Question

solve the above problem ​

Answer 1

Answer:

Don't mind if t he answer is wrong

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• The mean of the given frequency distribution is 62.8 .

• The sum of all the frequencies is 50.

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

1. The missing frequency $\bf{f_1}$ .
2. And the missing frequency $\bf{f_2}$ .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

$\begin{tabular}{| c | c | c | c |} \cline{1 - 4} Class interval & Frequency (f_i) & x_i & f_i x_i \\ \cline{1 - 4} 0 - 20 & 5 & 10 & 50 \\ 20 - 40 & f_1 & 30 & 30 f_1 \\ 40 - 60 & 10 & 50 & 500 \\ 60 - 80 & f_2 & 70 & 70 f_2 \\ 80 - 100 & 7 & 90 & 630 \\ 100 - 120 & 8 & 110 & 880 \\ \cline{1 - 4} & \Sigma{f_i} = 30 + f_1 + f_2 & & \Sigma{(f_i x_i)} = 2060 + 30f_1 + 70f_2 \\ \cline{1 - 4} \end{tabular}$

$\Big[\:{\bf{NOTE\:\longrightarrow}}\:\rm{x_i\:=\:\dfrac{Sum\:of\:class\:interval}{2}\:\Big]}$

☯︎ According to the question,

ᴄᴀsᴇ - 1 ;-

$\purple\bigstar\:\bf{\red{\overbrace{\underbrace{\green{The\:sum\:of\:all\:the\:frequencies\:=\:50\:}}}}}$

$\\ \bf{:\implies\:30\:+\:f_1\:+\:f_2\:=\:50\:}$

$\\ \bf{:\implies\:f_1\:+\:f_2\:=\:50\:-\:30\:}$

$\\ \bf{:\implies\:f_1\:+\:f_2\:=\:20\:}----(1)$

ᴄᴀsᴇ - 2 ;-

${\color{aqua}\bigstar}\:\bf{\pink{\overbrace{\underbrace{\blue{The\:mean\:of\:frequency\:distribution\:=\:62.8\:}}}}}$

$\\ \bf{:\implies\:\dfrac{\Sigma{f_i\:x_i}}{\Sigma{f_i}}\:=\:62.8\:}$

$\\ \bf{:\implies\:\dfrac{2060\:+\:30f_1\:+\:70f_2}{30\:+\:f_1\:+\:f_2}\:=\:62.8\:}$

$\\ \bf{:\implies\:2060\:+\:30f_1\:+\:70f_2\:=\:1884\:+\:62.8\:(f_1\:+\:f_2)\:}$

☞︎︎︎ Putting the value of $\bf{f_1\:+\:f_2}$ in the above equation from equation (1), we get

$\\ \bf{:\implies\:2060\:+\:30f_1\:+\:70f_2\:=\:1884\:+\:62.8\times{20}\:}$

$\\ \bf{:\implies\:2060\:+\:30f_1\:+\:70f_2\:=\:1884\:+\:1256\:}$

$\\ \bf{:\implies\:30f_1\:+\:70f_2\:=\:3140\:-\:2060\:}$

$\\ \bf{:\implies\:10\:(3f_1\:+\:7f_2)\:=\:1080\:}$

$\\ \bf{:\implies\:3f_1\:+\:7f_2\:=\:108\:}----(2)$

➪ Multiple '3' in equation (1), we get

$\huge\red\checkmark$ $\\ \bf{:\implies\:3f_1\:+\:3f_2\:=\:60\:}----(3)$

✈︎ Now substracting equation (3) from equation (2), we get

$\\ \bf{:\implies\:3f_1\:+\:7f_2\:-\:(3f_1\:+\:3f_2)\:=\:108\:-\:60\:}$

$\\ \bf{:\implies\:3f_1\:+\:7f_2\:-\:3f_1\:-\:3f_2\:=\:48\:}$

$\\ \bf{:\implies\:7f_2\:-\:3f_2\:=\:48\:}$

$\\ \bf{:\implies\:4f_2\:=\:48\:}$

$\\ \bf{:\implies\:f_2\:=\:\dfrac{48}{4}\:}$

$\\ \bf\green{:\implies\:f_2\:=\:12\:}$

✈︎ Now putting the value of $\bf{f_2}$ in the equation (1), we get

$\\ \bf{:\implies\:f_1\:+\:12\:=\:20\:}$

$\\ \bf{:\implies\:f_1\:=\:20\:-\:12\:}$

$\\ \bf\green{:\implies\:f_1\:=\:8\:}$

__________________________

$\huge{\color{orange}\therefore}$ [1] The missing frequency $\bf{f_1}$ is $\bf\gray{8}$ .

$\huge{\color{orange}\therefore}$ [2] The missing frequency $\bf{f_2}$ is $\bf\gray{12}$ .