Question

Mouse weights. Find the mean and median for the data in the following table. Interval 41.5 - 43.5 43.5 - 45.5 45.5 - 47.5 47.5 - 49.5 49.5-51.5 51.5 - 53.5 53.5-55.5 55.5-57.5 57.5 -59.5 Frequency 4 5 13 15 20 16 15 9 3 mean = (Round to two decimal places if needed.)

please find the mean, median and mode!

Answer 1

Date Page class A= 50.5, 6:2 d=a-A h lfd cf -4 3 -2 HG FIS +26 1141.5-4.3.5 143.5-45.5 145.5-47.5 147.5-49.5 49:5-51:5 51.5-53.5 153.5-56.5 158.5-56.5 1575-59.5 Frequency Midvalue (f) (9) 4 142.5 5 44.5 13 46.5 15 48.5 20 SOSEA 16 52.5 15 54.5 56.5 3 53.5 HIS 4 9 12 37 57 73 88 o - 2 16 30 27 97 3 4 12 lefd=13 Efd mean = At .b n = 50.5 & 13 100 - 2 50.5 +0.265 50.76 Median medicin class s value of ln(2yth Obrervation {n=lto = value of 50th Observation therefoue, median dan is $9.5-515 ^2-cf Median Lt c f

L = lower boundary point of mediandas: 249.5 n = Total frequeny = 100 Ef = cumulative feequency of class preceding the medianclan. =37 fa freequeny of medianclass. -20 2) C-class length of median class=2.

Mediana 50-57 49.5+ 20 Cau value teuken from 2 frequency =49.5+ 13/2012 table 49.5 +1.3 - 50.8. Mode Mode class = maximum frequency clas max. frequency therefore, mode classe 49.5-515 Mode = L+ ta fi-fo 2.4, -o-£ loc value taken from frequency table c49-5+ 20 -15 2.20 15-16 =ug.st 5 سلوک 12 49.5+ lolll - So. Ő Ill 20 L Lower mode class boundaries. = 49.5 fue fuequency of mode of mode class = 12: fuequency of succedding class = 16 fo: frequeny of preceding clay. = 15 C: Claus Length of mode clau = 2