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Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsMean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive2 Marks
Question
Find the mean of the following frequency distribution using step-deviation method:
Class
$84-90$
$90-96$
$96-102$
$102-108$
$108-114$
$114-120$
Frequency
$15$
$22$
$20$
$18$
$20$
$25$
✓

Answer

Class interval
Frequency $f_i$
Mid values $x_i$
 $\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{A}}{\text{h}}=\frac{\text{x}_i-99}{6}$
$f_i× u_i$
$84-90$
$15$
$87$
 $-2$
$-30$
$90-96$
 $22$
$93$
 $-1$
$-22$
$96-102$
 $20$
$99 = A$
 $0$
$0$
$102-108$
$18$
$105$
 $1$
$18$
$108-114$
$20$
$111$
 $2$
$40$
$114-120$ $25$ $117$ $3$ $75$
  $\sum\text{f}_\text{i}=120$     $\sum\text{f}_\text{i}\text{u}_\text{i}=81$
Thus, $\text{A}=99,\ \text{h}=6,\ \sum\text{f}_\text{i}=120$ and $\sum\text{f}_\text{i}\text{u}_\text{i}=81$
Mean $=\text{A}+\Big\{\text{h}\times\frac{\sum\text{f}_\text{i}\text{u}\text{i}}{\sum\text{f}_\text{i}}\Big\}$
$=99+\Big\{6\times\frac{81}{120}\Big\}$
$=99+4.05$
$=103.05$

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