Question

Grade-point averages. Find the mean and median for the grouped data in the following table. Interval 1.95 -2.15 2.15-2.35 2.35 -2.55 Frequency 2.75 -2.95 10 2.95 - 3.15 3.15 - 3.35 2.55 - 2.75 12 3.35 - 3.55 3.75 -3.95 3.55 -3.75 2 15 mean = (Round to two decimal places.) median (Round to two decimal places.)

Answer 1

I fixi of class solution the formula for sean is Mean x. - Mid-ralue SHEN Totat Frequency X; - (lower limit of class telpper limit of class ) > 2 class 11-95-2015 2-15 -2.35 12.35 -2.35 12-55-2.75 12-75-295 12.95-3-15 13-15-3.35 3-35-3-55 13-55-3-75 13.75-3.95 Frequency OLX 12 2-05 2-25 17 2.45 12 2-65 10 2.85 f;*X: 18* 2.05 = 36.9 15* 2:25 - 33.15 - 41.65 = 31.8 - 28.5 5 21:35 = 16.25 = 13.8 = 7.3 =7*7 {fixX; = 239 3.05 5 4 3-25 3-45 3.65 2 2 fi 292 3.85 1-95 - 2015 is Mid ralyerx;) for class Mid-rolue 9-95t 2015 CX;) 2 412-05 on these way we can find remaining mid-values for each class as shown in table. Neon = 239 g2 20597826 2 2.60 cround to 2 decimal place) Mean - 2.60

Med the formula foo median is median - l + (112)-(of h F Cofil Islapeo boundooy of median class, Me Tote foequench = less than cumulative forquency of the te class just poereeding the median class fofoequency of the median class h = class width class logs-2-15 2:15-2-35 12-35-2-55 2-55-2.75 2-75-2.95 2.95-3.15 13.15-3.35 13-35-3-55 3-55.3-75 3-75-3.95 Frequency sti) lless than cumulative frequency 18 18 15 6 15418 = 33 17 :17+33 = 50 median class 12 -12 +50 = 62 10 =10+62 = 72 = 7+72 = 79 5 - 5+79 = 84 4 = 4+84-88 2 88+2 = 690 2 - 90+2 = 92 EM th - 14edion = The value of (4) obsequation th The value of (92) observation The value of 1467th observation Median class= 2:35 -2-55, because 46 exreess the less than cumulative frequenry for the fiost time in this class ... e= 2.35, cof. = 33, f - 17 ho 2.55 -2.35 - 0:2 Medion - 2.35 + (46-33) 17 2.35+ 0.15 Median = 2:50 .. (Pound to 1 decimal place) * 0.2 .