JEE Main 2024 · 30 January · evening · Numerical answer · +4 / −1

The variance σ² of the data
is _________.

xᵢ0156101217
fᵢ3232633
Answer and solution

Answer 29

Solution

Maths sir · 3 steps

Find the mean first, then add up the frequency-weighted squared distances from it.

  1. 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.
    xᵢfᵢ(xᵢ - 8)²fᵢ (xᵢ - 8)²
    0364192
    124998
    53927
    6248
    106424
    1231648
    17381243
    Total22not added640

    σ² = (Σ fᵢ (xᵢ - x̄)²)/N = 640/22 ≈ 29.09

  2. Check. The shortcut gives the same result. Σ fᵢ xᵢ² = 2048, so σ² = 2048/22 - 8² = 93.09 - 64 = 29.09.
  3. 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.

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Asked about 100 times in JEE Main since 2020. This question is from 2024.