Learning Objectives
By the end of this chapter, you will be able to:
- Calculate arithmetic mean for individual, discrete, and continuous data
- Compute weighted mean when items have different importance
- Determine median for both ungrouped and grouped data
- Identify mode in various types of distributions
- Choose the appropriate measure for different situations
What is Central Tendency?
A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set. It represents the “typical” or “average” value.
Public administrators need that single summary constantly: a municipality briefing may report “typical” citizen wait time, a ward office may quote “average” monthly service applications, or a ministry may cite a representative salary figure in NPR. One number is easy to communicate in memos and exams—but choose carefully. A few extreme values (one very high contract, one very long complaint delay) can pull the mean away from what most citizens or staff experience.
mindmap
root("(Measures of Central Tendency"))
Arithmetic Mean
Simple Mean
Weighted Mean
Median
Middle Value
Position Based
Mode
Most Frequent
Can be Multiple
1. Arithmetic Mean
The arithmetic mean (commonly called “average”) is the sum of all values divided by the number of values.
Formula for Individual Data (Ungrouped)
\[\bar{x} = \frac{\sum x_i}{n} = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}\]Where:
- $\bar{x}$ = sample mean
- $\sum x_i$ = sum of all observations
- $n$ = number of observations
Step-by-Step Example 1: Simple Mean
Problem: Calculate the mean salary (in thousands) of 8 government officers: 45, 52, 48, 55, 50, 47, 53, 50
Solution:
Step 1: List all values
- $x_1 = 45$, $x_2 = 52$, $x_3 = 48$, $x_4 = 55$
- $x_5 = 50$, $x_6 = 47$, $x_7 = 53$, $x_8 = 50$
Step 2: Calculate the sum \(\sum x = 45 + 52 + 48 + 55 + 50 + 47 + 53 + 50 = 400\)
Step 3: Count observations \(n = 8\)
Step 4: Apply the formula \(\bar{x} = \frac{400}{8} = 50\)
Answer: The mean salary is 50 thousand (NPR 50,000)
Mean for Discrete Frequency Distribution
When data is presented with frequencies, use:
\[\bar{x} = \frac{\sum fx}{n} = \frac{\sum fx}{\sum f}\]Where:
- $f$ = frequency of each value
- $x$ = value
- $n = \sum f$ = total frequency
Step-by-Step Example 2: Discrete Data
Problem: Calculate the mean number of training programs attended by employees:
| Training Programs (x) | Number of Employees (f) |
|---|---|
| 0 | 5 |
| 1 | 12 |
| 2 | 18 |
| 3 | 10 |
| 4 | 5 |
Solution:
Step 1: Create a calculation table
| $x$ | $f$ | $fx$ |
|---|---|---|
| 0 | 5 | 0 |
| 1 | 12 | 12 |
| 2 | 18 | 36 |
| 3 | 10 | 30 |
| 4 | 5 | 20 |
| Total | $\sum f = 50$ | $\sum fx = 98$ |
Step 2: Apply the formula
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{98}{50} = 1.96\]Answer: The mean number of training programs is 1.96 ≈ 2 programs
Mean for Continuous (Grouped) Frequency Distribution
For grouped data with class intervals:
\[\bar{x} = \frac{\sum fm}{n}\]Where $m$ = mid-point of each class = $\frac{\text{Lower limit + Upper limit}}{2}$
Step-by-Step Example 3: Grouped Data
Problem: Calculate the mean monthly income of government employees:
| Monthly Income (NPR ‘000) | Number of Employees |
|---|---|
| 20-30 | 8 |
| 30-40 | 15 |
| 40-50 | 25 |
| 50-60 | 18 |
| 60-70 | 10 |
| 70-80 | 4 |
Solution:
Step 1: Find mid-points and create calculation table
| Class | $f$ | Mid-point ($m$) | $fm$ |
|---|---|---|---|
| 20-30 | 8 | 25 | 200 |
| 30-40 | 15 | 35 | 525 |
| 40-50 | 25 | 45 | 1125 |
| 50-60 | 18 | 55 | 990 |
| 60-70 | 10 | 65 | 650 |
| 70-80 | 4 | 75 | 300 |
| Total | 80 | 3790 |
Step 2: Calculate mid-points
- Class 20-30: $m = \frac{20+30}{2} = 25$
- Class 30-40: $m = \frac{30+40}{2} = 35$
- And so on…
Step 3: Apply the formula
\[\bar{x} = \frac{\sum fm}{\sum f} = \frac{3790}{80} = 47.375\]Answer: The mean monthly income is NPR 47,375
2. Weighted Mean
The weighted mean is used when different values have different levels of importance (weights).
Formula
\[\bar{x}_w = \frac{\sum wx}{\sum w}\]Where:
- $w$ = weight assigned to each value
- $x$ = value
When to Use Weighted Mean
flowchart TD
A{"Are all items<br/>equally important?"}
A -->|Yes| B[Use Simple Mean]
A -->|No| C[Use Weighted Mean]
C --> D["Examples:"]
D --> D1["GPA calculation"]
D --> D2["Price indices"]
D --> D3["Composite scores"]
D --> D4["Performance ratings"]
Step-by-Step Example 4: Weighted Mean
Problem: Calculate the weighted average performance score for a department:
| Performance Area | Score (x) | Weight (w) |
|---|---|---|
| Work Quality | 85 | 3 |
| Punctuality | 90 | 2 |
| Teamwork | 75 | 2 |
| Initiative | 80 | 1 |
| Communication | 78 | 2 |
Solution:
Step 1: Create calculation table
| Area | $x$ | $w$ | $wx$ |
|---|---|---|---|
| Work Quality | 85 | 3 | 255 |
| Punctuality | 90 | 2 | 180 |
| Teamwork | 75 | 2 | 150 |
| Initiative | 80 | 1 | 80 |
| Communication | 78 | 2 | 156 |
| Total | $\sum w = 10$ | $\sum wx = 821$ |
Step 2: Apply the formula
\[\bar{x}_w = \frac{\sum wx}{\sum w} = \frac{821}{10} = 82.1\]Answer: The weighted average performance score is 82.1
Note: Simple mean would be $\frac{85+90+75+80+78}{5} = 81.6$, which doesn’t account for the higher importance of Work Quality.
3. Median
The median is the middle value when data is arranged in order. It divides the distribution into two equal halves.
Key Properties
- Not affected by extreme values (outliers)
- Best measure for skewed distributions
- Requires data to be ordered
Formula for Ungrouped Data
Step 1: Arrange data in ascending order
Step 2: Find position of median
\[\text{Median Position} = \frac{n+1}{2}\]Step 3:
- If $n$ is odd: Median is the middle value
- If $n$ is even: Median is average of two middle values
Step-by-Step Example 5: Median (Odd n)
Problem: Find the median of these efficiency scores: 72, 85, 68, 91, 76, 80, 74
Solution:
Step 1: Arrange in ascending order \(68, 72, 74, 76, 80, 85, 91\)
Step 2: Find position ($n = 7$, odd) \(\text{Position} = \frac{7+1}{2} = 4^{th} \text{ position}\)
Step 3: Identify the median Counting: 68(1), 72(2), 74(3), 76(4), 80(5), 85(6), 91(7)
Answer: Median = 76
Step-by-Step Example 6: Median (Even n)
Problem: Find the median: 45, 52, 38, 61, 55, 49, 58, 42
Solution:
Step 1: Arrange in ascending order \(38, 42, 45, 49, 52, 55, 58, 61\)
Step 2: Find position ($n = 8$, even) \(\text{Position} = \frac{8+1}{2} = 4.5^{th} \text{ position}\)
Step 3: Average of 4th and 5th values
- 4th value = 49
- 5th value = 52
Answer: Median = 50.5
Median for Grouped Data
\[\text{Median} = L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\]Where:
- $L$ = Lower boundary of median class
- $n$ = Total frequency
- $cf$ = Cumulative frequency before median class
- $f$ = Frequency of median class
- $h$ = Class width
Step-by-Step Example 7: Median for Grouped Data
Problem: Find median from this salary distribution:
| Salary (NPR ‘000) | Frequency |
|---|---|
| 20-30 | 5 |
| 30-40 | 10 |
| 40-50 | 20 |
| 50-60 | 15 |
| 60-70 | 10 |
Solution:
Step 1: Calculate cumulative frequency
| Class | $f$ | Cumulative Frequency ($cf$) |
|---|---|---|
| 20-30 | 5 | 5 |
| 30-40 | 10 | 15 |
| 40-50 | 20 | 35 |
| 50-60 | 15 | 50 |
| 60-70 | 10 | 60 |
| Total | 60 |
Step 2: Find $\frac{n}{2}$ \(\frac{n}{2} = \frac{60}{2} = 30\)
Step 3: Identify median class
- $cf$ just exceeds 30 at class 40-50 (cf = 35)
- Median class = 40-50
Step 4: Identify values
- $L = 40$ (lower boundary)
- $cf = 15$ (cumulative frequency before median class)
- $f = 20$ (frequency of median class)
- $h = 10$ (class width)
Step 5: Apply the formula
\[\text{Median} = 40 + \left(\frac{30 - 15}{20}\right) \times 10\] \[= 40 + \left(\frac{15}{20}\right) \times 10\] \[= 40 + 0.75 \times 10 = 40 + 7.5 = 47.5\]Answer: Median salary = NPR 47,500
4. Mode
The mode is the value that occurs most frequently in a dataset.
Types of Modal Distributions
flowchart LR
A[Mode Types] --> B["Unimodal<br/>One mode"]
A --> C["Bimodal<br/>Two modes"]
A --> D["Multimodal<br/>More than two"]
A --> E["No Mode<br/>All values equal frequency"]
Step-by-Step Example 8: Mode (Ungrouped Data)
Problem: Find the mode: 5, 7, 8, 5, 9, 5, 10, 7, 5, 8
Solution:
Step 1: Count frequency of each value
| Value | Frequency |
|---|---|
| 5 | 4 |
| 7 | 2 |
| 8 | 2 |
| 9 | 1 |
| 10 | 1 |
Step 2: Identify highest frequency
- Highest frequency = 4 (for value 5)
Answer: Mode = 5
Mode for Grouped Data
\[\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\]Where:
- $L$ = Lower boundary of modal class
- $f_1$ = Frequency of modal class
- $f_0$ = Frequency of class before modal class
- $f_2$ = Frequency of class after modal class
- $h$ = Class width
Step-by-Step Example 9: Mode for Grouped Data
Problem: Find the mode from this age distribution of employees:
| Age Group | Frequency |
|---|---|
| 20-25 | 5 |
| 25-30 | 12 |
| 30-35 | 28 |
| 35-40 | 20 |
| 40-45 | 10 |
| 45-50 | 5 |
Solution:
Step 1: Identify modal class
- Highest frequency = 28 (for class 30-35)
- Modal class = 30-35
Step 2: Identify values
- $L = 30$
- $f_1 = 28$ (modal class frequency)
- $f_0 = 12$ (frequency before)
- $f_2 = 20$ (frequency after)
- $h = 5$
Step 3: Apply the formula
\[\text{Mode} = 30 + \left(\frac{28 - 12}{2(28) - 12 - 20}\right) \times 5\] \[= 30 + \left(\frac{16}{56 - 32}\right) \times 5\] \[= 30 + \left(\frac{16}{24}\right) \times 5\] \[= 30 + 0.667 \times 5 = 30 + 3.33 = 33.33\]Answer: Mode = 33.33 years
Relationship Between Mean, Median, and Mode
In a Symmetrical Distribution
\[\text{Mean} = \text{Median} = \text{Mode}\]In a Skewed Distribution
flowchart LR
subgraph "Positively Skewed (Right)"
A1[Mode] --> A2[Median] --> A3[Mean]
end
subgraph "Negatively Skewed (Left)"
B1[Mean] --> B2[Median] --> B3[Mode]
end
Empirical Relationship (for moderately skewed distributions):
\[\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}\]Or equivalently:
\[\text{Mean} - \text{Mode} = 3(\text{Mean} - \text{Median})\]Choosing the Right Measure
| Situation | Best Measure | Reason |
|---|---|---|
| Symmetrical distribution | Mean | Uses all data |
| Skewed distribution | Median | Not affected by outliers |
| Categorical/nominal data | Mode | Only applicable measure |
| Presence of extreme values | Median | Robust to outliers |
| Need for further calculations | Mean | Algebraic properties |
| Open-ended classes | Median or Mode | Mean cannot be calculated precisely |
PA decision tip: Use the mean for roughly symmetric salary scales or balanced performance scores when you need further calculations. Prefer the median for household income, landholding size, or service times that are right-skewed—one wealthy household or one extreme delay should not redefine “typical.” Use the mode for the most common complaint category or preferred service channel, not as a substitute for median income.
Summary of Formulas
Mean Formulas
| Data Type | Formula |
|---|---|
| Individual | $\bar{x} = \frac{\sum x}{n}$ |
| Discrete | $\bar{x} = \frac{\sum fx}{\sum f}$ |
| Grouped | $\bar{x} = \frac{\sum fm}{\sum f}$ |
| Weighted | $\bar{x}_w = \frac{\sum wx}{\sum w}$ |
Median Formula (Grouped)
\[\text{Median} = L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\]Mode Formula (Grouped)
\[\text{Mode} = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\]Summary
Central tendency answers “what is typical?” with one number. The mean uses every observation and supports further algebra; the median resists outliers and suits skewed PA data such as incomes and wait times; the mode highlights the most frequent category or peak. Match the measure to the distribution and to the audience of your report.
Common Mistakes
- Reporting the mean for heavily skewed data. A few high NPR salaries or one long service delay can inflate the mean so it no longer describes most cases—use the median.
- Treating mode as “middle.” Mode is the most frequent value, not the halfway point; it can sit far from the median in skewed distributions.
- Forgetting to order data before finding the median. Median requires sorted values (or a proper cumulative-frequency setup for grouped data).
Practice Problems
Problem 1
Calculate the mean, median, and mode for: 15, 18, 22, 18, 25, 18, 30, 28, 22
Answer / solution sketch
Sorted: 15, 18, 18, 18, 22, 22, 25, 28, 30. - Mean $= 196/9\approx 21.78$ - Median $= 22$ (5th value) - Mode $= 18$ (appears 3 times)Problem 2
The following table shows weekly wages of workers. Calculate mean and median:
| Wages (NPR) | Workers |
|---|---|
| 100-200 | 8 |
| 200-300 | 14 |
| 300-400 | 23 |
| 400-500 | 20 |
| 500-600 | 10 |
Answer / solution sketch
$N=75$; mid-points 150, 250, 350, 450, 550. $$\bar{x}=\frac{8(150)+14(250)+23(350)+20(450)+10(550)}{75}=\frac{27{,}250}{75}\approx 363.33$$ Median class 300–400 ($N/2=37.5$; cf before = 22): $$Med=300+\frac{37.5-22}{23}\times 100\approx 367.39$$ (Mode, if needed: modal class 300–400 → $Mode\approx 375$.)Problem 3
A civil servant’s performance is evaluated on three criteria with different weights. Calculate the weighted mean:
| Criterion | Score | Weight |
|---|---|---|
| Technical Skills | 88 | 40% |
| Leadership | 75 | 35% |
| Communication | 82 | 25% |
Answer / solution sketch
$$\bar{x}_w=0.40(88)+0.35(75)+0.25(82)=35.2+26.25+20.5=\mathbf{81.95}$$Next Chapter
In the next chapter, we will study Measures of Dispersion - understanding how spread out our data is using variance, standard deviation, and coefficient of variation.
