Find the mean of the distribution: Class | vedclass

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(D) To find the mean,we first calculate the class mark $(x_i)$ for each class interval using the formula: $x_i = \frac{	ext{Upper limit} + 	ext{Lowe

Vedclass · Accepted Answer

$5.5$

Vedclass · Answer

(D) To find the mean,we first calculate the class mark $(x_i)$ for each class interval using the formula: $x_i = \frac{	ext{Upper limit} + 	ext{Lower limit}}{2}$.						Class			Class mark $(x_i)$			Frequency $(f_i)$			$f_i x_i$							$1-3$			$2$			$9$			$18$							$3-5$			$4$			$22$			$88$							$5-7$			$6$			$27$			$162$							$7-10$			$8.5$			$17$			$144.5$							Total			-			$\Sigma f_i = 75$			$\Sigma f_i x_i = 412.5$			The mean $(\bar{x})$ is calculated as:$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{412.5}{75} = 5.5$.Thus,the mean of the given distribution is $5.5$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Frequency	$5$	$x$	$20$	$y$	$2$

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	$2$	$10$	$40$	$25$	$13$	$10$

Find the mean of the distribution:

Class $1-3$ $3-5$ $5-7$ $7-10$

Frequency $9$ $22$ $27$ $17$

Similar Questions

For a given frequency distribution,$n=100, A=20$ and $\bar{x}=20$. Then,$\Sigma f_{i} d_{i}=\ldots \ldots \ldots . .$

For some given data,if $M + \bar{x} = 165$ and $M - \bar{x} = 1$,then $Z = \dots$

$Z - \bar{x} = \dots \times (M - \bar{x})$

The median of the following frequency distribution is $50$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.

Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$

Frequency $5$ $x$ $20$ $y$ $2$

Find the mean of the following frequency distribution:

Class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$

Frequency $2$ $10$ $40$ $25$ $13$ $10$

(in $.7$)

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Find the mean of the distribution: Class $1-3$ $3-5$ $5-7$ $7-10$ Frequency $9$ $22$ $27$ $17$