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Example 6 (Shortcut Method) - Chapter 15 Class 11 Statistics (Term 1)
Last updated at Sept. 6, 2021 by Teachoo
Examples
Example 1
Example 2
Example 3 Important
Example 4
Example 5 Important
Example 6 (Normal Method)
Example 6 (Shortcut Method) Important You are here
Example 7 Important
Example 8
Example 9 Important
Example 10 Important
Example 11
Example 12 Important
Example 13 Deleted for CBSE Board 2022 Exams
Example 14 Important
Example 15 Important Deleted for CBSE Board 2022 Exams
Example 16
Example 17 Important
Example 18
Example 19 Important
Chapter 15 Class 11 Statistics (Term 1)
Serial order wise
- Ex 15.1
- Ex 15.2
- Ex 15.3
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Examples
- Miscellaneous
Transcript
Example 6 (shortcut method) Find the mean deviation about the mean for the following data. Mean(π₯ Μ ) = a + h Γ (ββππππ)/(ββππ) Here, a = assumed mean = 45 h = width of class = 10 π_π= (π₯_πβ π)/β Marks obtained Number of students(fi) Mid-point (xi) xi β a = (xi β 45) di = "xi β a" /h fidi 10 β 20 2 20 β 30 3 30 β 40 8 40 β 50 14 50 β 60 8 60 β 70 3 70 β 80 2 40 Marks obtained Number of students(fi) Mid-point (xi) xi β a = (xi β 45) di = "xi β a" /h fidi 10 β 20 2 20 β 30 3 30 β 40 8 40 β 50 14 50 β 60 8 60 β 70 3 70 β 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 β 20 2 20 β 30 3 30 β 40 8 40 β 50 14 50 β 60 8 60 β 70 3 70 β 80 2 Mean deviation (π₯ Μ ) = ( β128βππ|π₯π β π₯ Μ | )/(β128βππ) = 400/10 = 10
