# Ex 15.1, 12

Last updated at Dec. 12, 2016 by Teachoo

Last updated at Dec. 12, 2016 by Teachoo

Transcript

Ex15.1, 12 Calculate the mean deviation about median age for the age distribution of 100 persons given below: Converting the given data in continuous frequency by subtracting 0.5 from lower age limit adding and 0.5 in upper limit N = ∑128▒𝑓𝑖 = 100 Median Class = (𝑁/2)^𝑡ℎterm = (100/2)^𝑡ℎ term = 50th term In above data, cumulative frequency of class 35.5 − 40.5 is 63 which is greater than 50. ∴ Median class = 35.5 − 40.5 Median = 𝑙 + ( 𝑁/2 − 𝐶)/𝑓 ×ℎ Where, 𝑙 = lower limits of median class N = sum of frequencies 𝑓 = frequency of median class C = Cumulative frequency of class before median class Here, 𝑙 = 35.5, N = 100, C = 37, ℎ = 5, 𝑓 = 26 Median = 35.5 + (100/2 −37)/26 × 5 = 35.5 + (50 − 37)/26 × 5 = 20 + 13/26 × 5 = 35.5 + 2.5 = 38 Now, ∑128▒𝑓𝑖 = 100 ∑128▒𝑓𝑖 |𝑥𝑖 −𝑀| = 735 ∴ Mean Deviation (M) = (∑▒〖 𝑓_𝑖 |𝑥𝑖 − M| 〗)/𝑓_𝑖 = 735/100 = 7.35

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.