Pearson’s Coefficient of Skewness Calculator
Use this unified calculator to find Pearson’s coefficient of skewness (also called Karl Pearson’s skewness) for both ungrouped (raw) data and grouped (frequency distribution) data. Measure distribution asymmetry using mean, median, and standard deviation.
Quick Start
Choose your data type, enter your values, and click Calculate:
| Pearson's Coefficient of Skewness 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): | |
| Ascending order of X values: | |
| Sample Mean: | |
| Sample Median: | |
| Sample Std. Deviation: | |
| Pearson's Coefficient of Skewness: | |
Understanding Pearson’s Coefficient of Skewness
Pearson’s coefficient (also called Karl Pearson’s skewness) is the most commonly used skewness measure because:
- ✅ Uses all data points (mean and SD incorporate all values)
- ✅ Easier to interpret than moment coefficient
- ✅ Ranges from approximately -3 to +3
- ✅ Well-established in statistical practice
- ✅ Less sensitive to outliers than moment coefficient
The Relationship Between Mean and Median
Pearson’s coefficient reveals the relationship between mean and median:
- Symmetric: Mean ≈ Median → Sk ≈ 0
- Right-skewed: Mean > Median → Sk > 0
- Left-skewed: Mean < Median → Sk < 0
Interpretation Scale
| Pearson’s Coefficient | Distribution | Interpretation |
|---|---|---|
| -1 to -0.5 | Left-skewed | Moderate negative skewness |
| -0.5 to 0 | Slightly left-skewed | Mild negative skewness |
| -0.5 to 0.5 | Approximately symmetric | Very mild or no skewness |
| 0 to 0.5 | Slightly right-skewed | Mild positive skewness |
| 0.5 to 1 | Right-skewed | Moderate positive skewness |
| > ±1 | Highly skewed | Strong asymmetry |
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)
- Sorted values
- Sample mean
- Sample median
- Sample standard deviation
- Pearson’s coefficient of skewness
For Grouped (Frequency Distribution) Data
Step 1: Select “Grouped (Frequency Distribution)” as your data type
Step 2: Choose distribution type (Discrete or Continuous)
Step 3: Enter classes and frequencies
Step 4: Click “Calculate”
Results will show:
- All calculated values above
Formula
Pearson’s Coefficient of Skewness
$$S_k = \frac{3(\overline{x} - M)}{s_x}$$
Where:
- $\overline{x}$ = Sample mean
- $M$ = Median
- $s_x$ = Sample standard deviation
Why Multiply by 3?
The factor of 3 is empirical and comes from Karl Pearson’s research. For many distributions, the relationship:
$$\text{Mean} - \text{Mode} \approx 3(\text{Mean} - \text{Median})$$
holds approximately, so Pearson standardized the formula using this factor.
Alternative Form
Some texts use:
$$S_k = \frac{\overline{x} - M_o}{s_x}$$
Where $M_o$ is the mode. This form is less common because mode is harder to define and less stable than median.
Why Pearson’s Coefficient?
Advantages:
- ✅ Uses all data (mean and SD consider every value)
- ✅ Most commonly reported in practice
- ✅ Stable and well-defined measure
- ✅ Easy to interpret and compare
- ✅ Suitable for all distribution shapes
Disadvantages:
- ❌ More affected by outliers than Bowley’s or Kelly’s
- ❌ Can be affected by extreme values
- ❌ May be misleading for multimodal distributions
Worked Examples
Example 1: Ungrouped Data - Symmetric Distribution
Data: Scores: 20, 30, 40, 50, 60, 70, 80
Find Pearson’s coefficient of skewness.
Solution:
Step 1: Calculate Mean
$$\overline{x} = \frac{20+30+40+50+60+70+80}{7} = \frac{350}{7} = 50$$
Step 2: Find Median
With N = 7 (odd), median is the 4th value = 50
Step 3: Calculate Standard Deviation
Sum of squares = 400+900+1600+2500+3600+4900+6400 = 20,300
$$s_x = \sqrt{\frac{20300 - \frac{350^2}{7}}{6}} = \sqrt{\frac{20300 - 17500}{6}} = \sqrt{\frac{2800}{6}} = \sqrt{466.67} = 21.6$$
Step 4: Calculate Pearson’s coefficient
$$S_k = \frac{3(50-50)}{21.6} = \frac{0}{21.6} = 0$$
Answer: Sk = 0
Interpretation: Distribution is perfectly symmetric; no skewness present.
Example 2: Ungrouped Data - Right-Skewed Distribution
Data: Income (thousands): 20, 25, 30, 35, 40, 45, 100
Find Pearson’s coefficient of skewness.
Solution:
Step 1: Calculate Mean
$$\overline{x} = \frac{20+25+30+35+40+45+100}{7} = \frac{295}{7} = 42.14$$
Step 2: Find Median
With N = 7, median is 4th value = 35
Step 3: Calculate Standard Deviation
Sum of X² = 400+625+900+1225+1600+2025+10000 = 16,775
$$s_x = \sqrt{\frac{16775 - \frac{295^2}{7}}{6}} = \sqrt{\frac{16775 - 12433}{6}} = \sqrt{721.33} = 26.86$$
Step 4: Calculate Pearson’s coefficient
$$S_k = \frac{3(42.14-35)}{26.86} = \frac{3(7.14)}{26.86} = \frac{21.42}{26.86} = 0.797$$
Answer: Sk ≈ 0.797
Interpretation: Strong right skewness; the distribution has a pronounced tail toward higher values (the outlier 100). Mean > Median indicates rightward asymmetry.
Example 3: Grouped Data (Discrete) - Test Scores
Problem: Grade distribution for 50 students. Find Pearson’s coefficient of skewness.
| Grade | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Frequency | 3 | 8 | 20 | 15 | 4 |
Solution:
Step 1: Calculate Mean
| Grade | Frequency | f×x |
|---|---|---|
| 2 | 3 | 6 |
| 3 | 8 | 24 |
| 4 | 20 | 80 |
| 5 | 15 | 75 |
| 6 | 4 | 24 |
| Total | 50 | 209 |
$$\overline{x} = \frac{209}{50} = 4.18$$
Step 2: Find Median
Cumulative frequencies: 3, 11, 31, 46, 50
- N/2 = 25
- Cumulative ≥ 25 is at grade 4
- Median = 4
Step 3: Calculate Standard Deviation
| Grade | f | f×x² |
|---|---|---|
| 2 | 3 | 12 |
| 3 | 8 | 72 |
| 4 | 20 | 320 |
| 5 | 15 | 375 |
| 6 | 4 | 144 |
| Total | 50 | 923 |
$$s_x = \sqrt{\frac{923 - \frac{209^2}{50}}{49}} = \sqrt{\frac{923 - 872.42}{49}} = \sqrt{\frac{50.58}{49}} = 1.016$$
Step 4: Calculate Pearson’s coefficient
$$S_k = \frac{3(4.18-4)}{1.016} = \frac{3(0.18)}{1.016} = \frac{0.54}{1.016} = 0.531$$
Answer: Sk ≈ 0.531
Interpretation: Moderate positive (right) skewness; the distribution leans slightly toward higher grades.
Example 4: Grouped Data (Continuous) - Age Distribution
Problem: Age distribution of 60 customers. Find Pearson’s coefficient of skewness.
| Age Group | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|
| Frequency | 5 | 15 | 20 | 15 | 5 |
Solution:
Step 1: Calculate Mean
| Midpoint | Frequency | f×x | f×x² |
|---|---|---|---|
| 25 | 5 | 125 | 3,125 |
| 35 | 15 | 525 | 18,375 |
| 45 | 20 | 900 | 40,500 |
| 55 | 15 | 825 | 45,375 |
| 65 | 5 | 325 | 21,125 |
| Total | 60 | 2,700 | 128,500 |
$$\overline{x} = \frac{2700}{60} = 45$$
Step 2: Find Median
N/2 = 30
- Cumulative frequencies: 5, 20, 40, 55, 60
- Cumulative ≥ 30 is 40 (class 40-50)
- Using median formula: M = 40 + ((30-20)/20)×10 = 40 + 5 = 45
Step 3: Calculate Standard Deviation
$$s_x = \sqrt{\frac{128500 - \frac{2700^2}{60}}{59}} = \sqrt{\frac{128500 - 121500}{59}} = \sqrt{\frac{7000}{59}} = \sqrt{118.64} = 10.89$$
Step 4: Calculate Pearson’s coefficient
$$S_k = \frac{3(45-45)}{10.89} = \frac{0}{10.89} = 0$$
Answer: Sk = 0
Interpretation: Distribution is perfectly symmetric; mean equals median, indicating no skewness. The distribution is evenly balanced around the center.
Comparison: All Skewness Measures
| Measure | Formula Basis | Range | Data Sensitivity | Use Case |
|---|---|---|---|---|
| Pearson’s | Mean, Median, SD | -3 to +3 | All data | Standard measure |
| Bowley’s | Q₁, Q₂, Q₃ | -1 to +1 | Middle 50% | Robust to outliers |
| Kelly’s | D₁, D₅, D₉ | -1 to +1 | Outer 80% | Tail behavior |
| Moment | All deviations | -2 to +2 | All data | Detailed analysis |
When to Use Pearson’s Coefficient
✅ Use when:
- Standard, comparable measure is needed
- Publishing results (most recognized measure)
- Data is approximately normal or near-normal
- Outliers are not extreme
❌ Avoid when:
- Data has severe outliers (use Bowley’s instead)
- Only tail behavior matters (use Kelly’s instead)
- Complete moment information is needed (use moment coefficient)
Key Differences: Ungrouped vs. Grouped Data
| Aspect | Ungrouped Data | Grouped Data |
|---|---|---|
| Data Format | Individual raw values | Classes with frequencies |
| Mean Calc. | Direct summation | Weighted by frequencies |
| Median Finding | Direct from sorted values | Class-based formula |
| SD Calculation | From individual values | From class midpoints |
| Accuracy | Exact from raw data | Approximate from classes |
Common Mistakes to Avoid
❌ WRONG: Forgetting to multiply by 3 in the formula ✓ RIGHT: Always use Sk = 3(Mean - Median)/SD
❌ WRONG: Using population formula (divide by N) for sample data ✓ RIGHT: Use n-1 in denominator for sample standard deviation
❌ WRONG: Mixing Pearson with other skewness interpretations ✓ RIGHT: Remember Pearson’s range is roughly -3 to +3, while Bowley’s is -1 to +1
❌ WRONG: Assuming Mean > Median always means right skew ✓ RIGHT: Check that Sk is positive; mathematically confirmed by formula
Practical Interpretation Examples
Financial Returns (Sk = 0.85)
- Moderately right-skewed
- Positive outliers (extreme gains) possible
- Downside risk may be more frequent
Student Test Scores (Sk = -0.45)
- Slightly left-skewed
- Easier test; scores cluster at higher end
- Negative outliers (very low scores) possible
Customer Ages (Sk ≈ 0)
- Symmetric
- Evenly distributed across age range
- Mean and median are similar
Related Calculators & Tutorials
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