What is the median of an ordered series of n | vedclass

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(D) For a series with an even number of observations $n$,the data is first arranged in ascending or descending order. Since $n$ is even,there are two

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Mean of the $(\frac{n}{2})$-th and $(\frac{n}{2}+1)$-th terms

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(D) For a series with an even number of observations $n$,the data is first arranged in ascending or descending order. Since $n$ is even,there are two middle terms located at the $(\frac{n}{2})$-th position and the $(\frac{n}{2}+1)$-th position. The median is calculated as the arithmetic mean (average) of these two middle terms. Therefore,$	ext{Median} = \frac{(\frac{n}{2}	ext{-th term}) + ((\frac{n}{2}+1)	ext{-th term})}{2}$.

Class	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
$f_i$	$8$	$30$	$40$	$12$	$10$

Marks obtained	No. of students
$0-10$	$2$
$10-20$	$18$
$20-30$	$30$
$30-40$	$45$
$40-50$	$35$
$50-60$	$20$
$60-70$	$6$
$70-80$	$3$

What is the median of an ordered series of $n$ even observations?

Similar Questions

Find the median of the following frequency distribution.

Class $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$

$f_i$ $8$ $30$ $40$ $12$ $10$

Compute the median from the following table:

Marks obtained No. of students

$0-10$ $2$

$10-20$ $18$

$20-30$ $30$

$30-40$ $45$

$40-50$ $35$

$50-60$ $20$

$60-70$ $6$

$70-80$ $3$

Consider a set of observations ${x_1}, {x_2}, {x_3}, ..., {x_{101}}$ such that ${x_1} < {x_2} < {x_3} < ... < {x_{100}} < {x_{101}}$. The mean deviation of this set of observations about a point $k$ is minimum when $k$ equals:

The average of $n$ numbers $x_1, x_2, x_3, ..., x_n$ is $M$. If $x_n$ is replaced by $x'$,then the new average is:

The mean of discrete observations $y_1, y_2, \ldots, y_n$ is given by:

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