Math, asked by sairishikr, 6 months ago

)The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is

Rs 20.2. Find the missing frequency f.

Daily pocket allowance in Rs 10 15 20 25 30

Number of children 6 8 P 10 6​

Answers

Answered by DeerghayuThandar
1

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

\begin{gathered}\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}\end{gathered}

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

ToFind:

The missing frequency P.

\begin{gathered}\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}\end{gathered} \bf{\Large{\underline{\tt{\purple{Explanation\::}}}}}

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}} }}}}

Mean=

∑f

i

∑f

i

x

i

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

44+P

940+35P

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

★Cross−Multiplication:

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

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

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

\implies\sf{-10P\:=\:-160}⟹−10P=−160

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

−10

−160

\implies\sf{\purple{P\:=\:16}}⟹P=16

Thus,

\bf{\Large{\boxed{\sf{The\:missing

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