Q75 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $M_1$ is the mean deviation from the mean of the discrete data $44,5,27,20,8,54,9,14,35$ and $M_2$ is the mean deviation from the median of the same data, then $M_1-M_2=$
- A$\frac{7}{9}$
- B$\frac{2}{3}$
- C$\frac{5}{9}$
- Correct answerD$\frac{4}{9}$
Solution
Arrange the data in ascending order, $5,8,9,14,20,27,35,44,54$ (ungrouped data)
Mean $_3-\vec{x}=\frac{216}{9}=24$.
Mean deviation about mean, i.e. $M_1=\frac{\Sigma\left|x_1-\bar{x}\right|}{N}$
$M_1=\frac{\begin{array}{c}
1-|9|+|-16|+|-15|+|-10| \\
+|-4|+3+11+20+30
\end{array}}{9}=\frac{128}{9}$
Median $=\left(\frac{N+1}{2}\right)^{\text {th }}$ term as $N$ is odd $\Rightarrow$ Median $=5$ th term $\Rightarrow$ Median $(M)=20$ Mean deviation about Median, i.e. $M_2=\frac{\Sigma\left|x_i-M\right|}{N}=\frac{124}{9} \quad \therefore M_1-M_2=\frac{4}{9} .$
Median $=\left(\frac{N+1}{2}\right)^{\text {th }}$ term as $N$ is odd $\Rightarrow$ Median $=5$ th term $\Rightarrow$ Median $(M)=20$ Mean deviation about Median, i.e. $M_2=\frac{\Sigma\left|x_i-M\right|}{N}=\frac{124}{9} \quad \therefore M_1-M_2=\frac{4}{9} .$
Hence, the correct answer is option: 'D'.