The mean deviation from the mean of the discr | vedclass

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(D) Step $1$: Find the mean $(\bar{x})$ of the data.$\bar{x} = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 22}{9} = \frac{99}{9} = 11$.Step $2$: Calcula

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$6$

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(D) Step $1$: Find the mean $(\bar{x})$ of the data.$\bar{x} = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 22}{9} = \frac{99}{9} = 11$.Step $2$: Calculate the absolute deviations from the mean $|x_i - \bar{x}|$.$|2 - 11| = 9$$|3 - 11| = 8$$|5 - 11| = 6$$|7 - 11| = 4$$|11 - 11| = 0$$|13 - 11| = 2$$|17 - 11| = 6$$|19 - 11| = 8$$|22 - 11| = 11$Step $3$: Calculate the mean of these absolute deviations.$	ext{Mean Deviation} = \frac{9 + 8 + 6 + 4 + 0 + 2 + 6 + 8 + 11}{9} = \frac{54}{9} = 6$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq	$6$	$7$	$15$	$16$	$4$	$2$

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$

Class Interval	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$
Frequency	$1$	$2$	$3$	$2$	$1$

The mean deviation from the mean of the discrete data $2, 3, 5, 7, 11, 13, 17, 19, 22$ is

Similar Questions

Calculate the mean deviation about the median for the following data:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Freq $6$ $7$ $15$ $16$ $4$ $2$

Find the mean deviation about the median for the given data: $340, 150, 210, 240, 300, 310, 320$.

The mean deviation from the median for the following distribution (corrected to $2$ decimals) is
$x_i$ $3$ $6$ $9$ $12$ $13$ $15$ $21$ $22$
$f_i$ $3$ $4$ $5$ $2$ $4$ $5$ $4$ $3$

The approximate value of the mean deviation about the mean for the following data is
Class Interval $0-2$ $2-4$ $4-6$ $6-8$ $8-10$
Frequency $1$ $2$ $3$ $2$ $1$

If the mean deviation of the numbers $1, 1 + d, . . . , 1 + 100d$ from their mean is $255,$ then a value of $d$ is

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