Variance and standard deviation: ungrouped data
Variance is the mean of the squared deviations from the mean. Standard deviation (σ) is its square root. Four methods give the same answer.
Deviation, variance and standard deviation
The deviation of a value is its distance from the mean: d = x − x̄. Deviations from the mean always add up to zero, so we square them first. Variance (σ²) is the mean of the squared deviations. Standard deviation (σ) is the positive square root of the variance, and it has the same unit as the data. A small σ means the values stay close to the mean.
Four methods for ungrouped data
| Method | Formula | Use it when |
|---|---|---|
| Direct method | σ = √(Σx²/n − (Σx/n)²) | The values are small numbers |
| Mean method | σ = √(Σd²/n), d = x − x̄ | The mean is a whole number |
| Assumed mean method | σ = √(Σd²/n − (Σd/n)²), d = x − A | The mean is a decimal; take A near the middle |
| Step deviation method | σ = c × √(Σd²/n − (Σd/n)²), d = (x − A)/c | The deviations share a common factor c |
Example 1: mean method, checked by the direct method
The runs scored by a batsman in 8 overs are 2, 4, 4, 4, 5, 5, 7, 9. Find the variance and standard deviation.
- n = 8 and Σx = 40, so x̄ = 40/8 = 5. The mean is a whole number, so use the mean method.
- Deviations d = x − 5: −3, −1, −1, −1, 0, 0, 2, 4
- Squares d²: 9, 1, 1, 1, 0, 0, 4, 16, so Σd² = 32
- Variance σ² = Σd²/n = 32/8 = 4
- Standard deviation σ = √4 = 2
- Check by the direct method: Σx² = 232, so σ = √(232/8 − 5²) = √(29 − 25) = √4 = 2.
Variance = 4, standard deviation = 2 runs
Example 2: assumed mean method
The weights (in kg) of 5 students are 44, 47, 50, 53, 59. Find the standard deviation using the assumed mean method.
- The mean is 253/5 = 50.6, a decimal, so take the assumed mean A = 50.
- d = x − 50: −6, −3, 0, 3, 9, so Σd = 3
- d²: 36, 9, 0, 9, 81, so Σd² = 135
- σ = √(Σd²/n − (Σd/n)²) = √(135/5 − (3/5)²)
- = √(27 − 0.36) = √26.64
- ≈ 5.16 kg
σ ≈ 5.16 kg
Example 3: step deviation method
The time (in minutes) five students spent on homework one evening is 20, 30, 40, 50, 80. Find the standard deviation using the step deviation method.
- Take A = 40. Every deviation from 40 is a multiple of c = 10.
- d = (x − 40)/10: −2, −1, 0, 1, 4, so Σd = 2
- d²: 4, 1, 0, 1, 16, so Σd² = 22
- σ = c × √(Σd²/n − (Σd/n)²) = 10 × √(22/5 − (2/5)²)
- = 10 × √(4.4 − 0.16) = 10 × √4.24
- = 10 × 2.059 ≈ 20.59 minutes
σ ≈ 20.59 minutes
Your marks in five unit tests: a small SD means you score steadily, and a large SD means some very high and some very low scores. Coaches use SD to judge a player's consistency.
2 marks: “Find the variance and standard deviation of 5, 10, 15, 20, 25.” Mean = 15. d = −10, −5, 0, 5, 10, so Σd² = 250. Variance = 250/5 = 50. σ = √50 ≈ 7.07.
5 marks: “A student reads 40, 43, 45, 47 and 52 pages on five days. Find the standard deviation using the assumed mean method.” Take A = 45. d = −5, −2, 0, 2, 7; Σd = 2, Σd² = 82. σ = √(82/5 − (2/5)²) = √(16.4 − 0.16) = √16.24 ≈ 4.03 pages.
5 marks: “The mean and standard deviation of 10 values are 20 and 3. Later it is found that one value was copied as 15 instead of 25. Find the correct mean and standard deviation.” Σx = 10 × 20 = 200. Since σ² = Σx²/n − x̄², Σx² = 10 × (9 + 400) = 4090. Correct Σx = 200 − 15 + 25 = 210, so the correct mean = 21. Correct Σx² = 4090 − 225 + 625 = 4490. Correct σ² = 449 − 441 = 8, so σ = √8 ≈ 2.83.
Square, average, root: square the deviations, take their mean (the variance), then take the square root (the SD).