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Ex 13.3, 1 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Finding Mode Mode = l + (π’‡πŸ βˆ’ π’‡πŸŽ)/(πŸπ’‡πŸ βˆ’ π’‡πŸŽ βˆ’ π’‡πŸ) Γ— h Modal class = Interval with highest frequency = 125 – 145 where l = lower limit of modal class h = class-interval f1 = frequency of the modal class f0 = frequency of class before modal class f2 = frequency of class after modal class Putting values in formula Mode = l + (𝑓1 βˆ’π‘“0)/(2𝑓1 βˆ’π‘“0 βˆ’π‘“2) Γ— h = 125 + (20 βˆ’ 13)/(2(20) βˆ’ 13 βˆ’ 14) Γ— 20 = 125 + 7/(40 βˆ’ 27) Γ— 20 = 125 + 7/13 Γ— 20 = 125 + 10.77 = 135.77 Finding Median Median = l + (𝑁/2 βˆ’π‘π‘“)/𝑓 Γ— h Here, 𝑡/𝟐 ∴ 125 – 145 is the median class And, l = lower limit of median class h = class-interval cf = cumulative frequency of the class before median class f = frequency of the median class Putting values in formula Median = l + (𝑁/2 βˆ’π‘π‘“)/𝑓 Γ— h = 125 + (34 βˆ’ 22)/20 Γ— 20 = 125 + 12/20 Γ— 20 = 125 + 12 = 137 Now, let’s find Mean Mean(π‘₯ Μ…) = a + h Γ— (βˆ‘β–’π’‡π’Šπ’–π’Š)/(βˆ‘β–’π’‡π’Š) Where a = assumed mean h = Class interval Also, βˆ‘β–’π’‡π’Š = 68 βˆ‘β–’π’‡π’Šπ’–π’Š = 7 Putting values in formula Mean(𝒙 Μ…) = a + h Γ— (βˆ‘β–’π’‡π’Šπ’–π’Š)/(βˆ‘β–’π’‡π’Š) π‘₯ Μ… = 135 + 20 Γ— 7/68 π‘₯ Μ… = 135 + 2.05 𝒙 Μ… = 137.05 Therefore, Mean is 137.05 So, Mean = 137.05 , Median = 137, Mode = 135.77 ∴ Mean, Median, Mode are approximately the same

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo