Q75 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
The mean deviation about the mean for the following data is
\begin{array}{|c|c|c|c|c|c|}\hline Class interval & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\\hline Frequency & 1 & 3 & 5 & 3 & 1 \\\hline\end{array}
- A2
- B$\frac{15}{13}$
- C$\frac{22}{13}$
- Correct answerD$\frac{20}{13}$
Solution
$\begin{array}{|c|c|c|c|c|c|}\hline \text{Class interval} & \text{Class marks} & \text{Frequency} & \text{xifi} & \left(x_i-\bar{x}\right) & f_i\left(x_i-\bar{x}\right) \\\hline 0-2 & 1 & 1 & 1 & 4 & 4 \\\hline 2-4 & 3 & 3 & 9 & 2 & 6 \\\hline 4-6 & 5 & 5 & 25 & 0 & 0 \\\hline 6-8 & 7 & 3 & 21 & 2 & 6 \\\hline 8-10 & 9 & 1 & 9 & 4 & 4 \\\hline & & & \sum_{x_i f_i} =65 & & 20 \\\hline\end{array}$
$\sum f_i=13 \Rightarrow \mu=\frac{\sum x_i f_i}{\sum f_i}=\frac{65}{13}=5$
Mean deviation about mean $=\frac{\sum f_i\left(x_i-\bar{x}\right)}{\sum f_i}=\frac{20}{13}$.
$\sum f_i=13 \Rightarrow \mu=\frac{\sum x_i f_i}{\sum f_i}=\frac{65}{13}=5$
Mean deviation about mean $=\frac{\sum f_i\left(x_i-\bar{x}\right)}{\sum f_i}=\frac{20}{13}$.
Hence, the correct answer is option: 'D'.