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

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

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.