--- title: "Bulb life of two companies: grouped data" subject: "Probability and Statistics" chapter: 6 key: "semester_5/probability_and_statistics:bulbs-two-companies-grouped-data" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bulbs-two-companies-grouped-data site: BCT Engineering --- # Bulb life of two companies: grouped data Probability and Statistics · Chapter 6: Inference concerning mean - Asked in: 2075 Asoj, 8 marks; 2071 Shrawan, 8 marks (asked 2 times) - Video: [Life of bulbs produced by two factories A and B are given below:](https://www.youtube.com/watch?v=MJk23GNrX2U) — PW Solutions - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bulbs-two-companies-grouped-data - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6 ## Question The samples of length of life of bulbs from two companies are given below. | Length of life (hours) | Company A | Company B | |---|---|---| | 500-600 | 10 | 3 | | 600-700 | 21 | 8 | | 700-800 | 6 | 15 | | 800-900 | 8 | 12 | | 900-1000 | 21 | 4 | | 1000-1100 | 10 | 5 | | 1100-1200 | 2 | 15 | | 1200-1300 | 12 | 13 | | 1300-1400 | 19 | 7 | | 1400-1500 | 9 | 7 | | 1500-1600 | 3 | 4 | | 1600-1700 | 7 | 6 | | 1700-1800 | 5 | 3 | | 1800-1900 | 4 | 2 | | 1900-2000 | 1 | 3 | a) Calculate mean length of life of bulbs for company A and company B. b) Calculate sample standard deviation and sample variance for given data. c) Which company's bulbs are more uniform? ## Short answer Means: A = 1095.7 h, B = 1134.1 h. Sample std dev: A = 376.1 h, B = 362.2 h. Coefficients of variation: A = 34.3%, B = 31.9%, so company B’s bulbs are more uniform. ## Answer Use class mid-points m for the grouped data. Mean = Σfm/N; sample variance s² = Σf(m − x̄)²/(N − 1) = [Σfm² − N·x̄²]/(N − 1). | Class | m | f (A) | f (B) | |---|---|---|---| | 500–600 | 550 | 10 | 3 | | 600–700 | 650 | 21 | 8 | | 700–800 | 750 | 6 | 15 | | 800–900 | 850 | 8 | 12 | | 900–1000 | 950 | 21 | 4 | | 1000–1100 | 1050 | 10 | 5 | | 1100–1200 | 1150 | 2 | 15 | | 1200–1300 | 1250 | 12 | 13 | | 1300–1400 | 1350 | 19 | 7 | | 1400–1500 | 1450 | 9 | 7 | | 1500–1600 | 1550 | 3 | 4 | | 1600–1700 | 1650 | 7 | 6 | | 1700–1800 | 1750 | 5 | 3 | | 1800–1900 | 1850 | 4 | 2 | | 1900–2000 | 1950 | 1 | 3 | | Total | | 138 | 107 | Σfm: A = 151200, B = 121350. Σfm²: A = 185045000, B = 151527500. #### a) Mean life - Company A: x̄ = 151200/138 = **1095.65 hours** - Company B: x̄ = 121350/107 = **1134.11 hours** #### b) Sample variance and standard deviation - Company A: s² = (185045000 − 138×1095.65²)/137 = **141477.3**, s = **376.13 hours** - Company B: s² = (151527500 − 107×1134.11²)/106 = **131160.3**, s = **362.16 hours** #### c) Uniformity Coefficient of variation CV = (s/x̄) × 100: - A: 376.13/1095.65 × 100 = **34.33%** - B: 362.16/1134.11 × 100 = **31.93%** **Company B's bulbs are more uniform** (smaller CV, less relative variation in life). **Answer:** x̄_A = 1095.7 h, x̄_B = 1134.1 h; s_A = 376.1 h, s_B = 362.2 h; B is more uniform. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the mean for grouped data found? A. Σfm/N using class mid-points B. Σf/N C. Sum of class limits / N D. Σm/N Answer: A. Mean = Σfm/N with m the mid-point. (Key point: Grouped mean) 2. Which are the means for A and B? A. 1134.1 h and 1095.7 h B. 1095.7 h and 1134.1 h C. 1095.7 h and 1095.7 h D. 1200 h and 1134 h Answer: B. A = 151200/138 = 1095.65; B = 121350/107 = 1134.11. (Key point: Means) 3. What divisor is used for the sample variance? A. N B. N + 1 C. N − 1 D. N² Answer: C. Sample variance s² = [Σfm² − N·x̄²]/(N − 1). (Key point: Sample variance) 4. Which measure compares uniformity? A. Coefficient of variation (s/x̄)×100 B. The larger standard deviation C. The mean D. The range of mid-points Answer: A. CV gives relative variation; smaller CV means more uniform. 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