--- title: "Mean, variance, CV of preschool ages" subject: "Probability and Statistics" chapter: 1 key: "semester_5/probability_and_statistics:ages-preschool-mean-cv" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ages-preschool-mean-cv site: BCT Engineering --- # Mean, variance, CV of preschool ages Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability - Asked in: 2070 Chaitra, 8 marks - Video: [Mean, Variance & Standard Deviation of Grouped Data │ Statistics](https://www.youtube.com/watch?v=yvKrZfvZWe0) — GoTutor Math - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/ages-preschool-mean-cv - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1 ## Question The following data are the ages (in months) at which n = 50 children were first enrolled in a preschool. 38 40 30 35 39 47 35 34 43 41 32 34 41 30 46 55 39 33 32 32 42 50 37 39 33 40 48 36 31 36 36 41 43 48 40 35 40 30 46 37 45 42 41 36 50 45 38 46 36 31 a) Find sample mean, sample variance and sample standard deviation. b) Compute a value that measures the amount of variability relative to the value of mean. ## Short answer Sample mean x̄ = 39.08 months, variance s² = 35.83 months², standard deviation s ≈ 5.99 months. The coefficient of variation CV = (s/x̄)×100 ≈ 15.3% measures relative variability. ## Answer The sample mean is x̄ = Σx/n, the sample variance is s² = Sxx/(n − 1) with Sxx = Σx² − (Σx)²/n, and the coefficient of variation is CV = (s / x̄) × 100%. #### a) Mean, variance and standard deviation n = 50. From the data, Σx = 1954 and Σx² = 78,118. ```text x̄ = 1954 / 50 = 39.08 months Sxx = 78,118 − (1954)²/50 = 78,118 − 76,362.32 = 1755.68 s² = 1755.68 / 49 = 35.83 months² s = √35.83 = 5.99 months ``` #### b) Relative variability The coefficient of variation measures the spread relative to the mean: ```text CV = (5.986 / 39.08) × 100 = 15.32 % ``` **Answer:** x̄ = 39.08 months, s² ≈ 35.83 months², s ≈ 5.99 months, CV ≈ 15.3 %. The spread is about 15% of the mean age, which is moderate. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the sample variance computed with n - 1 = 49? A. Because n = 49 B. It is the sample estimate, Sxx divided by n - 1 C. To get the CV D. Because the mean is zero Answer: B. For sample data s^2 = Sxx/(n - 1) = 1755.68/49 = 35.83. (Key point: Sample variance divisor) 2. What is the sample mean? A. 40.2 B. 39.08 C. 38.5 D. 35.83 Answer: B. 1954/50 = 39.08 months. (Key point: Mean) 3. How is Sxx found? A. Sum x^2 + (Sum x)^2/n B. Sum x^2 - (Sum x)^2/n C. Sum x - n D. (Sum x^2)/n Answer: B. Sxx = 78,118 - 76,362.32 = 1755.68. (Key point: Sxx) 4. Which measure is variability relative to the mean, and what is its value? A. SD, 5.99 months B. Range, 25 C. Variance, 35.83 D. CV = s/mean x 100, about 15.3% Answer: D. CV = 5.986/39.08 x 100 = 15.32%. 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