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Answers Given that The following da qivy the weight Ciokg) full a sample of 7-18 white rhino horns. 1.) 2.1 13 2.0 3.1 1.6 2.3 1.3 2.2 1.8 27 30 1.6 1.5 0.9 2.1 cas from the given dia mean (T) = sau 51.99 deviation (s) ) {(xi-tj2 standard and deviation (srl - s= {(ai-1.9992 - 18-1 S = 0.66. (6) for median, first arrange data is increasing sdu then medias is 272.1 22.05 first quartile (Q.) = 1.5 cic from above median find median, Third reedtile (Q3) = 2.4 cie from below the median, food median] Upper helt 0:9,1.1,1.3, 1.3,15,116,1.6, 1.8, 2, 2.1, 2.1, 2.2, 2.3,2.4, 2.7,2.9,3,3-1 Q3. Medion = 242.) – 2.05 dowel hilt Scanned with CamScanner
IQR - Q2-Q, = 2.4-1.5 = 0.9 C for outliers 1.5*IQR = 1.5*0.9 = 1.35 so if any value smaller than Q -1.5* IQR of langer than Q3 +1.54 IQR as cuttier. i-e 1.5-1.35 20.5 ,2.471.35 3.95 There is no ledue smaller than 0.5 and larger than 3.75 So there is no outliel. Scanned with CamScanner