Question

Solve this in attachment :::: ​

Answer 1

$\large{\underline{\underline{\red{\sf{\hookrightarrow Given:- }}}}}$

• A frequency distribution table is given to us .
• The mean is 62.8

$\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find:-}}}}}$

• The value of x .

$\large{\underline{\underline{\red{\sf{\hookrightarrow Solution:-}}}}}$

Now , we may prepare it like this ,

$\boxed{\begin{tabular}{|c|c|c|c|}\cline{1-4} Class Interval & Frequency &Class Mark & \sf f_i\times x_i \\ \cline{1-4} 0-20 &5&10&50 \\ \cline{1-4} 20-40 &8&30&240 \\ \cline{1-4} 40-60 &x&50&50x \\ \cline{1-4} 60-80 &12&70&840 \\ \cline{1-4} 80-100 &7&90&630 \\ \cline{1-4} 100-120 &8&110&880 \\ \cline{1-4} & \Sigma f_i=(40+x) && \Sigma(f_i\times x_i) =2640+50x \end{tabular}}$

[ Note - If LaTeX doesn't work on app , see attachment ]

$\tt:\implies Mean=\dfrac{\Sigma(f_i\times x_i)}{\Sigma f_i}=\dfrac{2640+50x}{40+x}$

$\tt:\implies \dfrac{2640+50x}{40+x}=62.8$

$\tt:\implies 2640+50x=(62.8\times 40)+62.8x$

$\tt:\implies 2640 + 50x = 2512 + 62.8x$

$\tt:\implies 62.8x-50x=2640-2512$

$\tt:\implies 12.8x = 128$

$\tt:\implies x =\dfrac{128}{12.8}$

$\underline{\boxed{\red{\tt{\longmapsto x = 10}}}}$

$\pink{\boxed{\purple{\tt{\dag Hence\:value\:of\:x\:is\:10.}}}}$

Answer 2

Answer:

$Hence,\:the\:value\:of\:x\:is\:10.$

Step-by-step explanation:

It is the correct answer.

Hope this attachment helps you.