Measures of Central Tendency – Full Chapter Guide for Ethiopian Students
Hello dear students! 🙂 Are you preparing for your Statistics exam? Welcome! In this lesson, we’ll walk through Chapter 3: Measures of Central Tendency from your Stat 173 lecture notes. We’ll go step by step—nothing rushed—so you understand every idea deeply. This chapter is super important for your exams, so stick with me! 😊
If you’re using notes like Ethio Temari, you’ll find this matches perfectly. Also, you can join our study group on Telegram or follow us on YouTube and TikTok for daily tips.
What Is Central Tendency?
Imagine you and your friends took a quiz. One friend scored 90, another 30, and you got 65. If someone asks, “How did your group do?” you wouldn’t list all three numbers—you’d probably say something like, “About 60.” That single number that best represents the group? That’s what we call a measure of central tendency.
There are three main types:
- Mean – the average
- Median – the middle value
- Mode – the most frequent value
Each one has special situations where it works best. Let’s explore them in detail!
The Arithmetic Mean
If your scores in five quizzes are: 82, 76, 90, 85, and 77, then your mean score is:
\[ \bar{X} = \frac{82 + 76 + 90 + 85 + 77}{5} = \frac{410}{5} = 82 \]
In general, for values \(X_1, X_2, …, X_n\), the mean is:
\[ \bar{X} = \frac{\sum_{i=1}^{n} X_i}{n} \]
A) 65
B) 68.33
C) 70
D) 72
Calculation: (58 + 72 + 64 + 80 + 66 + 70) ÷ 6 = 410 ÷ 6 ≈ 68.33
Mean for Grouped Data
When data is grouped into classes (like 10–20, 20–30), we use the class mark (midpoint) and frequency.
Formula:
\[ \bar{X} = \frac{\sum f_i X_i}{\sum f_i} \]
| Age Group | Frequency (f) | Class Mark (X) | f × X |
|---|---|---|---|
| 6–10 | 35 | 8 | 280 |
| 11–15 | 23 | 13 | 299 |
| 16–20 | 15 | 18 | 270 |
| 21–25 | 12 | 23 | 276 |
| 26–30 | 9 | 28 | 252 |
| 31–35 | 6 | 33 | 198 |
| Total | 100 | 1575 |
A) 30
B) 40
C) 50
D) 60
Because \(\bar{X} = 3200 ÷ 80 = 40\)
Special Properties of the Arithmetic Mean
- The sum of deviations from the mean is always zero: \(\sum (X_i – \bar{X}) = 0\)
- It’s affected by extreme values (like one student scoring 10 and others 80–90)
- It’s used in further calculations (like variance)
The Mode
Example: In the list 5, 3, 5, 8, 9 → mode = 5 (appears twice).
But what if no number repeats? → No mode.
What if two numbers repeat equally? → Bimodal (e.g., 2, 2, 5, 5, 7).
Mode for Grouped Data
Use this formula:
\[ \hat{X} = L + \left( \frac{\Delta_1}{\Delta_1 + \Delta_2} \right) w \]
Where:
- \(L\) = lower boundary of modal class (class with highest frequency)
- \(\Delta_1 = f_m – f_{m-1}\)
- \(\Delta_2 = f_m – f_{m+1}\)
- \(w\) = class width
| Size (hectares) | Frequency |
|---|---|
| 5–15 | 8 |
| 15–25 | 12 |
| 25–35 | 17 |
| 35–45 | 29 |
| 45–55 | 31 |
| 55–65 | 5 |
| 65–75 | 3 |
\(L = 45\), \(f_m = 31\), \(f_{m-1} = 29\), \(f_{m+1} = 5\), \(w = 10\)
\(\Delta_1 = 31 – 29 = 2\), \(\Delta_2 = 31 – 5 = 26\)
\[ \hat{X} = 45 + \left( \frac{2}{2 + 26} \right) \times 10 = 45 + \frac{20}{28} \approx 45.71 \]
A) 22.5
B) 24.0
C) 25.0
D) 26.25
Because:
\(\Delta_1 = 18 – 12 = 6\), \(\Delta_2 = 18 – 10 = 8\), \(w = 10\), \(L = 20\)
\(\hat{X} = 20 + (6 / (6 + 8)) × 10 = 20 + 60/14 ≈ 24.29\)
Wait! That’s not matching? Let’s check again…
Actually, 60/14 ≈ 4.29 → 20 + 4.29 = 24.29 → so none match perfectly. But if options are approximate, B (24.0) is closest. However, if the question had different numbers…
Correction: If modal class is 20–30, \(L = 20\), not 25. So answer ≈ 24.3 → **B) 24.0** is best.
The Median
If there are odd values: median = middle number.
If even: median = average of two middle numbers.
Sorted: 2, 4, 5, 6, 8, 9 → n = 6 (even)
Median = (5 + 6)/2 = 5.5
Median for Grouped Data
Formula:
\[ \tilde{X} = L + \left( \frac{\frac{n}{2} – c}{f} \right) w \]
Where:
- \(L\) = lower boundary of median class
- \(n\) = total frequency
- \(c\) = cumulative frequency before median class
- \(f\) = frequency of median class
- \(w\) = class width
| Class | Frequency | Cumulative Freq |
|---|---|---|
| 40–44 | 7 | 7 |
| 45–49 | 10 | 17 |
| 50–54 | 22 | 39 |
| 55–59 | 15 | 54 |
| 60–64 | 12 | 66 |
| 65–69 | 6 | 72 |
| 70–74 | 3 | 75 |
Median class = 50–54 → \(L = 49.5\), \(c = 17\), \(f = 22\), \(w = 5\)
\[ \tilde{X} = 49.5 + \left( \frac{37.5 – 17}{22} \right) \times 5 = 49.5 + \frac{20.5}{22} \times 5 \approx 49.5 + 4.66 = 54.16 \] Wait! But class is 50–54, so should be around 54? Actually, **lower boundary is 49.5**, yes.
Correct: \(\tilde{X} \approx 54.16\) → but that’s above class! Mistake? No—median can be estimated beyond if distribution is skewed.
**Actual calculation from notes**: \(L = 49.5\), so:
\[ \tilde{X} = 49.5 + \left( \frac{37.5 – 17}{22} \right) \times 5 = 49.5 + \left( \frac{20.5}{22} \right) \times 5 \approx 49.5 + 4.66 = 54.16 \] But class 50–54 goes up to 54.5, so 54.16 is valid.
A) 62
B) 64
C) 66
D) 68
\(n/2 = 50\)
\(\tilde{X} = 60 + ((50 – 35)/25) × 10 = 60 + (15/25)×10 = 60 + 6 = 66\)? Wait—
15/25 = 0.6 → 0.6 × 10 = 6 → 60 + 6 = 66 → **C) 66**.
Correction: Answer is C) 66
Quantiles: Quartiles, Deciles, Percentiles
These divide data into equal parts:
- Quartiles (Q): 4 parts → Q₁ (25%), Q₂ (50% = median), Q₃ (75%)
- Deciles (D): 10 parts → D₁ to D₉
- Percentiles (P): 100 parts → P₁ to P₉₉
Formula for grouped data (e.g., Q₁):
\[ Q_1 = L + \left( \frac{\frac{n}{4} – c}{f} \right) w \]
Q₁ class: where cumulative freq ≥ 493/4 = 123.25 → class 170–180
\(L = 170\), \(c = 88\), \(f = 72\), \(w = 10\)
\[ Q_1 = 170 + \left( \frac{123.25 – 88}{72} \right) \times 10 = 170 + \frac{35.25}{72} \times 10 \approx 170 + 4.90 = 174.90 \]
Weighted Mean
When some values matter more than others (like exam scores with different weights).
Formula:
\[ \bar{X}_w = \frac{\sum W_i X_i}{\sum W_i} \]
English 60 (weight 1), Biology 75 (weight 2), Math 63 (1), Physics 59 (3), Chemistry 55 (3)
\[ \bar{X}_w = \frac{(1×60) + (2×75) + (1×63) + (3×59) + (3×55)}{1+2+1+3+3} = \frac{615}{10} = 61.5 \]
Geometric and Harmonic Means
Geometric Mean (GM)
Used for growth rates, ratios:
\[ GM = \sqrt[n]{X_1 \cdot X_2 \cdot \ldots \cdot X_n} \]
Or using logs:
\[ \log(GM) = \frac{\sum \log X_i}{n} \]
Harmonic Mean (HM)
Used for average speed, rates:
\[ HM = \frac{n}{\sum \frac{1}{X_i}} \]
\[ HM = \frac{2}{\frac{1}{10} + \frac{1}{15}} = \frac{2}{\frac{3+2}{30}} = \frac{2}{5/30} = \frac{2 \times 30}{5} = 12 \text{ km/h} \]
Exercises with Answers
Exercise 1 (From Notes)
Marks of 75 students:
| Marks | No. of Students |
|---|---|
| 40–44 | 7 |
| 45–49 | 10 |
| 50–54 | 22 |
| 55–59 | f₄ |
| 60–64 | f₅ |
| 65–69 | 6 |
| 70–74 | 3 |
Given: 20% of students scored between 55–59.
(i) Find f₄ and f₅
20% of 75 = 15 → f₄ = 15
Total = 75 → 7 + 10 + 22 + 15 + f₅ + 6 + 3 = 75 → 63 + f₅ = 75 → f₅ = 12
(ii) Find the mean
| Class | f | X | fX |
|---|---|---|---|
| 40–44 | 7 | 42 | 294 |
| 45–49 | 10 | 47 | 470 |
| 50–54 | 22 | 52 | 1144 |
| 55–59 | 15 | 57 | 855 |
| 60–64 | 12 | 62 | 744 |
| 65–69 | 6 | 67 | 402 |
| 70–74 | 3 | 72 | 216 |
| Total | 75 | 4125 |
\[ \bar{X} = \frac{4125}{75} = 55 \]
Exercise 2
Mean weight of 10 students = 65 kg. One weight was misread as 40 instead of 80. Find correct mean.
\[ \text{Correct Mean} = 65 + \frac{80 – 40}{10} = 65 + 4 = 69 \text{ kg} \]
Why This Matters
Understanding central tendency helps you summarize data quickly—whether it’s farm yields, exam scores, or business profits. On your exam, expect questions on:
- Calculating mean, median, mode
- Interpreting which measure is best
- Solving word problems with real data
Need more practice? Visit Ethio Temari or watch our YouTube channel. And hey—share this with a friend who’s struggling! 😃
Keep practicing, and you’ll ace this chapter! 💪