Statistics • Topic 1 of 3

Mean, Median, Mode of Grouped Data — Bimodal Excluded

What are mean, median, and mode for grouped data? When data is organized into frequency distributions (class intervals with frequencies), we use special formulas to find the three measures of central tendency.

1. Mean (Arithmetic Average) for grouped data: There are three methods to find the mean:

Direct method: Mean = (Σ fᵢxᵢ) / (Σ fᵢ), where xᵢ is the class mark (midpoint) = (upper limit + lower limit)/2
Assumed mean method: Mean = a + (Σ fᵢdᵢ) / (Σ fᵢ), where dᵢ = xᵢ − a (a = assumed mean)
Step deviation method: Mean = a + (Σ fᵢuᵢ) / (Σ fᵢ) × h, where uᵢ = (xᵢ − a)/h, h = class width

2. Median for grouped data: Median is the middle value. For grouped data:

Median = L + [ (N/2 − cf) / f ] × h

where:

L = lower limit of median class
N = total frequency (Σ fᵢ)
cf = cumulative frequency of the class before median class
f = frequency of median class
h = class width

3. Mode for grouped data (bimodal excluded): Mode is the most frequent value. For grouped data:

Mode = L + [ ($f_{1} - f_{0}$) / ($2f_{1} - f_{0} - f_{2}$) ] × h

where:

L = lower limit of modal class (highest frequency 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

Empirical relationship: For moderately skewed data: Mode ≈ 3 Median − 2 Mean

┌─────────────────────────────────────────────────────────────┐
│      MEAN, MEDIAN, MODE FOR GROUPED DATA - FORMULA MAP       │
└─────────────────────────────────────────────────────────────┘

SAMPLE DATA TABLE:

┌──────────────┬─────────┬──────────┬────────────┬──────────┐
│ Class Interval│ Midpoint│Frequency│  fᵢ × xᵢ   │Cumulative│
│               │   (xᵢ)  │   (fᵢ)  │            │ Frequency│
├──────────────┼─────────┼──────────┼────────────┼──────────┤
│    0-10      │    5    │    2     │     10     │    2     │
│   10-20      │   15    │    5     │     75     │    7     │
│   20-30      │   25    │    8     │    200     │   15     │
│   30-40      │   35    │    4     │    140     │   19     │
│   40-50      │   45    │    1     │     45     │   20     │
├──────────────┼─────────┼──────────┼────────────┼──────────┤
│    Total     │         │   N=20   │   Σfᵢxᵢ=470 │         │
└──────────────┴─────────┴──────────┴────────────┴──────────┘

MEAN = 470/20 = 23.5


FINDING MEDIAN CLASS (N/2 = 10):

    cf just greater than 10 is 15 → median class = 20-30
    L = 20, cf = 7, f = 8, h = 10
    
    Median = 20 + [(10-7)/8] × 10 = 20 + (3/8)×10 = 20 + 3.75 = 23.75


FINDING MODAL CLASS (highest frequency = 8 → 20-30):

    L = 20, f₁=8, f₀=5, f₂=4, h=10
    
    Mode = 20 + [(8-5)/(16-5-4)] × 10 = 20 + [3/7]×10 = 20 + 4.29 = 24.29


COMPARISON ON NUMBER LINE:

    <--- Mean=23.5 --- Median=23.75 --- Mode=24.29 --->
    
    For this positively skewed distribution: Mean < Median < Mode


DIRECT METHOD FLOWCHART:

    Class Intervals → Find Midpoints (xᵢ)
              │
              ▼
    Multiply each xᵢ by frequency fᵢ → fᵢxᵢ
              │
              ▼
    Sum all fᵢxᵢ and sum all fᵢ
              │
              ▼
    Mean = Σfᵢxᵢ / Σfᵢ

Solved Examples

Worked Example

Find the mean for the following data using the direct method: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |-------|------|-------|-------|-------|-------| | Frequency | 3 | 5 | 7 | 4 | 1 |

Solution

Step 1: Find midpoints (xᵢ): 5, 15, 25, 35, 45
Step 2: Calculate fᵢxᵢ: 3×5=15, 5×15=75, 7×25=175, 4×35=140, 1×45=45
Step 3: Σfᵢ = 3+5+7+4+1 = 20
Step 4: Σfᵢxᵢ = 15+75+175+140+45 = 450
Step 5: Mean = 450/20 = 22.5

Answer: Mean = 22.5

Worked Example

Find the median for the data in Example 1.

Solution

Step 1: N = 20, N/2 = 10
Step 2: Cumulative frequencies: 3, 8, 15, 19, 20
Step 3: Median class = class where cf ≥ 10 → 20-30
Step 4: L = 20, cf = 8, f = 7, h = 10
Step 5: Median = 20 + [(10-8)/7] × 10 = 20 + (2/7)×10 = 20 + 2.86 = 22.86

Answer: Median = 22.86

Worked Example

Find the mode for the data in Example 1.

Solution

Step 1: Highest frequency = 7 → modal class = 20-30
Step 2: L = 20, $f_{1}$=7, $f_{0}$=5, $f_{2}$=4, h=10
Step 3: Mode = 20 + [(7-5)/(2×7 - 5 - 4)] × 10
Step 4: = 20 + [2/(14-9)] × 10 = 20 + (2/5)×10 = 20 + 4 = 24

Answer: Mode = 24

Key Points

Mean = Σfᵢxᵢ / Σfᵢ (xᵢ = class midpoint)
Median uses cumulative frequency: Median = L + [(N/2 − cf)/f] × h
Mode uses frequencies of modal and adjacent classes: Mode = L + [($f_{1}-f_{0}$)/($2f_{1}-f_{0}-f_{2}$)] × h
Modal class = class with highest frequency
Median class = class where cumulative frequency first exceeds N/2
Empirical relation: Mode ≈ 3 Median − 2 Mean (not exact but useful for checking)

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The mean of grouped data is:

Explanation: Weighted average.

Q2.The empirical relation is:

Explanation: $\text{Mode}=3\,\text{Median}-2\,\text{Mean}$.

Q3.The class with the highest frequency is the:

Explanation: Modal class.

Q4.The mid-value of a class is the:

Explanation: Class mark $=$ midpoint.

Practice more MCQs → 📝 Assignment · 35 marks