# Example 8 - Chapter 14 Class 10 Statistics

Last updated at Sept. 25, 2018 by Teachoo

Last updated at Sept. 25, 2018 by Teachoo

Transcript

Example 8 The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class interval Frequency 0 – 100 2 100 – 200 5 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 1000 4 ∑▒𝑓𝑖 = 100 Cumulative frequency 2 2 + 5=7 7 + x 7 + x + 12 = 19 + x 19 + x + 17 = 36 + x 36 + x + 20 = 56 + x 56 + x + y 56 + x + y + 9 = 65 + x + y 65 + x + y + 7 = 72 + x + y 72 + x + y + 4 = 76+ x + y Since Median = 525 500 – 600 is median class Median = l + (𝑛/2 −𝑐𝑓)/𝑓 × h Where l = lower limit of median class = 500 h = class-interval= 100 – 0 = 100 n = ∑▒𝑓𝑖= 100 cf = cumulative frequency of the class before median class= 36 + x f = frequency of the median class =20 Median = l + (𝑛/2 −𝑐𝑓)/𝑓 × h 525 = 500 + (100/2 −(36 + 𝑥))/20 × 100 525 = 500 + (50 – (36 + x)) × 5 525 – 500 = (50 – 36 – x) × 5 25 = (14 – x) × 5 25 = 14(5) – 5x 25 = 70 – 5x 5x = 70 – 25 5x = 45 x = 45/5 x = 9 Also, ∑▒𝑓𝑖 = 76 + x + y 100 = 76 + x + y 100 – 76 = 9 + y 24 = 9 + y y = 24 – 9 y = 15

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.