Find the missing frequencies in the following frequency distribut…

Question

Find the missing frequencies in the following frequency distribution whose mean is $50.$

$x$	$10$	$30$	$50$	$70$	$90$	Total
$f$	$17$	$f_1$	$32$	$f_2$	$19$	$120$

✓

Answer

Let $f_1$ and $f_{2 }$ be the missing frequencies.
We prepare the following frequency distribution table.

$(x_i)$	$(f_i)$	$f_ix_i$
$10$	$17$	$170$
$30$	$f_1$	$30f_1$
$50$	$32$	$1600$
$70$	$f_2$	$70f_2$
$90 $	$19$	$1710$
Total	120	$3480+30f_1+70f_2$

Here,
$\sum\text{f}_\text{i}=68+\text{f}_1+\text{f}_2$
But
$68+\text{f}_1+\text{f}_2=120 ($Given$)$
Therefore,
$68+\text{f}_1+\text{f}_2=120$
$\Rightarrow\text{f}_1+\text{f}_2=120-68=52$
$\text{f}_2=52-\text{f}_1\ \dots(1)$
Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{3480+30\text{f}_1+70\text{f}_2}{120}$
$\Rightarrow\frac{3480+30\text{f}_1+70(52-\text{f}_1)}{120}$ Using equation $1$
$\Rightarrow\frac{3480+30\text{f}_1+3640+70\text{f}_1}{120}$
$=\frac{7120-40\text{f}_1}{120}$
But mean $= 50 ($given$)$
Therefore,
We have,
$50=\frac{7120-40\text{f}_1}{120}$
$6000=7120-40\text{f}_1$
$40\text{f}_1=1120$
$\text{f}_1=\frac{1120}{40}=28$
Substituting the value of $f_1$ in equation $1,$
We have,
$f_2 = 52 - 28 = 24$
Thus, the missing frequencies are $f_1 = 28$ and $f_2 = 24$ respectively.

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