Question

Calculate the Missing Frequency from the following distribution, it is given that the Median of the Distribution is 24.
Age In Yrs : 0-10, 10-20, 20-30, 30-40, 40-50
No. of Persons : 5, 25, ?, 18, 7​

Answer 1

AnswEr :

$\begin{tabular}{|c|c|c|c|c|c|}\cline{1-2}\cline{1-3}\cline{1-6}Age(In Yrs) & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\\cline{1-6}No. of Persons&5&25&?&18&7\cline{1-3}\cline{1-6}\end{tabular}$

$\begin{tabular}{|c|c|c|}\cline{1-2}\cline{1-3}\cline{1-4}Class Interval & Frequency & Cumulative Frequency \\\cline{1-3}0 - 10 & 5 & 5 \\\cline{1-3}10 - 20 & 25 & 5+25 = 30\cline{1-3}\cline{1-3}20 - 30 & x & x+30 = x+30\cline{1-3}\cline{1-3}30 - 40 & 18 & (x+30)+18 = x+48\cline{1-3}\cline{1-3}40 - 50 & 7 & (x+48)+7 = x+55\cline{1-3}\cline{1-3}&n = x+55&&\cline{1-3}\cline{1-3}\end{tabular}$

⋆ Given Median = 24

• Now, Some Important Points :

• L = Lower boundary of median class = 20
• n = Toral Frequency = (x + 55)
• cf = Cumulative Frequency of the class preceding the Median Class = 30
• f = frequency of the Median Class = x
• h = class length of Median Class = 10

• According to the Question Now :

$\implies \sf Median = L + \bigg(\dfrac{\frac{n}{2}-cf}{f}\bigg)\times h\\\\\implies \sf 24 = 20 + \bigg(\dfrac{\frac{(x+55)}{2}-30}{x}\bigg)\times 10\\\\\implies \sf 24 -20 = \bigg(\dfrac{\frac{(x+55)}{2}-30}{x}\bigg)\times 10\\\\\implies \sf 4x = \bigg(\dfrac{(x+55)}{2}-30\bigg)\times 10\\\\\implies \sf 4x \times 2=[(x+55) - 60]\times 10\\\\\implies \sf 8x = [x - 5]\times 10\\\\\implies \sf 8x = 10x - 50$

$\implies \sf 50 = 10x - 8x\\\\\implies \sf \cancel\dfrac{50}{2} = x\\\\\implies \large\boxed{\sf x = 25}$

⠀

∴ Missing Frequency of distribution is 25.

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\green{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\sf{Given\::}}}}$

The median of the Distribution is 24.

$\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6} Age \:in\: years & 0-10 & 10-20 & 20-30 & 30-40 & 40-50\\ \cline{1-6} No.\:of\: person & 5 & 25 & ? & 18 & 17\\ \cline{1-6}\end{tabular}$

$\bf{\Large{\underline{\sf{\red{To\:find\::}}}}}$

The missing frequency from the following distribution.

$\bf{\Large{\underline{\sf{\pink{Explanation\::}}}}}$

Here, we have the following table:

$\begin{tabular}{|c|c|c|}\cline{1-3}\multicolumn{3}{|c|}{Frequency\:Table}\\\cline{1-3} Age\:(In\:years) &Number\:of\:person & Cumulative\:frequency\\\cline{1-3}0-10 & 5 & 5\\10-20 & 25 & 25+5=30\:(C.F)\\20-30 & R\:(f) & 30+R\\30-40 & 18 & 30+18+R=48+R\\40-50 & 7 & 48+7+R=55+R\\\cline{1-3}\end{tabular}$

The sum of total frequency is n = R+55.

We know that formula of the median:

Median = $\bold{l+\frac{h}{f} (\frac{N}{2} -c)}$

$\bf{Where}\begin{cases}\sf{l=lower\:class\:boundary\:of\:middle\:class}\\ \sf{h=size\:of\:median\:class\:interval}\\ \sf{f=frequency\:corresponding\:to\:the\:median\:class}\\ \sf{N=sum\:of\:frequency}\\ \sf{c=cumulative\:frequency\:preceding\:median\:class}\end{cases}$

We have,

The median= 24.

Now,

Putting the value of given in formula of median;

→ $\bf{Median=l+\frac{h}{f} (\frac{N}{2} -c)}$

→ $\bf{24=20+\frac{10}{R} (\frac{R+55}{2} -30)}$

→ $\bf{24=20+\frac{10}{R} (\frac{R+55-60}{2} )}$

→ $\bf{24=20+\frac{10R+550-600}{2R}}$

→ 48R = 40R + 10R + 550 - 600

→ 48R = 50R - 50

→ 48R - 50R = -50

→ -2R = -50

→ R= $\bf{\cancel{\frac{-50}{-2} }}$

→ R = 25

Thus,

The Missing frequency from the distribution is 25.