Measures of Central Tendency: Mean, Median, and Mode
In statistics, measures of central tendency are single values that attempt to describe a set of data by identifying the central position within that set of data. These measures are crucial for understanding the typical or average value of a dataset. The three most common measures of central tendency are the mean, the median, and the mode. We will explore how to calculate these for both ungrouped (individual observations) and grouped (data presented in frequency tables) data.
1. Mean
The mean, often referred to as the average, is calculated by summing all the values in a dataset and then dividing by the total number of values. It is sensitive to extreme values (outliers).
1.1 Mean for Ungrouped Data
For a set of n individual observations x1, x2, ..., xn, the mean (denoted by $\bar{x}$) is calculated as:
$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$
Where:
- $\sum_{i=1}^{n} x_i$ represents the sum of all the observations.
- n is the total number of observations.
Example: Consider the marks obtained by 5 students in a quiz: 10, 15, 12, 18, 15.
Here, $n = 5$.
Sum of marks = $10 + 15 + 12 + 18 + 15 = 70$.
Mean ($\bar{x}$) = $\frac{70}{5} = 14$.
1.2 Mean for Grouped Data
For grouped data, where observations are presented in classes with corresponding frequencies, we assume that all values within a class interval are equal to the midpoint of that interval.
Let the class intervals be $c_1, c_2, ..., c_k$ and their corresponding frequencies be $f_1, f_2, ..., f_k$. The midpoint of the i-th class is denoted by $x_i$.
The mean ($\bar{x}$) for grouped data is calculated as:
$\bar{x} = \frac{\sum_{i=1}^{k} f_i x_i}{\sum_{i=1}^{k} f_i}$
Where:
- $x_i$ is the midpoint of the i-th class.
- $f_i$ is the frequency of the i-th class.
- $\sum_{i=1}^{k} f_i$ is the total number of observations (N).
Calculating Class Midpoints: The midpoint of a class interval is calculated as:
Midpoint ($x_i$) = $\frac{\text{Upper limit of the class} + \text{Lower limit of the class}}{2}$
Example: Consider the following frequency distribution of the heights of 50 students:
| Height (cm) | Number of Students (f) |
|---|---|
| 150-155 | 5 |
| 155-160 | 10 |
| 160-165 | 15 |
| 165-170 | 12 |
| 170-175 | 8 |
Calculate the midpoints ($x_i$) for each class:
- 150-155: Midpoint = $\frac{150+155}{2} = 152.5$
- 155-160: Midpoint = $\frac{155+160}{2} = 157.5$
- 160-165: Midpoint = $\frac{160+165}{2} = 162.5$
- 165-170: Midpoint = $\frac{165+170}{2} = 167.5$
- 170-175: Midpoint = $\frac{170+175}{2} = 172.5$
Now, calculate $f_i x_i$ for each class:
| Height (cm) | Midpoint ($x_i$) | Frequency ($f_i$) | $f_i x_i$ |
|---|---|---|---|
| 150-155 | 152.5 | 5 | $5 \times 152.5 = 762.5$ |
| 155-160 | 157.5 | 10 | $10 \times 157.5 = 1575$ |
| 160-165 | 162.5 | 15 | $15 \times 162.5 = 2437.5$ |
| 165-170 | 167.5 | 12 | $12 \times 167.5 = 2010$ |
| 170-175 | 172.5 | 8 | $8 \times 172.5 = 1380$ |
| Total | 50 | 8165 |
Total frequency ($\sum f_i$) = 50.
Sum of ($f_i x_i$) = 8165.
Mean ($\bar{x}$) = $\frac{8165}{50} = 163.3$ cm.
When the values of $x_i$ and $f_i$ are large, direct calculation can be tedious. The Assumed Mean method simplifies this.
1. Choose an assumed mean (a), preferably one of the $x_i$ values or a value close to the center of the data. 2. Calculate the deviation $d_i = x_i - a$ for each class. 3. Calculate $f_i d_i$ for each class. 4. The mean is given by: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$
Example (using the same height data): Let's assume $a = 162.5$.
| Height (cm) | Midpoint ($x_i$) | Frequency ($f_i$) | $d_i = x_i - 162.5$ | $f_i d_i$ |
|---|---|---|---|---|
| 150-155 | 152.5 | 5 | $152.5 - 162.5 = -10$ | $5 \times (-10) = -50$ |
| 155-160 | 157.5 | 10 | $157.5 - 162.5 = -5$ | $10 \times (-5) = -50$ |
| 160-165 | 162.5 | 15 | $162.5 - 162.5 = 0$ | $15 \times 0 = 0$ |
| 165-170 | 167.5 | 12 | $167.5 - 162.5 = 5$ | $12 \times 5 = 60$ |
| 170-175 | 172.5 | 8 | $172.5 - 162.5 = 10$ | $8 \times 10 = 80$ |
| Total | 50 | 40 |
$\sum f_i d_i = 40$
Mean ($\bar{x}$) = $162.5 + \frac{40}{50} = 162.5 + 0.8 = 163.3$ cm.
This method is useful when the class sizes are equal and the deviations ($d_i$) have a common factor.
1. Choose an assumed mean (a). 2. Calculate the class size (h). 3. Calculate the step-deviations $u_i = \frac{x_i - a}{h}$. 4. Calculate $f_i u_i$ for each class. 5. The mean is given by: $\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Example (using the same height data): $a = 162.5$, $h = 5$.
| Height (cm) | Midpoint ($x_i$) | Frequency ($f_i$) | $d_i = x_i - 162.5$ | $u_i = d_i / 5$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 150-155 | 152.5 | 5 | -10 | -2 | $5 \times (-2) = -10$ |
| 155-160 | 157.5 | 10 | -5 | -1 | $10 \times (-1) = -10$ |
| 160-165 | 162.5 | 15 | 0 | 0 | $15 \times 0 = 0$ |
| 165-170 | 167.5 | 12 | 5 | 1 | $12 \times 1 = 12$ |
| 170-175 | 172.5 | 8 | 10 | 2 | $8 \times 2 = 16$ |
| Total | 50 | 8 |
$\sum f_i u_i = 8$
Mean ($\bar{x}$) = $162.5 + 5 \left( \frac{8}{50} \right) = 162.5 + 5 \times 0.16 = 162.5 + 0.8 = 163.3$ cm.
The mean is a powerful measure but can be skewed by extreme values.
2. Median
The median is the middle value in a dataset that has been ordered from least to greatest. It divides the dataset into two equal halves, meaning 50% of the data points are below the median and 50% are above it. The median is not affected by outliers.
2.1 Median for Ungrouped Data
To find the median for ungrouped data:
- Arrange the data in ascending (or descending) order.
- If the number of observations (n) is odd, the median is the middle value, which is the $\left(\frac{n+1}{2}\right)$-th observation.
- If the number of observations (n) is even, the median is the average of the two middle values, which are the $\left(\frac{n}{2}\right)$-th and $\left(\frac{n}{2}+1\right)$-th observations.
Example 1 (Odd number of observations): Consider the dataset: 5, 2, 8, 1, 9.
1. Arrange in order: 1, 2, 5, 8, 9.
2. Here, $n=5$ (odd). The median is the $\left(\frac{5+1}{2}\right) = 3$rd observation.
3. Median = 5.
Example 2 (Even number of observations): Consider the dataset: 6, 2, 9, 5, 7, 4.
1. Arrange in order: 2, 4, 5, 6, 7, 9.
2. Here, $n=6$ (even). The median is the average of the $\left(\frac{6}{2}\right) = 3$rd and $\left(\frac{6}{2}+1\right) = 4$th observations.
3. The 3rd observation is 5, and the 4th observation is 6.
4. Median = $\frac{5 + 6}{2} = \frac{11}{2} = 5.5$.
2.2 Median for Grouped Data
For grouped data, the median lies in the class interval where the cumulative frequency first exceeds $\frac{N}{2}$, where $N = \sum f_i$ is the total frequency. This class is called the median class.
The formula for the median of grouped data is:
Median = $L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h$
Where:
- $L$ is the lower limit of the median class.
- $N$ is the total frequency ($\sum f_i$).
- $CF$ is the cumulative frequency of the class preceding the median class.
- $f$ is the frequency of the median class.
- $h$ is the class size (width of the median class).
Example: Using the same height data distribution:
| Height (cm) | Frequency ($f_i$) | Cumulative Frequency ($CF$) |
|---|---|---|
| 150-155 | 5 | 5 |
| 155-160 | 10 | 15 |
| 160-165 | 15 | 30 |
| 165-170 | 12 | 42 |
| 170-175 | 8 | 50 |
| Total | 50 |
1. Total frequency $N = 50$.
2. Calculate $\frac{N}{2} = \frac{50}{2} = 25$.
3. Find the median class: We look for the class where the cumulative frequency first exceeds 25. This is the 165-170 class, as its cumulative frequency is 42.
4. Identify the values for the formula:
- $L$ (lower limit of the median class) = 165
- $N = 50$
- $CF$ (cumulative frequency of the class before 165-170, which is 160-165) = 30
- $f$ (frequency of the median class 165-170) = 12
- $h$ (class size) = $160-155 = 5$ (or $165-160=5$, etc.)
5. Apply the formula:
Median = $165 + \left( \frac{25 - 30}{12} \right) \times 5$
Median = $165 + \left( \frac{-5}{12} \right) \times 5$
Median = $165 - \frac{25}{12}$
Median = $165 - 2.0833...$
Median $\approx 162.917$ cm.
The median is a robust measure, unaffected by extreme values, making it suitable for skewed distributions.
3. Mode
The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), two modes (bimodal), or more (multimodal). For ungrouped data, it's simply the most repeated value. For grouped data, we find the modal class and then calculate the mode.
3.1 Mode for Ungrouped Data
For ungrouped data, the mode is the observation that occurs with the highest frequency.
Example 1 (Unimodal): Dataset: 2, 5, 3, 5, 7, 5, 8.
The value 5 appears 3 times, which is more than any other value. So, the mode is 5.
Example 2 (Bimodal): Dataset: 1, 2, 2, 3, 4, 4, 5.
The values 2 and 4 both appear twice, which is the highest frequency. So, the modes are 2 and 4.
Example 3 (No Mode): Dataset: 1, 2, 3, 4, 5.
All values appear once. There is no mode.
3.2 Mode for Grouped Data
For grouped data, the modal class is the class interval with the highest frequency. The formula to calculate the mode is:
Mode = $L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
- $L$ is the lower limit of the modal class.
- $f_1$ is the frequency of the modal class.
- $f_0$ is the frequency of the class preceding the modal class.
- $f_2$ is the frequency of the class succeeding the modal class.
- $h$ is the class size.
Example: Using the same height data distribution:
| Height (cm) | Frequency ($f_i$) |
|---|---|
| 150-155 | 5 |
| 155-160 | 10 |
| 160-165 | 15 |
| 165-170 | 12 |
| 170-175 | 8 |
1. Identify the modal class: The highest frequency is 15, which corresponds to the 160-165 class. So, the modal class is 160-165.
2. Identify the values for the formula:
- $L$ (lower limit of the modal class) = 160
- $f_1$ (frequency of the modal class 160-165) = 15
- $f_0$ (frequency of the preceding class 155-160) = 10
- $f_2$ (frequency of the succeeding class 165-170) = 12
- $h$ (class size) = 5
3. Apply the formula:
Mode = $160 + \left( \frac{15 - 10}{2 \times 15 - 10 - 12} \right) \times 5$
Mode = $160 + \left( \frac{5}{30 - 10 - 12} \right) \times 5$
Mode = $160 + \left( \frac{5}{8} \right) \times 5$
Mode = $160 + \frac{25}{8}$
Mode = $160 + 3.125$
Mode = $163.125$ cm.
For moderately asymmetrical distributions, there is an approximate relationship:
Mean - Mode $\approx$ 3 (Mean - Median)
This relationship can be useful for estimating one measure if the other two are known. Rearranging it, we can also write:
Mode $\approx$ 3 Median - 2 Mean
Median $\approx$ $\frac{2 \text{Mean} + \text{Mode}}{3}$
Note: This is an approximation and holds best for unimodal, moderately skewed distributions. It may not be accurate for highly skewed or multimodal distributions.
Summary Table
| Measure | Ungrouped Data | Grouped Data | Sensitivity to Outliers |
|---|---|---|---|
| Mean | $\bar{x} = \frac{\sum x_i}{n}$ | $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ (or Assumed Mean/Step-Deviation methods) | High |
| Median | Middle value of ordered data ($\frac{n+1}{2}$-th or average of $\frac{n}{2}$-th and $(\frac{n}{2}+1)$-th) | $L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h$ | Low |
| Mode | Most frequent value | $L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$ | Very Low / None |
Understanding these measures of central tendency is fundamental to descriptive statistics. They provide a concise summary of a dataset's typical value, allowing for easier comparison and interpretation of data.