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Ex 15.2, 9 - Find mean, variance, standard deviation using short - Standard deviation and variance - Continuous frequency (grouped data)

Ex 15.2,  9 - Chapter 15 Class 11 Statistics - Part 2
Ex 15.2,  9 - Chapter 15 Class 11 Statistics - Part 3


Transcript

Ex15.2, 9 Find the mean, variance and standard deviation using short-cut method Mean(π‘₯ Μ…) = A + (βˆ‘β–’π‘“π‘–π‘¦π‘–)/(βˆ‘β–’π‘“π‘–) Γ—β„Ž where A = assumed mean = 92.5 𝑦_𝑖=(π‘₯_𝑖 βˆ’ 𝐴)/β„Ž h = class size = 75 βˆ’ 70 = 5 Variance (𝜎)2 = β„Ž2/N2 [π‘βˆ‘β–’π‘“π‘–π‘¦π‘–2 βˆ’ (βˆ‘β–’π‘“π‘–π‘¦π‘–)^2 ] = 52/602 [N Γ— (254) – (6)2] = 25/3600 [60 Γ—254 βˆ’36] = 25/3600 Γ— 15204 = 105.58 ∴ Standard deviation (𝜎) = √Variance = √105.58 = 10.27

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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.