For a statistical data x₁, x₂, …, x₁₀ of 10 values, a student obtained the mean as 5.5 and Σᵢ₌₁¹⁰ xᵢ²=371. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

Solution

Maths sir · 3 steps

Fix the sum and the sum of squares first, then use variance equals mean of squares minus square of mean.

1. Correct the sum. The wrong sum is 10 × 5.5 = 55. Take out the wrong values 4 and 5 and put in 6 and 8: 55 - 9 + 14 = 60. So the corrected mean is 60/10 = 6.
2. Correct the squares. Do the same swap on the squares: remove 4² + 5² = 41 and add 6² + 8² = 100. This gives 371 - 41 + 100 = 430.

   σ² = (Σ xᵢ²)/n - x̄² = 430/10 - 6² = 43 - 36 = 7

3. So. The variance of the corrected data is 7, option (b).

The idea to keep. When values are miscopied, correct Σ xᵢ and Σ xᵢ² separately by removing the wrong values and adding the right ones. Then recompute the mean before using the variance formula.

Answer from the NTA final answer key. Solution by PrepWiser.