begin{aligned} &text{Find the mean deviation | vedclass

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(B) Step $1$: Find the class midpoints $(x_i)$ : The midpoint for a class interval is calculated as: $x_i = \frac{	ext{Lower limit} + 	ext{Upper lim

Vedclass · Accepted Answer

$\frac{16}{3}$

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(B) Step $1$: Find the class midpoints $(x_i)$ : The midpoint for a class interval is calculated as: $x_i = \frac{	ext{Lower limit} + 	ext{Upper limit}}{2}$ - $x_1 = \frac{0+6}{2} = 3$ - $x_2 = \frac{6+12}{2} = 9$ - $x_3 = \frac{12+18}{2} = 15$ - $x_4 = \frac{18+24}{2} = 21$ - $x_5 = \frac{24+30}{2} = 27$ Thus,$x_i = 3, 9, 15, 21, 27$. Step $2$: Compute the mean $(\bar{x})$ : The mean is calculated as: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ - $\sum f_i = 1+2+3+2+1 = 9$ - $\sum f_i x_i = (1 \cdot 3) + (2 \cdot 9) + (3 \cdot 15) + (2 \cdot 21) + (1 \cdot 27) = 3 + 18 + 45 + 42 + 27 = 135$ - $\bar{x} = \frac{135}{9} = 15$ Step $3$: Find $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$ : - For $x_1 = 3: |3-15| = 12, f_1 |x_i - \bar{x}| = 1 \cdot 12 = 12$ - For $x_2 = 9: |9-15| = 6, f_2 |x_i - \bar{x}| = 2 \cdot 6 = 12$ - For $x_3 = 15: |15-15| = 0, f_3 |x_i - \bar{x}| = 3 \cdot 0 = 0$ - For $x_4 = 21: |21-15| = 6, f_4 |x_i - \bar{x}| = 2 \cdot 6 = 12$ - For $x_5 = 27: |27-15| = 12, f_5 |x_i - \bar{x}| = 1 \cdot 12 = 12$ Step $4$: Calculate mean deviation about the mean: - $	ext{Mean deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{12+12+0+12+12}{9} = \frac{48}{9} = \frac{16}{3}$

$x_i$	$10$	$30$	$50$	$70$	$90$
$f_i$	$4$	$24$	$28$	$16$	$8$

$\begin{aligned} &\text{Find the mean deviation about the mean for the following data:} \\ &\begin{array}{|l|c|c|c|c|c|} \hline \text{Class Interval} & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \hline \text{Frequency} & 1 & 2 & 3 & 2 & 1 \\ \hline \end{array} \end{aligned}$

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