Question

-point averages. Find the mean and median for the grouped data in the following table 26 -Round to two decimal places)

please find mean, median and mode!

interval= 1.95-2.15, 2.15-2.35, 2.35-2.55, 2.55-2.75, 2.75-2.95, 2.95-3.15, 3.15-3.35, 3.35-3.55, 3.55-3.75, 3.75-3.95

frequency= 21, 18, 13, 14, 10, 6, 5, 3, 2, 1

please find the mean, median AND mode

Answer 1

Date 30 7 2020 > solution- class frequency mid value fox cof f 1.95-2.15 21 2.05 43.05 21 2 18 2.15-2.35 2.35 2.55 2:25 4005 39 3 - 13 2.45 31.85 52 4 2.55 2.75 14 2.65 37.1 66 S 2.75 -2.95 10 28.5 76 6 2.95 3.15 6 2.85 3.05 3.25 18.3 82 7 3.15 3.35 S 16.25 87 3.55 3 3.45 10.35 go 9 13.35 13.55 3.75 103.75 -3.95 7.3 92 3.65 3.85 3.85 93 193 - 237. OS mean n = Efx 2.5489 = 237.os 93 median value of th ( m ) abservation value of fon ( 93 ) observation Ucelue of 46 observation from the column of cumulative frequency (Cf) we find that the observation lies in the class 2.35-2.55 The median class is 2.35 2.55 Now .. L = lower boundary point of median class = 2.35 n = 93 cf = 30 cumulative frequency of the Predicting the median class a39 Class

Date 'fa frequency of the median class=13 class length of median class - 0.2 ..c- n -cf median Ma 2 Lt cos f = 2.35 + 46.5-39 13 A 0.2 22.35 + 705 13 A 002 > ) 2.35 to 1154 median - 2.4654 mode - class length is not equal for each data. so mode can not be obtained directly Here we have given mean ( A ) = 2.5489 median (M) = 2.4654 mode (z ) = ? z = 3M - 2 3 A (2.4654) - 2462.548g) = 7.3962-5.0978 2 = 2.2983 Mode = 2.2983