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Corrected mean and SD after wrong entry

Short answer

Corrected mean = (20×20 − 13 + 30)/20 = 20.85 minutes. Original Σx² = 20(25+400)=8500, corrected Σx² = 8500 − 169 + 900 = 9231, so corrected variance = 9231/20 − (20.85)² = 26.8275, SD ≈ 5.18 minutes.

A professor of statistics, in his class with 20 students, had conducted a survey about time spent by students on social media, during college hours, and found average time of 20 minutes and standard deviation of 5 minutes. In subsequent verification, it was found that observation 30 was entered as 13. Then find corrected mean and standard deviation of time spent by the students on social media during college hours.

When one observation is entered wrongly, the old totals Σx and Σx² are recovered from the given mean and SD. The wrong value is then removed and the right value added.

Given: n = 20, mean = 20, SD σ = 5 (variance σ² = Σx²/n − x̄²). Wrong value 13, correct value 30.

Step 1: Original totals

Σx  = n × mean = 20 × 20 = 400
Σx² = n (σ² + x̄²) = 20 (25 + 400) = 8500

Step 2: Corrected totals

Σx  (corrected) = 400 − 13 + 30 = 417
Σx² (corrected) = 8500 − 13² + 30² = 8500 − 169 + 900 = 9231

Step 3: Corrected mean and SD

Mean = 417 / 20 = 20.85 minutes
σ² = 9231/20 − (20.85)² = 461.55 − 434.7225 = 26.8275
σ  = √26.8275 = 5.18 minutes

Answer: corrected mean = 20.85 minutes, corrected standard deviation ≈ 5.18 minutes. (If the given SD was a sample SD with divisor n − 1, the corrected SD would be 5.31 minutes.)

✓ Answer checked by Claude Sonnet 5.5

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