Ex 14.3, 1 Class 10 Maths - Monthly consumption of electricity of 68

Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 2
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 3
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 4
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 5
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 6
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 7
Ex 14.3, 1 - Chapter 14 Class 10 Statistics - Part 8

  1. Chapter 14 Class 10 Statistics (Term 2)
  2. Serial order wise

Transcript

Ex 14.3, 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 = 85 – 65 = 20 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.