Question

The following distribution shows the daily pocket allowance of children of a locality The mean pocket allowance is Rs 25 .Find
the missing frequency P.

Class Interval : 0-10 10-20 20-30 30-40 40-50

Frequency : 5 18 15 P 6​
​

Answer 1

AnswEr :

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-2}\cline{1-3}\cline{1-7} Pocket Money Allowance & 0-10 & 10-20 & 20-30 & 30-40 & 40-50\\\cline{1-7}No. of Children&5&18&15&P&6&\cline{1-7}\cline{1-6}\end{tabular}$

$\rule{120}{1}$

$\begin{array}{|c|c|c|c|}\cline{1-4}\sf Daily\: Pocket\: Allowance& \sf No.\:of\: Children(f_{i})&\sf \sf x_{i}=\dfrac{(Upper+Lower)Limits}{2}&\sf f_{i}x_{i}\\\cline{1-4}0-10&5&5&25\\10-20&18&15&270\\20-30&15&25&375\\30-40&\sf P&35&\sf35P\\40-50&6&45&270\\\cline{1-4} \sf Total&\sf\sum\limits\: f_{i}=44+P&&\sf \sum\limits\: f_{i}x_{i}=940+35P\\\cline{1-4}\end{array}$

$\rule{200}{2}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\sf Mean=\dfrac{\sum\limits\: f_{i}\:x_{i}}{\sum\limits\: f_{i}}\\\\\\:\implies\sf 25= \dfrac{940+35P}{44 + P}\\\\\\:\implies\sf 25(44 + P) = 940 + 35P\\\\\\:\implies\sf 1100 + 25P = 940 + 35P\\\\\\:\implies\sf 1100 - 940 = 35P - 25P\\\\\\:\implies\sf160 = 10P\\\\\\:\implies\sf \dfrac{160}{10} = P\\\\\\:\implies \boxed{ \blue{\sf P = 16}}$

⠀

$\therefore\:\underline{\textsf{Missing Frequency of distribution is \textbf{16}}}$

Answer 2

ANSWER :

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

The following distrubution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.25.

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-2}\cline{1-3}\cline{1-7}Pocket Money Allownace & 0-10 & 10-20 & 20 - 30 & 30 - 40 & 40 - 50\\ \cline{1-7}Number of Children& 5 & 18 & 15 & P& 6& \cline{1-7}\cline{1-6}\end{tabular}$

$\bf{\Large{\underline{\bf{To\:Find\::}}}}$

The missing frequency P.

$\begin{array}{|c|c|c|c|}\cline{1-4}\sf Daily\: Pocket \:Allownace\:(Rs.) & \sf Frequency(f_{i})&\sf Class-Mark\:(x_{i})&\sf{f_{i}x_{i}\\ \cline{1-4}0-10&5&5&25\\10-20&18&15&270\\20-30&15&25&375\\30-40&\sf P&35&35P\\40-50&6&45&270\\ \cline{1-4}\sf Total &\sf \sum \limits\:f_{i}=44+P}&&\sf \sum\limits\:f_{i}x_{i}=940+35P\\ \cline{1-4}\end{array}$$\bf{\Large{\underline{\tt{\purple{Explanation\::}}}}}$

We have mean is Rs.25

$\blacksquare$ We know that formula of the mean:

$\bf{\Large{\boxed{\sf{Mean\:=\:\frac{\sum\limits\:f_{i}x_{i}}{\sum\limits\:f_{i}} }}}}$

$\implies\sf{25\:=\:\dfrac{940+35P}{44+P} }$

$\bf{\underline{\sf{\bigstar{\sf{Cross-Multiplication\::}}}}}$

$\implies\sf{25(44+P)=940+35P}$

$\implies\sf{1100+25P\:=\:940+35P}$

$\implies\sf{25P\:-\:35P\:=\:940-1100}$

$\implies\sf{-10P\:=\:-160}$

$\implies\sf{P\:=\:\cancel{\dfrac{-160}{-10} }}$

$\implies\sf{\purple{P\:=\:16}}$

Thus,

$\bf{\Large{\boxed{\sf{The\:missing\:frequency\:is\:\:16}}}}}$