--- title: "Central tendency vs dispersion; mode 58" subject: "Probability and Statistics" chapter: 1 key: "semester_5/probability_and_statistics:central-tendency-vs-dispersion-mode-58" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/central-tendency-vs-dispersion-mode-58 site: BCT Engineering --- # Central tendency vs dispersion; mode 58 Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability - Asked in: 2071 Shrawan, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/central-tendency-vs-dispersion-mode-58 - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1 ## Question Write difference between measure of central tendency and measure of dispersion and their importance. The following table represents the marks of 100 students. | Marks | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | |---|---|---|---|---|---| | No. of Students | 14 | ? | 27 | ? | 15 | If the mode value is 58, find the missing frequencies and the mean of all 100 students. ## Short answer Central tendency gives a typical value; dispersion shows spread. Both are needed to describe data. Missing frequencies: 18 (20-40) and 26 (60-80). Mean = 5200/100 = 52 marks. ## Answer #### Difference and importance | Basis | Central tendency | Dispersion | |---|---|---| | Meaning | A typical or central value of the data | The spread of the values about the centre | | Examples | Mean, median, mode | Range, SD, variance, mean deviation | | Importance | Summarises the whole data by one value; allows comparison of groups | Shows reliability of the average, consistency and risk | | Limitation | Hides how scattered the data is | Does not tell the location of data | Both are needed: two sets can have the same mean but different spread. #### Missing frequencies Let the frequency of 20-40 be x and of 60-80 be y. The total is 100, so ```text 14 + x + 27 + y + 15 = 100 → x + y = 44 ``` Mode = 58 lies in the class 40-60 (L = 40, h = 20, f₁ = 27, f₀ = x, f₂ = y). ```text Mode = L + (f1 − f0) / (2f1 − f0 − f2) × h 58 = 40 + (27 − x) / (54 − x − y) × 20 ``` Since x + y = 44, the denominator is 54 − 44 = 10: ```text 18 = (27 − x)/10 × 20 = 2(27 − x) 27 − x = 9 → x = 18, y = 44 − 18 = 26 ``` #### Mean | Marks | Mid x | f | fx | |---|---|---|---| | 0-20 | 10 | 14 | 140 | | 20-40 | 30 | 18 | 540 | | 40-60 | 50 | 27 | 1350 | | 60-80 | 70 | 26 | 1820 | | 80-100 | 90 | 15 | 1350 | | Total | | 100 | 5200 | ```text Mean = 5200 / 100 = 52 ``` **Answer:** the missing frequencies are 18 (for 20-40) and 26 (for 60-80); the mean mark is 52. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the role of dispersion that central tendency cannot give? A. It always equals the mean B. It gives the most typical value C. It locates the middle item D. It shows the spread, the reliability of the average and consistency Answer: D. Central tendency gives a typical value; dispersion shows how scattered the data is, so two sets can share a mean but differ in spread. (Key point: Central vs dispersion) 2. With mode 58 in class 40-60, which pair of equations finds the missing frequencies? A. x = y and x + y = 100 B. x - y = 44 and median = 58 C. x + y = 56 and mean = 58 D. x + y = 44 and the mode formula with f0 = x, f2 = y Answer: D. The total gives 14 + x + 27 + y + 15 = 100, so x + y = 44; the mode formula supplies the second equation. (Key point: Modal class equation) 3. What are the missing frequencies for 20-40 and 60-80? A. 18 and 26 B. 14 and 30 C. 26 and 18 D. 20 and 24 Answer: A. 18 = 2(27 - x) gives x = 18, hence y = 26. (Key point: Missing frequencies) 4. What is the mean of the 100 students? A. 58 B. 54 C. 52 D. 50 Answer: C. Sum fx = 5200, so mean = 5200/100 = 52. 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