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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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 + (𝒇𝟏 − 𝒇𝟎)/(𝟐𝒇𝟏 − 𝒇𝟎 − 𝒇𝟐) × 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 −𝑓0)/(2𝑓1 −𝑓0 −𝑓2) × h = 125 + (20 − 13)/(2(20) − 13 − 14) × 20 = 125 + 7/(40 − 27) × 20 = 125 + 7/13 × 20 = 125 + 10.77 = 135.77 Finding Median Median = l + (𝑁/2 −𝑐𝑓)/𝑓 × h Here, 𝑵/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 −𝑐𝑓)/𝑓 × h = 125 + (34 − 22)/20 × 20 = 125 + 12/20 × 20 = 125 + 12 = 137 Now, let’s find Mean Mean(𝑥 ̅) = a + h × (∑▒𝒇𝒊𝒖𝒊)/(∑▒𝒇𝒊) Where a = assumed mean h = Class interval Also, ∑▒𝒇𝒊 = 68 ∑▒𝒇𝒊𝒖𝒊 = 7 Putting values in formula Mean(𝒙 ̅) = a + h × (∑▒𝒇𝒊𝒖𝒊)/(∑▒𝒇𝒊) 𝑥 ̅ = 135 + 20 × 7/68 𝑥 ̅ = 135 + 2.05 𝒙 ̅ = 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.