Question

3rd entry is 16!

Grade-point averages. Find the mean and median for the grouped data in the following table. 165-215 255275 Frequency 275-206 13 315635 375-75 3 mean-(Round to two decimal places) median (Round to two decimal places.)

Answer 1

PAGE NO.: 0 DATE: # SOLUTION - interval 1.95-2.15 ef= frequency (f) 21 18 f= 16 13 7.15 - 2.35 meolian-class 8 midpoint mut.) (X) xf 2.05 43.05 2.25 40.50 2.45 39.20 2.65 34.45 2.85 22.80 3.05 18.30 3.25 16.25 3.45 17.25 3.65 7.30 3.85 7.70 Exf = 246.8 2.35 -2.55 2.55 - 7.75 2.75 -2.95 2.95 - 3.15 3.15 3.35 3.35- 3.55 3.55 -3.75 3.75 -3.95 6 5 2 2 n= Ef = 96 Mean We know that, Mean = E( midpoint).(frequency) E frequency Σχ-4 Ef 246.8 96 Mean = 2.5708 Round-off Mean = 2.57 Ans.

O PAGE NO.: DATE: 1 Now To find median class value of (2. th Observation = value of ( 26 ) 1th observation Valeee of 48th observation : that is let's in the class mterval (2.35-9.55 (median class 235- 2.55 5) now L = lower boundary point of mediom class = 2.35 = Ef = 96 n cf = Cumulative frequency of the class preceding the median class = 39 f frequency of median class = 16 C class length of medion class = (0-2) then Median (M) 1+ 4-cf C f = 2.35 + +(18-11 - 39 9:(0-2) (02) = 2.35+ 9 76 = 2.35 to 1125 - 2.4625 Rouncing off Medion = 2.46