Standard deviation of grouped data
For a frequency table, weight each deviation by its frequency. For class intervals use mid-values and the step deviation method with class width c.
Formulas for grouped data
| Data type | Method | Formula |
|---|---|---|
| Discrete (values with frequencies) | Mean method | σ = √(Σfd²/N), d = x − x̄ |
| Discrete | Assumed mean method | σ = √(Σfd²/N − (Σfd/N)²), d = x − A |
| Continuous (class intervals) | Step deviation method | σ = c × √(Σfd²/N − (Σfd/N)²), d = (x − A)/c, x = mid-value, c = class width |
How to set up the table
- 1Write the columns: class (or x), f, mid-value x, d, fd, fd².
- 2For classes, find each mid-value: x = (lower limit + upper limit)/2.
- 3Take A as a mid-value near the middle of the table, and c as the class width.
- 4Find d, then fd (f × d), then fd² (fd × d).
- 5Add the f, fd and fd² columns to get N, Σfd and Σfd².
- 6Put the totals into the formula. Take the square root last.
Example 1: discrete data, mean method
In a 10-mark quiz, 16 students scored marks x = 2, 4, 6, 8, 10, with numbers of students f = 1, 4, 6, 4, 1. Find the standard deviation.
- N = Σf = 16 and Σfx = 2 + 16 + 36 + 32 + 10 = 96, so x̄ = 96/16 = 6. It is a whole number, so use the mean method.
- d = x − 6: −4, −2, 0, 2, 4
- fd² = f × d²: 1 × 16, 4 × 4, 6 × 0, 4 × 4, 1 × 16, that is 16, 16, 0, 16, 16
- Σfd² = 64
- σ = √(Σfd²/N) = √(64/16) = √4 = 2
σ = 2 marks
Example 2: class intervals, step deviation method
The marks of 50 students in a 50-mark test are 0-10: 4, 10-20: 9, 20-30: 14, 30-40: 13, 40-50: 10. Find the standard deviation.
- Mid-values x: 5, 15, 25, 35, 45. Take A = 25 and c = 10.
- d = (x − 25)/10: −2, −1, 0, 1, 2
- fd: −8, −9, 0, 13, 20, so Σfd = 16
- fd²: 16, 9, 0, 13, 40, so Σfd² = 78
- σ = c × √(Σfd²/N − (Σfd/N)²) = 10 × √(78/50 − (16/50)²)
- = 10 × √(1.56 − 0.1024) = 10 × √1.4576
- = 10 × 1.207 ≈ 12.07 marks
σ ≈ 12.07 marks
Properties of standard deviation
- Adding or subtracting the same number to every value does not change σ or the variance.
- Multiplying or dividing every value by a positive number k multiplies or divides σ by k, and the variance by k².
- The standard deviation of the first n natural numbers is σ = √((n² − 1)/12). For 1 to 7: σ = √(48/12) = 2.
- σ is never negative. σ = 0 only when all the values are equal.
Surveys such as the heights of students in a district or their daily travel time to school are grouped into classes. The SD of the grouped table shows how widely the values are spread.
1 marks: “The standard deviation of a data set is 3. Each value is multiplied by 4. What is the new variance?” New σ = 3 × 4 = 12, so the new variance = 12² = 144.
2 marks: “Find the standard deviation of the first 13 natural numbers.” σ = √((n² − 1)/12) = √((169 − 1)/12) = √14 ≈ 3.74.
5 marks: “The time (in minutes) taken by 40 students to reach school is 0-10: 6, 10-20: 10, 20-30: 13, 30-40: 7, 40-50: 4. Find the standard deviation.” Mid-values 5 to 45, A = 25, c = 10. Σfd = −7, Σfd² = 57. σ = 10 × √(57/40 − (−7/40)²) = 10 × √1.394375 ≈ 11.81 minutes.
Grouped SD table: x, f, d, fd, fd². Shortcut for the last column: fd² = fd × d.