Example 6 - Normal & Shortcut Method - Chapter 15 Class 11 Statistics - Part 3

Advertisement

Example 6 - Normal & Shortcut Method - Chapter 15 Class 11 Statistics - Part 4

Advertisement

 

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

Transcript

Example 6 (shortcut method) Find the mean deviation about the mean for the following data. Mean(π‘₯ Μ…) = a + h Γ— (βˆ‘β–’π‘“π‘–π‘‘π‘–)/(βˆ‘β–’π‘“π‘–) Here, a = assumed mean = 45 h = width of class = 10 𝑑_𝑖= (π‘₯_π‘–βˆ’ π‘Ž)/β„Ž Marks obtained Number of students(fi) Mid-point (xi) xi – a = (xi – 45) di = "xi – a" /h fidi 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2 40 Marks obtained Number of students(fi) Mid-point (xi) xi – a = (xi – 45) di = "xi – a" /h fidi 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2 40 Mean(π‘₯ Μ…) = 45 + 10 Γ— 0/40 π‘₯ Μ… = 45 + 0 π‘₯ Μ… = 45 Marks obtained Number of students(fi) Mid-point (xi) |xi-x Μ… | fi|xi-x Μ… | 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2 Mean deviation (π‘₯ Μ…) = ( βˆ‘128▒𝑓𝑖|π‘₯𝑖 βˆ’ π‘₯ Μ… | )/(βˆ‘128▒𝑓𝑖) = 400/10 = 10

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.