The median of a set of 9 distinct observation | vedclass

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(D) Given that the number of observations $n = 9$.The median of $9$ observations is the $\left( \frac{9+1}{2} \right)^{th} = 5^{th}$ observation.Let t

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Remains the same as that of the original set

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(D) Given that the number of observations $n = 9$.The median of $9$ observations is the $\left( \frac{9+1}{2} \right)^{th} = 5^{th}$ observation.Let the ordered observations be $x_1 The median is $x_5 = 20.5$.If the largest $4$ observations $(x_6, x_7, x_8, x_9)$ are increased by $2$,the new set of observations is $x_1, x_2, x_3, x_4, x_5, (x_6+2), (x_7+2), (x_8+2), (x_9+2)$.Since $x_5$ is smaller than $x_6$,and $x_6 Therefore,the $5^{th}$ observation remains $x_5$,which is $20.5$.Thus,the median remains the same as that of the original set.

Height (in $cm$)	$160$	$150$	$152$	$161$	$156$	$154$	$155$
No of students	$12$	$8$	$4$	$4$	$3$	$3$	$7$

The median of a set of $9$ distinct observations is $20.5$. If each of the largest $4$ observations of the set is increased by $2$,then the median of the new set:

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The following data gives the distribution of height of students:

Height (in $cm$) $160$ $150$ $152$ $161$ $156$ $154$ $155$

No of students $12$ $8$ $4$ $4$ $3$ $3$ $7$

The median of the distribution is:

$x_1, x_2, \dots, x_{34}$ are numbers such that $x_i = 150$ for all $i \in \{1, 2, \dots, 10\}$ and $x_{i+1} - x_i = -2$ for all $i \in \{10, 11, \dots, 33\}$. Find the median of $x_1, x_2, \dots, x_{34}$.

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