The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________

Solution

Maths sir · 4 steps

Rebuild Σ x and Σ x² from the mean and SD, swap the wrong marks, then recompute.

1. Recover the totals. With n = 10 and mean 50, Σ x = 500. From σ² = (Σ x²)/n - x̄², we get 144 = (Σ x²)/10 - 2500, so Σ x² = 26440.
2. Fix the sum. Remove 45 and 50 and put in 20 and 25. The correct Σ x = 500 - 95 + 45 = 450, so the correct mean is 45.
3. Fix the squares. Do the same swap on the squares. The correct Σ x² = 26440 - 2025 - 2500 + 400 + 625 = 22940.

   σ² = 22940/10 - 45² = 2294 - 2025 = 269

4. So. The correct variance is 269.

The common slip: 619. Correcting the mean to 45 but leaving Σ x² at 26440 gives 2644 - 2025 = 619, not 269. The squares must be corrected too.

The idea to keep. When data values are misread, mean and SD cannot be corrected directly. Go back to Σ x and Σ x², swap the wrong values in both, then recompute.

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