JEE Main 2024 · 30 January · evening · Numerical answer · +4 / −1
The variance σ² of the data
is _________.
| xᵢ | 0 | 1 | 5 | 6 | 10 | 12 | 17 |
|---|---|---|---|---|---|---|---|
| fᵢ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer and solution
Answer 29
Solution
Maths sir · 3 steps
Find the mean first, then add up the frequency-weighted squared distances from it.
- Find the mean. N = Σ fᵢ = 3+2+3+2+6+3+3 = 22 and Σ fᵢ xᵢ = 0+2+15+12+60+36+51 = 176. So x̄ = 176/22 = 8.
xᵢ fᵢ (xᵢ - 8)² fᵢ (xᵢ - 8)² 0 3 64 192 1 2 49 98 5 3 9 27 6 2 4 8 10 6 4 24 12 3 16 48 17 3 81 243 Total 22 not added 640 σ² = (Σ fᵢ (xᵢ - x̄)²)/N = 640/22 ≈ 29.09
- Check. The shortcut gives the same result. Σ fᵢ xᵢ² = 2048, so σ² = 2048/22 - 8² = 93.09 - 64 = 29.09.
- So. To the nearest integer, the variance is 29.
The common slip: 30. Dividing by N - 1 = 21 instead of N = 22 gives 640/21 ≈ 30.5, which rounds to 30. That is the sample variance. The variance of the data divides by N.
The idea to keep. For a frequency table, multiply every squared deviation by its frequency and divide by N = Σ fᵢ. A whole-number mean makes the deviation method quick.
Answer from the NTA final answer key. Solution by PrepWiser.