--- title: "Bus ticket counter: mean, SD, CV, CI" subject: "Probability and Statistics" chapter: 6 key: "semester_5/probability_and_statistics:bus-ticket-passengers-mean-cv-ci" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bus-ticket-passengers-mean-cv-ci site: BCT Engineering --- # Bus ticket counter: mean, SD, CV, CI Probability and Statistics · Chapter 6: Inference concerning mean - Asked in: 2072 Chaitra, 8 marks - Video: [Standard Deviation and Coefficient of Variation](https://www.youtube.com/watch?v=PsJ_NgDUayI) — The Organic Chemistry Tutor - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bus-ticket-passengers-mean-cv-ci - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6 ## Question Following data gives the sample records of number of passenger take ticket at the counter of Bus during one hour period. 22 58 32 36 62 57 25 45 23 37 64 56 46 60 29 49 63 36 26 58 60 26 58 58 29 43 53 36 45 22 52 43 45 31 45 39 35 38 30 60 58 42 54 62 52 42 65 58 51 60 53 45 31 53 22 53 51 52 47 59 Find the i. Sample mean of number of passenger ii. Sample standard deviation and coefficient of variation. iii. Standard error of the sample mean. iv. Find the 95% and 99% confidence limit of sample mean. ## Short answer Sample mean = 45.70 passengers, sample standard deviation = 12.75, coefficient of variation = 27.90%, standard error = 1.65. The 95% confidence limits are (42.47, 48.93) and 99% limits (41.46, 49.94) passengers per hour. ## Answer The data are ungrouped, n = 60. #### i. Sample mean ```text Σx = 2742 x̄ = Σx/n = 2742/60 = 45.700 ``` **x̄ ≈ 45.70 passengers per hour.** #### ii. Standard deviation and coefficient of variation ```text Σx² = 134904 Σ(x − x̄)² = Σx² − n·x̄² = 134904 − 60×45.700² = 9594.600 s² = Σ(x − x̄)²/(n−1) = 9594.600/59 = 162.620 s = 12.752 CV = (s/x̄) × 100 = (12.752/45.700) × 100 = 27.90% ``` **s ≈ 12.75, CV ≈ 27.90%.** #### iii. Standard error of the mean ```text SE = s/√n = 12.752/√60 = 1.646 ``` **SE ≈ 1.65.** #### iv. Confidence limits Since n = 60 (large), the normal critical values are used: x̄ ± z·SE. ```text 95%: z = 1.960, margin = 1.960 × 1.646 = 3.227 limits = 45.700 ± 3.227 = (42.47, 48.93) 99%: z = 2.576, margin = 2.576 × 1.646 = 4.241 limits = 45.700 ± 4.241 = (41.46, 49.94) ``` **Answer: 95% limits: (42.47, 48.93); 99% limits: (41.46, 49.94) passengers per hour.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is s calculated? A. sqrt(sum(x-xbar)^2/(n-1)) B. sum(x-xbar)^2/(n-1) C. sum x^2 / n D. sqrt(sum(x-xbar)^2 / n) Answer: A. s^2 = 9594.6/59 = 162.62, s = 12.75. (Key point: Sample SD) 2. What is CV? A. s/sqrt(n) B. s^2/xbar C. (s/xbar) x 100 = 27.9% D. (xbar/s) x 100 Answer: C. CV = 12.752/45.70 x 100 = 27.90%. (Key point: Coefficient of variation) 3. What is SE? A. s^2/60 B. s x sqrt(60) C. s/60 D. s/sqrt(60) = 1.646 Answer: D. 12.752/7.746 = 1.646. (Key point: Standard error) 4. Which values are used since n = 60? A. t values 1.96 and 2.576 from tables with df 59 exactly B. chi-square values C. z = 1.645 and 2.326 D. Normal z = 1.960 (95%) and 2.576 (99%) Answer: D. For large n, normal critical values apply. 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