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Mean, Median and Mode of Ungrouped Data — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsMean, Median and Mode of Ungrouped Data5 Marks
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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