Answer to Question #215510 in Statistics and Probability for Naad

Question

A number of particular articles have been classified according to their weights.

After drying for two weeks, the same articles have again been weighed and

similarly classified. It is known that the median weight in the first weighing

was 20.83 gm while in the second weighing it was 17.35 gm. Some

frequencies a and b in the first weighing and x and y in the second are missing.

It is known that a = 1/3x and b = 1/2 y. Find out the values of the missing

frequencies.

Answer 1

$M_1=20.83 \\ M_2=17.35 \\ x=3a \\ y=2b \\ N_1=a+b+11+52+75+22=160+a+b \\ N_2=x+y+40+50+30+28 = 148 +x+y = 148 +3a+2b$

The median class are 20-25 and 10-15 respectively.

$Median = l + \frac{n}{f}(\frac{N}{2}+c) \\ M_1= l+ \frac{n}{f}(\frac{N_1}{2} -cf) \\ 20.83 = 20 + \frac{5}{75}[\frac{160+a+b}{2}-(63+a+b)] \\ 12.45 = 17 - \frac{a+b}{2} \\ a+b=9.1 \;(1)\\ M_2 = l + \frac{n}{f}(\frac{N_2}{2} -cf) \\ = 15 + \frac{5}{50}[\frac{148+3a+2b}{2}-(40+3a+2b)] \\ \frac{3a+2b}{2}=10.5 \\ 3a+2b=21 \;(2)$

Solving (1) and (2), we get

a=3

b=6

x=9

y=12