Question

How do u solve it with 3 variables

Answer 1

The missing frequency values are x=60, y=100 and z=40.

How to determine the missing frequency values?

The median and modal class in the given frequency distribution correspond to the interval 30-40 for wages in Rs.

The frequency distribution table is structured as follows:

Wages (in Rs) | Frequency | Cumulative Frequency

0-10 | 4 | 4

10-20 | 16 | 20

20-30 | x | 20 + x

30-40 | y | 20 + x + y

40-50 | z | 20 + x + y + z

50-60 | 6 | 26 + x + y + z

60-70 | 4 | 30 + x + y + z

Given the total frequency is 230:

We have,

`30+x+y+z=230`

`x+y+z=200 ....eq1`

Mode=34

`Mode= 1+ ((f_1-f_9)/(2f_1-f_o-f_2)) * h`

`34 =30 + ((x-y)/(2y-x-z) ) * 10`

`8y-4x-4z = 10y-10x`

`6x-2y4z=0`

`3x-y-2z0...eq2`

Median=33.5

`Median= 1 + (((n)/(2)-cf )/(f) ) * h`

`33.5 = 30+ ((115-20-x))/(y) * 10`

`3.5 = (95-x)/(y) * 10`

`(35y)/(100) = 95 - r`

`35y+100x = 9500...eq3`

from eq1 and eq2

`2x+2y+2z = 400`

`3x-y-2z=0`

`5x+y=400 ...eq4`

from equation 3 and equation 4

`(5x+y=400) * 20`

`100x + 20y = 8000`

`100x + 35y=9500`

`-15y-1500y = 100`

Now let us put the value of y in equation 3

`5x+100 = 400`

`5x = 300`

`x=(300)/(5)`

`x=60`

Now let us put the value of x and y in equation 1

`60+100+z=200`

`z=200-160`

`z = 40`

Therefore, the missing frequency values are x=60, y=100 and z=40.