Measures of Central Tendency Test - 25

$Since$ $n= 9,$ $the$ $median$ $will$ $be$ $the$ $5th(\frac{9 + 1}{2}) term.$
$If$ $the$ $largest$ $4$ $terms$ $are$ $increased$ $by$ $2,$ $there$ $is$ $no$ $effect$ $on$ $the$ $5th$ $term,$ $implying$ $median$ $will$ $remain$ $same.$

Average of marks of $70$  boys =$75$

Mode is the value which occurs most often in the data set of values.

Mean $= 6.5 = \dfrac{3 \times (x-2) + 5 \times (x + 2) + 7 \times (x - 3) + 9 \times (x + 3)}{x-2+x+2+x-3+x+3}$
$\therefore$ $6.5 = \dfrac{3x - 6 + 5x + 10 + 7x - 21 + 9x + 27}{4x}$
$\therefore$ $6.5 \times 4x = 24x + 10$
$\therefore$ $26x - 24x = 10$
$\therefore$ $x = 5$

Given Data: 

Terms are: $4, 4, 5, 7, 6, 7, 7, 12, 3$.
Arranging the terms in ascending order: $3,4,4,5,6,7,7,7,12$.

Since the number of terms is odd the median will be the middle term i.e. $5^{th}$ term which is $6$.

Mean of a frequency distribution =$\displaystyle \dfrac{\sum f_i x_i}{\sum f_i}$
Mean $= \dfrac {5520}{80}$
Mean $= 69$

Mean of combined three sets $=\dfrac{Total \space Sum}{Total \space Count}$
$=\dfrac{15\times6 + 22.5\times4 + 24\times5}{6+4+5}$
$=\dfrac{300}{15}$
$= 20$

$$
\text{Given Mean  weight of 20 students}=25\\ \Rightarrow \dfrac { \text{Sum of weight of 20 students} }{ 20 } =25\\
\Rightarrow \text{Sum of weight of 20 students}=25\times 20=500kg\\
\text{Given Mean weight of 10 students}=40\\ \Rightarrow \dfrac { \text{Sum of weight of 10 students} }{ 10 } =40\\
\Rightarrow \text{Sum of weight of 10 students}=40\times 10=400kg$$
Mean weight of all 30 students
$$\Rightarrow \dfrac { \text{Sum of weight of 30 students} }{ 30 } \\
\Rightarrow \dfrac { \text{Sum of weight of 20 students+Sum of weight of 10 students} }{ 30 } \\
=\dfrac { 500+400 }{ 30 } =\dfrac { 900 }{ 30 } =30kg
$$

$\textbf{Step -1: Finding the sum of first 10 observations.}$

                $\text{Average of first 10 observations}=4.5$

                $\mathbf{\because\text{Average}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$

                $\therefore4.5=\dfrac{\text{Sum of first 10 observations}}{10}$

                $\Rightarrow\text{Sum of first 10 observations}=4.5\times10=45$

$\textbf{Step -2: Finding the sum of remaining 30 observations.}$

                $\text{Average of remaining 30 observations}=3.5$

                $\therefore3.5=\dfrac{\text{Sum of remaining 30 observations}}{30}$

                $\Rightarrow\text{Sum of remaining 30 observations}=3.5\times30=105$

$\textbf{Step -3: Finding the average of the whole group.}$

                $\text{Sum of total 40 students}=45+105=150$

                $\text{Number of total observations}=40$

                $\text{Average of the whole group}=\dfrac{150}{40}$

                $=\dfrac{15}{4}$

$\textbf{ Hence, option B is correct.}$