The mean deviation about the mean for the obs | vedclass

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Question

(C) Given observations are $x_i = -1, 0, 4$.First,calculate the mean $(\bar{x})$:$\bar{x} = \frac{-1 + 0 + 4}{3} = \frac{3}{3} = 1$.Next,calculate the

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$2$

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(C) Given observations are $x_i = -1, 0, 4$.First,calculate the mean $(\bar{x})$:$\bar{x} = \frac{-1 + 0 + 4}{3} = \frac{3}{3} = 1$.Next,calculate the absolute deviations from the mean $|x_i - \bar{x}|$:$|-1 - 1| = |-2| = 2$$|0 - 1| = |-1| = 1$$|4 - 1| = |3| = 3$Sum of absolute deviations = $2 + 1 + 3 = 6$.Mean deviation about the mean = $\frac{1}{n} \sum |x_i - \bar{x}| = \frac{6}{3} = 2$.

Height in $cm$	Number of boys
$95-105$	$9$
$105-115$	$13$
$115-125$	$26$
$125-135$	$30$
$135-145$	$12$
$145-155$	$10$

The mean deviation about the mean for the observations $ -1, 0, 4 $ is:

Similar Questions

For a frequency distribution,the mean deviation from the mean is computed by:

The mean deviation from the median for the following data is
$x_i$ $2$ $3$ $5$ $7$ $9$
$f_i$ $8$ $6$ $4$ $2$ $1$

What is the mean deviation of a frequency distribution equal to?

Find the mean deviation about the mean for the following data:
Height in $cm$ Number of boys
$95-105$ $9$
$105-115$ $13$
$115-125$ $26$
$125-135$ $30$
$135-145$ $12$
$145-155$ $10$

Find the mean deviation about the median for the data $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$.

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