Question

please tell me the answer
quickkkkk​

Answer 1

Since the total frequency of the distribution is 40,

$\displaystyle\longrightarrow \sum f_i=40$

$\longrightarrow 6+f_1+10+f_2+9=40$

$\longrightarrow f_1+f_2+25=40$

$\longrightarrow f_1+f_2=15\quad\quad\dots(1)$

The same frequency distribution table with the middle of each class is given below.

$\begin{tabular}{|c|c|}\cline{1-2} x_i & f_i \\\cline{1-2}4&6\\\cline{1-2}12& f_1 \\\cline{1-2}20&10\\\cline{1-2}28& f_2 \\\cline{1-2}36&9\\\cline{1-2}\end{tabular}$

Since mean of this distribution is 21.4,

$\longrightarrow\bar x=\dfrac{\displaystyle\sum f_ix_i}{\displaystyle\sum f_i}$

$\longrightarrow 21.4=\dfrac{4\times6+12f_1+20\times10+28f_2+36\times9}{40}$

$\longrightarrow 24+12f_1+200+28f_2+324=21.4\times40$

$\longrightarrow 12f_1+28f_2+548=856$

$\longrightarrow 12f_1+28f_2=308$

$\longrightarrow 12f_1+12f_2+16f_2=308$

$\longrightarrow 12\left(f_1+f_2\right)+16f_2=308$

From (1),

$\longrightarrow 12\times15+16f_2=308$

$\longrightarrow 180+16f_2=308$

$\longrightarrow f_2=\dfrac{308-180}{16}$

$\longrightarrow\underline{\underline{f_2=8}}$

And, from (1),

$\longrightarrow f_1=15-f_2$

$\longrightarrow f_1=15-8$

$\longrightarrow\underline{\underline{f_1=7}}$

Hence the frequencies are 7 and 8 respectively.