The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If μ and σ² denote the mean and variance of the correct observations respectively, then 15(μ+μ²+σ²) is equal to __________.

Solution

Maths sir · 4 steps

Do not chase σ² itself: μ² + σ² is just the corrected Σ x² divided by 15.

1. Recover the totals. With mean 12, the wrong Σ x = 15 × 12 = 180. From σ² = (Σ x²)/15 - 12² = 9, the wrong Σ x² = 15 × 153 = 2295.
2. Correct both sums. Take out 10 and put in 12. The corrected Σ x = 180 - 10 + 12 = 182. The corrected Σ x² = 2295 - 100 + 144 = 2339.

   σ² = (Σ x²)/15 - μ² ⇒ μ² + σ² = 2339/15, μ = 182/15

3. Multiply by 15. 15(μ + μ² + σ²) = 15μ + 15(μ² + σ²) = 182 + 2339.
4. So. The expression equals 2521.

The common slip: 2481. Fixing Σ x² by adding (12-10)² = 4 instead of 12² - 10² = 44 gives 2481. You must remove 10² and add 12².

The idea to keep. To correct a wrong entry, rebuild Σ x and Σ x² from the given mean and variance, swap the wrong value for the right one, then recompute.

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