} - c.f.}{f} \times h = 35 + \frac{80 - 75}{30} \times 5 = 35 + \frac{5}{30} \times 5 = 35 + \frac{25}{30} = 35 + 0.833... \approx 35.83$
The median daily wage is Rs 35.83.
Quartiles
Quartiles divide data into four equal parts. Q1 (Lower Quartile) has 25% data below it, Q2 (Median) has 50% below, and Q3 (Upper Quartile) has 75% below. Q1 and Q3 define the range of the central 50% of the data.
Percentiles
Percentiles divide data into 100 equal parts, resulting in 99 dividing positions (P1 to P99). P50 is the median.
Calculation Of Quartiles
For individual and discrete series, the positions of Q1 and Q3 in an ordered series are $\frac{N+1}{4}^{th}$ item and $\frac{3(N+1)}{4}^{th}$ item respectively.
Example 9. Calculate the value of lower quartile from the data of the marks obtained by ten students in an examination.
22, 26, 14, 30, 18, 11, 35, 41, 12, 32.
Answer:
Arranging in ascending order:
11, 12, 14, 18, 22, 26, 30, 32, 35, 41. N=10.
Position of Q1 = $\frac{10+1}{4} = 2.75^{th}$ item.
Q1 = 2nd item + 0.75 (3rd item – 2nd item) = $12 + 0.75 (14 - 12) = 12 + 0.75 \times 2 = 12 + 1.5 = 13.5$ marks.
Mode
The **Mode** (Mo) is the most frequently occurring value in a dataset. It represents the typical or most fashionable value in a distribution. It's useful for understanding preferences or common occurrences (e.g., most popular shoe size).
Computation Of Mode
Methods differ for discrete and continuous series.
Discrete Series
Identify the value with the highest frequency. This value is the mode. A dataset can have one mode (unimodal), two modes (bi-modal), or more (multi-modal). Data with no repeating values has no mode.
Example 10. Look at the following discrete series: Variable: 10, 20, 30, 40, 50; Frequency: 2, 8, 20, 10, 5. Find the mode.
Answer:
The maximum frequency is 20, which corresponds to the variable value 30. Mode = 30.
Continuous Series
Identify the modal class (the class with the largest frequency). Use the formula: $Mode = L + \frac{D_1}{D_1 + D_2} \times h$. Where L=lower limit of modal class, $D_1$=difference between frequency of modal class and frequency of preceding class, $D_2$=difference between frequency of modal class and frequency of succeeding class, h=class interval width. Class intervals should be equal and exclusive for this formula.
Example 11. Calculate the value of modal worker family’s monthly income from the following data: Less than cumulative frequency distribution of income per month (in ’000 Rs): Less than 15 (4), Less than 20 (12), Less than 25 (30), Less than 30 (60), Less than 35 (80), Less than 40 (90), Less than 45 (95), Less than 50 (97).
Answer:
Convert cumulative frequency distribution to exclusive frequency distribution:
| Income Group (in ’000 Rs) | Frequency |
|---|---|
| 10–15 | 4 |
| 15–20 | 12 – 4 = 8 |
| 20–25 | 30 – 12 = 18 |
| 25–30 | 60 – 30 = 30 |
| 30–35 | 80 – 60 = 20 |
| 35–40 | 90 – 80 = 10 |
| 40–45 | 95 – 90 = 5 |
| 45–50 | 97 – 95 = 2 |
| Total | 97 (N) |
The highest frequency is 30, corresponding to the class 25–30. This is the modal class.
L = 25, $D_1 = 30 - 18 = 12$, $D_2 = 30 - 20 = 10$, h = 5.
$Mode = L + \frac{D_1}{D_1 + D_2} \times h = 25 + \frac{12}{12 + 10} \times 5 = 25 + \frac{12}{22} \times 5 = 25 + \frac{60}{22} \approx 25 + 2.727 = 27.273$
The modal worker family's monthly income is Rs 27.273 thousand.
Relative Position Of Arithmetic Mean, Median And Mode
For a symmetrical distribution, the Arithmetic Mean, Median, and Mode are equal. For skewed distributions, their positions differ. In a positively skewed distribution, the Mean is highest, followed by Median, then Mode ($\text{Mean} > \text{Median} > \text{Mode}$). In a negatively skewed distribution, the Mode is highest, followed by Median, then Mean ($\text{Mean} < \text{Median} < \text{Mode}$). The Median is always located between the Mean and the Mode.
Conclusion
Measures of central tendency (Arithmetic Mean, Median, Mode) are vital for summarizing data and finding a single representative value. The Arithmetic Mean is widely used, simple to calculate, and considers all observations but is sensitive to extremes. The Median is less affected by extremes and suitable for skewed data or open-ended distributions. The Mode represents the most frequent value and is useful for qualitative data or identifying popular items. Median and mode can be computed graphically (Median from ogives, Mode from histogram). The choice of the appropriate average depends on the data type, distribution nature, and analysis purpose.
Recap:
- Central tendency measures summarize data with a single representative value.
- Arithmetic mean is sum of values divided by number of observations.
- Sum of deviations from A.M. is zero.
- Weighted A.M. assigns importance to items.
- Median is the middle value in an ordered dataset.
- Quartiles divide data into four equal parts; Percentiles divide into one hundred.
- Mode is the most frequent value.