There are n observations and all of them are | vedclass

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(C) Given that $x_1, x_2, \ldots, x_n$ are negative numbers in ascending order. Since they are negative,we can write them as $-|x_1|, -|x_2|, \ldots,

Vedclass · Accepted Answer

$|x_1| - x_2$

Vedclass · Answer

(C) Given that $x_1, x_2, \ldots, x_n$ are negative numbers in ascending order. Since they are negative,we can write them as $-|x_1|, -|x_2|, \ldots, -|x_n|$.Because the order is ascending,we have $-|x_1| |x_2| > \ldots > |x_n|$.After changing the signs of the first term $(x_1)$ and the last term $(x_n)$,the new set of observations becomes $|x_1|, -|x_2|, -|x_3|, \ldots, -|x_{n-1}|, |x_n|$.In this new set,the largest value is $|x_1|$ and the smallest value is $-|x_2|$ (since $|x_1| > |x_2| > \ldots > |x_n|$).The range is defined as $ ext{Maximum} - ext{Minimum}$.Range $= |x_1| - (-|x_2|) = |x_1| + |x_2|$.Wait,re-evaluating the sequence: The original sequence is $x_1 < x_2 < \ldots < x_n < 0$. After changing signs,the new sequence is $-x_1, x_2, x_3, \ldots, x_{n-1}, -x_n$. Since $x_1$ is the most negative,$-x_1$ is the largest positive. The smallest value is $x_2$. Thus,the range is $-x_1 - x_2 = |x_1| - x_2$.

Expenditure in $Rs. (x)$	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
No. of families $(f)$	$24$	$33$	$37$	$b$	$25$

There are $n$ observations and all of them are negative numbers. The ascending order of these observations is $x_1, x_2, \ldots, x_n$. If the signs of the first term and last term in that order are changed,then the range of the data is

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