Question

SYTE The median and the mode of the following daily wage Listribention of 830 workers are known to be Rs. 33.60 and Rs. 84 Mospectively . Find missing beequencies. cuages (Rs.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 4 16 ? ? 40 ? 4.​

Answer 1

Given :

$\fbox{Data :}$

$\begin{gathered}\boxed{\begin{array}{c|c|c}\bf Wages(Rs.) & \bf Frequency & \bf Cumulative \:Frequency(cf) \\ \\ \frac{\qquad\qquad}{} & \frac{\qquad\qquad}{} & \frac{\qquad\qquad}{} \\ \sf 0-10 & \sf 4 & \sf4 \\ \\ \sf 10-20 & \sf 16 & \sf 20 \\ \\ \sf 20-30 & \sf (x) & \sf 20+x \\ \\ \sf 30-40 & \sf (y) & \sf 20+x+y \\ \\ \sf 40-50 & \sf 40 & \sf 60+x+y \\ \\ \sf 50-60 & \sf (z) & \sf 60+x+y+z \\ \\ \sf 60-70 & \sf 4 & \sf 64+x+y+z\\ \\ \frac{\qquad\qquad}{} & \frac{\qquad\qquad}{} & \frac{\qquad\qquad}{}\\ \sf Total & \sf 830 & \sf \:\:\:\: \end{array}} \\ \end{gathered}$

Median = 33.60

Mode = 34

To find :

Missing frequencies x, y, z

Solution :

We are given, that the sum of frequencies is 230.

Form as equation corresponding to the given condition.

the sum of frequencies in the form of an equation.

$\sf\implies 16+x+y+40+z+4 = 830$

Add the numbers on the left side of the equation.

$\sf\implies 64+x+y+z = 830$

After subtracting 64 from both sides, we get,

$\sf\implies x+y+z = 830-64$

$\sf\implies x+y+z = 766...(i)$

We are given that the median of the data is 33.6, this implies that the median class for the given data is 30-40.

total number of workers(n)=830

Lower class(L)=30

cf = 20-x [cf of preceding interval ]

h=40-30=10 [width of interval]

Substitute the values in the formula of median,

$\tt\red{ :⟶\: median = L + \frac{( \frac{n}{2} - cf)}{f} \times h}$

$\tt{ :⟶\: 33.60 = 30 + \frac{( \frac{850}{2} -(20-x) )}{y} \times 10}$

$\tt{ :⟶\: 33.60 - 30 = \frac{( 415 -20+x )}{y} \times 10}$

$\tt{ :⟶\: 3.60 = \frac{( 395+x )}{y} \times 10}$

$\tt{ :⟶\: 3.60y = (395+x) \times 10}$

$\tt{ :⟶\: 3.60y = (3950+10x) }$

$\tt :⟶10x = 3950 - 3.6y$

$\tt :⟶x = \frac{3950 - 3.6y}{10}...(ii)$

Also, it is given that the mode of the data is 34.

Substitute the values in the formula of mode,

lower class (L)=30

$\sf f_{1}$ = y [frequency of the modal class ]

$\sf f_{0}$ = x [frequency of the previous modal]

$\sf f_{2}$ = 40 [frequency of the succeeding modal class ]

h=40-30=10 [width of class interval]

$\tt\red {:⟶mode \: =L + \frac{ f_{1} - f_{0} }{2f_{1} - f_{0} - f_{2}} \times h }$

$\tt {:⟶34 \: =30 + \frac{ y- x }{2y-x-40}\times 10 }$

$\tt {:⟶34 -30 = \frac{ y- x }{2y-x-40}\times 10 }$

$\tt {:⟶4 = \frac{ y- x }{2y-x-40}\times 10 }$

$\tt {:⟶4 = \frac{10( y- x )}{2y-x-40} }$

$\tt :⟶8y - 4x - 160 = 10y - 10x$

$\tt :⟶8y - 10y - 4x + 10x - 160 = 0$

$\tt :⟶ - 2y + 6x - 160 = 0$

$\tt :⟶ - 2y + 6 \frac{(3950 - 3.6y)}{10} - 160 = 0$

$\tt :⟶ - 2y + \frac{23700 - 21.6y}{10} - 160 = 0$

$\tt :⟶ - 2y + 2370 - 2.16y - 160 = 0$

$\tt :⟶ - 4.16y + 2210 = 0$

$\tt :⟶ - 4.16y = - 2210$

$\tt :⟶y = \frac{ - 2210}{ - 4.16}$

$\tt {:⟶y=}$$\displaystyle{\tt { \cancel{ \frac{-2210}{-4.16} }}}$

$\tt :⟶y = 531.25$

$\tt :⟶y≈531$

Inserting value of y in equation (ii) to get value of x

$\tt :⟶x = \frac{3950 - 3.6y}{10}$

$\tt :⟶x = \frac{3950 - 3.6(531.25)}{10}$

$\tt :⟶x = \frac{3950 - 1912.5}{10}$

$\tt :⟶x = \frac{2037.5}{10}$

$\tt :⟶x = 203.75$

$\tt :⟶x ≈204$

[Note than values of x and y are taken as approximation because x and y are frequencies and they cannot be in decimal form, but only as natural numbers]

Inserting value of x and y in equation (I) to get value of z

$\sf\implies x+y+z = 766$

$\sf:⟶ 204+531+z = 766$

$\sf:⟶ 735+z = 766$

$\sf:⟶ z = 766-735$

$\sf:⟶ z = 31$

Missing frequencies are :

$\huge\purple{x=204}$

$\huge\purple{y=531}$

$\huge\purple{z=31}$