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Example 6 (Shortcut Method) - Chapter 15 Class 11 Statistics (Term 1)
Last updated at May 24, 2022 by Teachoo
Examples
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Example 6 (Normal Method)
Example 6 (Shortcut Method) Important You are here
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Example 10 Important
Example 11
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Example 13 Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 15 Important Deleted for CBSE Board 2023 Exams
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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
