--- title: "Corrected mean and SD after wrong entry" subject: "Probability and Statistics" chapter: 1 key: "semester_5/probability_and_statistics:corrected-mean-sd-20-students" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/corrected-mean-sd-20-students site: BCT Engineering --- # Corrected mean and SD after wrong entry Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability - Asked in: 2076 Chaitra, 6 marks - Video: [Calculation of Correct Standard Deviation form incorrect standard deviation](https://www.youtube.com/watch?v=aMPSm3gdtpM) — Commerce and Management - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/corrected-mean-sd-20-students - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1 ## Question 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. ## 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. ## Answer 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 ```text Σx = n × mean = 20 × 20 = 400 Σx² = n (σ² + x̄²) = 20 (25 + 400) = 8500 ``` #### Step 2: Corrected totals ```text Σx (corrected) = 400 − 13 + 30 = 417 Σx² (corrected) = 8500 − 13² + 30² = 8500 − 169 + 900 = 9231 ``` #### Step 3: Corrected mean and SD ```text 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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the original Σx^2 recovered from the mean and SD? A. Σx^2 = n x σ = 2000 B. Σx^2 = n x̄^2 = 8000 C. Σx^2 = n(σ + x̄)^2 D. Σx^2 = n(σ^2 + x̄^2) = 8500 Answer: D. From σ^2 = Σx^2/n - x̄^2, Σx^2 = 20(25 + 400). (Key point: Recover old totals) 2. What is the corrected Σx? A. 400 - 13 + 30 = 417 B. 400 + 13 + 30 = 443 C. 400 - 30 + 13 = 383 D. 400 + 30 = 430 Answer: A. Remove the wrong value 13 and add the correct 30. (Key point: Corrected sum) 3. What is the corrected Σx^2? A. 8500 - 13 + 30 = 8517 B. 8500 - 169 + 900 = 9231 C. 8500 + 169 + 900 = 9569 D. 8500 - 900 + 169 = 7769 Answer: B. Squares of the values must be swapped, not the values themselves. (Key point: Corrected Σx^2) 4. What are the corrected mean and SD? A. 20.85 and about 5.00 B. 20 and 5.18 C. 20.85 and about 5.18 D. 21.5 and 5.18 Answer: C. Mean = 417/20; σ^2 = 461.55 - 434.7225 = 26.8275. 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