EX 13.3 (Question: 1)

EX 13.3 (Question: 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. Monthly consumption (in units) Number of consumers 65 - 85 4 85 - 105 5 105 - 125 13 125 - 145 20 145 - 165 14 165 - 185 8 185 - 205 4

Given: Let ℎ=20,𝑎=135 From the table: ∑𝑓_𝑖=68 ∑𝑓_𝑖 𝑢_𝑖=7Mean 𝑥 ‾=135+(7/68)×20 =135+140/68 =135+2.05 =137.05Mean, 𝒙 ‾=𝒂+((∑1▒ 𝒇_𝒊 𝒖_𝒊)/(∑1▒ 𝒇_𝒊 ))×𝒉 Mode, =𝒍+[(𝒇_𝟏−𝒇_𝟎)/(𝟐𝒇_𝟏−𝒇_𝟎−𝒇_𝟐 )]×𝒉 Median, =𝒍+[(𝒏/𝟐−𝒄𝒇)/𝒇]×𝒉 Mode From table: Modal class =125−145 𝑙=125,𝑓_1=20,𝑓_0=13,𝑓_2=14 Mode =125+((20−13)/(2×20−13−14))×20 =125+(7/(40−27))×20 =125+(7/13)×20 =125+140/13 =125+10.76 =135.76 Median Given 𝑛=68⇒𝑛/2=34 From table: Cumulative frequency just > 34 is 42 (class: 125-145) So, median class =125−145 𝑙=125,𝑓=20,𝑐𝑓=22 " Median "=125+((34−22)/20)×20 =125+12/20×20 =125+12 =137So, Mean =137.05, Mode =135.76, Median = 137 Mean > Median > Mode

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