Five Number Summary Calculator
Use this unified calculator to find the five-number summary for both ungrouped (raw) data and grouped (frequency distribution) data. The five-number summary provides a quick overview of data distribution and outliers.
Quick Start
Choose your data type, enter your values, and click Calculate:
| Five Number Summary Calculator | |
|---|---|
| Data Type | Ungrouped (Raw Data) Grouped (Frequency Distribution) |
| Enter the X Values (Separated by comma,) | |
| Type of Frequency Distribution | DiscreteContinuous |
| Enter the Classes for X (Separated by comma,) | |
| Enter the frequencies (f) (Separated by comma,) | |
| Results | |
| Number of Observations (N): | |
| Minimum Value: | |
| First Quartile (Q₁): | |
| Median (Q₂): | |
| Third Quartile (Q₃): | |
| Maximum Value: | |
What is a Five-Number Summary?
A five-number summary is a quick and easy way to determine the center, spread, and outliers (if any) of a dataset.
The Five Numbers
The five-number summary consists of five values:
- Minimum (Min): The smallest value in the dataset
- First Quartile (Q₁): The 25th percentile; 25% of data falls below this
- Median (Q₂): The 50th percentile; middle value that divides data in half
- Third Quartile (Q₃): The 75th percentile; 75% of data falls below this
- Maximum (Max): The largest value in the dataset
Visual Representation: Box Plot
The five-number summary is typically displayed as a box plot (box-and-whisker plot):
- The “box” shows Q₁, median, and Q₃
- The “whiskers” extend to min and max values
- Outliers appear as individual points beyond the whiskers
How to Use This Calculator
For Ungrouped (Raw) Data
Step 1: Select “Ungrouped (Raw Data)” as your data type
Step 2: Enter your data values separated by commas (e.g., 10, 15, 20, 25, 30)
Step 3: Click “Calculate”
Results will show:
- Number of observations (N)
- Minimum value
- First quartile (Q₁)
- Median (Q₂)
- Third quartile (Q₃)
- Maximum value
For Grouped (Frequency Distribution) Data
Step 1: Select “Grouped (Frequency Distribution)” as your data type
Step 2: Choose frequency distribution type:
- Discrete: For individual values (e.g., 2, 3, 4, 5)
- Continuous: For class intervals (e.g., 10-20, 20-30, 30-40)
Step 3: Enter class values or intervals separated by commas
Step 4: Enter the corresponding frequencies separated by commas
Step 5: Click “Calculate”
Results will show:
- Number of observations (N)
- Minimum value
- First quartile (Q₁)
- Median (Q₂)
- Third quartile (Q₃)
- Maximum value
Formulas
For Ungrouped Data
The position of each quartile is calculated as:
$$Q_i = \text{Value of } \left(\frac{i(N+1)}{4}\right)^{th} \text{ observation}$$
Where:
- $i$ = 1, 2, or 3 (for Q₁, Q₂, Q₃ respectively)
- $N$ = number of observations
Minimum: The smallest value Maximum: The largest value
For Grouped Data
For Discrete Frequency Distribution
$$Q_i = l + \left(\frac{\frac{iN}{4} - F_<}{f}\right)$$
For Continuous Frequency Distribution
$$Q_i = l + \left(\frac{\frac{iN}{4} - F_<}{f}\right) \times h$$
Where:
- $l$ = lower limit of the quartile class
- $N$ = total number of observations
- $i$ = 1, 2, or 3
- $F_<$ = cumulative frequency before the quartile class
- $f$ = frequency of the quartile class
- $h$ = class width (for continuous distributions)
Minimum: Lower limit of the first class Maximum: Upper limit of the last class
Worked Examples
Example 1: Ungrouped Data - Test Scores
Data: Test scores for 9 students: 45, 52, 58, 63, 72, 81, 85, 90, 95
Find the five-number summary.
Solution:
Step 1: Arrange data in ascending order (already sorted) 45, 52, 58, 63, 72, 81, 85, 90, 95
Step 2: Find Minimum and Maximum
- Minimum = 45
- Maximum = 95
Step 3: Find Q₁ (First Quartile)
Position = (1 × 10) / 4 = 2.5 (between 2nd and 3rd values)
- 2nd value = 52
- 3rd value = 58
- Q₁ = 52 + 0.5(58 - 52) = 52 + 3 = 55
Step 4: Find Q₂ (Median)
Position = (2 × 10) / 4 = 5 (the 5th value)
- Q₂ = 72
Step 5: Find Q₃ (Third Quartile)
Position = (3 × 10) / 4 = 7.5 (between 7th and 8th values)
- 7th value = 85
- 8th value = 90
- Q₃ = 85 + 0.5(90 - 85) = 85 + 2.5 = 87.5
Five-Number Summary:
- Min = 45
- Q₁ = 55
- Median = 72
- Q₃ = 87.5
- Max = 95
Interpretation: The data is fairly well-distributed from 45 to 95, with the middle 50% of scores falling between 55 and 87.5.
Example 2: Grouped Data (Discrete) - Student Absences
Problem: A class teacher has data about the number of absences of 35 students. Find the five-number summary.
| Absences (days) | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Number of Students | 1 | 15 | 10 | 5 | 4 |
Solution:
Step 1: Create calculation table
| Days | Frequency | Cumulative Frequency |
|---|---|---|
| 2 | 1 | 1 |
| 3 | 15 | 16 |
| 4 | 10 | 26 |
| 5 | 5 | 31 |
| 6 | 4 | 35 |
Step 2: Find Minimum and Maximum
- Minimum = 2 days
- Maximum = 6 days
Step 3: Find Q₁ (First Quartile)
Position = (1 × 35) / 4 = 8.75
- Cumulative frequency ≥ 8.75 is 16 (class value = 3)
- Q₁ = 3 days
Thus, 25% of students had 3 or fewer absences.
Step 4: Find Q₂ (Median)
Position = (2 × 35) / 4 = 17.5
- Cumulative frequency ≥ 17.5 is 26 (class value = 4)
- Q₂ = 4 days
Thus, 50% of students had 4 or fewer absences.
Step 5: Find Q₃ (Third Quartile)
Position = (3 × 35) / 4 = 26.25
- Cumulative frequency ≥ 26.25 is 31 (class value = 5)
- Q₃ = 5 days
Thus, 75% of students had 5 or fewer absences.
Five-Number Summary:
- Min = 2 days
- Q₁ = 3 days
- Median = 4 days
- Q₃ = 5 days
- Max = 6 days
Interpretation: Most students had 3-5 absences (middle 50%), with the range from 2-6 days.
Example 3: Grouped Data (Continuous) - Internet Usage
Problem: Time (in minutes) spent on internet each evening by 56 students. Find the five-number summary.
| Time (minutes) | 10-12 | 13-15 | 16-18 | 19-21 | 22-24 |
|---|---|---|---|---|---|
| Number of Students | 3 | 12 | 15 | 24 | 2 |
Solution:
Step 1: Create calculation table with class boundaries
| Class | Boundaries | Frequency | Cumulative Frequency |
|---|---|---|---|
| 10-12 | 9.5-12.5 | 3 | 3 |
| 13-15 | 12.5-15.5 | 12 | 15 |
| 16-18 | 15.5-18.5 | 15 | 30 |
| 19-21 | 18.5-21.5 | 24 | 54 |
| 22-24 | 21.5-24.5 | 2 | 56 |
Step 2: Find Minimum and Maximum
- Minimum = 9.5 minutes
- Maximum = 24.5 minutes
Step 3: Find Q₁ (First Quartile)
Position = (1 × 56) / 4 = 14
- Cumulative frequency ≥ 14 is 15 (class 12.5-15.5)
- l = 12.5, F< = 3, f = 12, h = 3
$$Q_1 = 12.5 + \left(\frac{14 - 3}{12}\right) \times 3 = 12.5 + 2.75 = 15.25 \text{ minutes}$$
Step 4: Find Q₂ (Median)
Position = (2 × 56) / 4 = 28
- Cumulative frequency ≥ 28 is 30 (class 15.5-18.5)
- l = 15.5, F< = 15, f = 15, h = 3
$$Q_2 = 15.5 + \left(\frac{28 - 15}{15}\right) \times 3 = 15.5 + 2.6 = 18.1 \text{ minutes}$$
Step 5: Find Q₃ (Third Quartile)
Position = (3 × 56) / 4 = 42
- Cumulative frequency ≥ 42 is 54 (class 18.5-21.5)
- l = 18.5, F< = 30, f = 24, h = 3
$$Q_3 = 18.5 + \left(\frac{42 - 30}{24}\right) \times 3 = 18.5 + 1.5 = 20 \text{ minutes}$$
Five-Number Summary:
- Min = 9.5 minutes
- Q₁ = 15.25 minutes
- Median = 18.1 minutes
- Q₃ = 20 minutes
- Max = 24.5 minutes
Interpretation: Most students (middle 50%) spend 15.25 to 20 minutes on the internet, with an overall range from 9.5 to 24.5 minutes.
Using Five-Number Summary for Outlier Detection
The Interquartile Range (IQR) helps identify outliers:
$$\text{IQR} = Q_3 - Q_1$$
Outlier Rules:
- Lower outlier: Value < Q₁ - 1.5 × IQR
- Upper outlier: Value > Q₃ + 1.5 × IQR
Example: Test Scores (from Example 1)
IQR = 87.5 - 55 = 32.5
- Lower outlier threshold = 55 - 1.5(32.5) = 55 - 48.75 = 6.25
- Upper outlier threshold = 87.5 + 1.5(32.5) = 87.5 + 48.75 = 136.25
Any score below 6.25 or above 136.25 would be considered an outlier. In this dataset, there are no outliers.
Key Differences: Ungrouped vs. Grouped Data
| Aspect | Ungrouped Data | Grouped Data |
|---|---|---|
| Data Format | Individual raw values | Classes with frequencies |
| Information Loss | None - exact values known | Some - exact values not known |
| Calculation | Direct by position | Using class formula |
| Accuracy | Exact values | Approximate values |
| When to Use | Small datasets, all values available | Large datasets, summarized data |
| Min/Max | Actual minimum/maximum values | Class boundaries |
When to Use Five-Number Summary
Advantages
- ✅ Quick overview of data distribution
- ✅ Shows spread and skewness
- ✅ Identifies potential outliers
- ✅ Useful for comparing multiple datasets
- ✅ Forms basis for box plot visualization
- ✅ Requires minimal calculation
Disadvantages
- ❌ Less detailed than full distribution
- ❌ Only shows 5 points; misses detail between them
- ❌ Sensitive to extreme outliers (uses min/max)
- ❌ Doesn’t show actual data pattern
Common Mistakes to Avoid
❌ WRONG: Using ungrouped formula on grouped data ✓ RIGHT: Use grouped data formula that accounts for class boundaries and frequencies
❌ WRONG: Forgetting to sort data before finding quartiles (ungrouped) ✓ RIGHT: Always sort data in ascending order first
❌ WRONG: Misidentifying the quartile class ✓ RIGHT: Find the class where cumulative frequency first equals or exceeds (iN)/4
❌ WRONG: Using data min/max from class boundaries instead of actual boundaries ✓ RIGHT: For continuous data, use class boundaries (not midpoints) for min/max
Interpretation Guidelines
Understanding the Five-Number Summary
- Minimum: Best-case or lowest value in dataset
- Q₁: Typical value for lower 25% of data
- Median: Typical center value (robust to outliers)
- Q₃: Typical value for upper 25% of data
- Maximum: Worst-case or highest value in dataset
Distribution Patterns
Symmetric Distribution:
- Q₁ to Median ≈ Median to Q₃
- Min to Q₁ ≈ Q₃ to Max
Right-Skewed Distribution:
- Q₁ to Median < Median to Q₃
- Longer tail on right side
Left-Skewed Distribution:
- Q₁ to Median > Median to Q₃
- Longer tail on left side
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