Example 6 - Mean deviation Normal & Shortcut Method  - Mean deviation about mean - Continuous frequency distibution

Example 6 - Normal & Shortcut Method - Chapter 15 Class 11 Statistics - Part 2
 

  1. Chapter 15 Class 11 Statistics (Term 1)
  2. Serial order wise

Transcript

Example 6 - Chapter 15 Class 11 Statistics - NCERT (Normal method) Find the mean deviation about the mean for the following data. Marks obtained Number of students(fi) Mid-point (xi) fixi 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2 Mean(𝑥 ̅) = (∑▒〖𝑥𝑖 〗 𝑓𝑖)/(∑▒𝑓𝑖) = 1800/40 𝑥 ̅ = 45 ∑128▒𝑓𝑖|𝑥𝑖 − 𝑥 ̅ | = 400 ∴ Mean Deviation = (Σ 𝑓_𝑖 |𝑥_(𝑖 )− 𝑥 ̅ |)/𝑓_𝑖 Mean Deviation (𝑥 ̅) = 400/40 = 10

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.