If Z+M=34 and M+bar{x}=40, then M=ldots ldots | vedclass

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Question

(C) We know the empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by: $Z = 3M - 2\bar{x}$.
Given equations:
$1) Z

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(C) We know the empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by: $Z = 3M - 2\bar{x}$.
Given equations:
$1) Z + M = 34 \implies Z = 34 - M$
$2) M + \bar{x} = 40 \implies \bar{x} = 40 - M$
Substitute these into the empirical formula:
$34 - M = 3M - 2(40 - M)$
$34 - M = 3M - 80 + 2M$
$34 - M = 5M - 80$
$34 + 80 = 5M + M$
$114 = 6M$
$M = \frac{114}{6} = 19$
Therefore,the value of $M$ is $19$.

Height (in $cm$)	Number of students
$150-155$	$15$
$155-160$	$13$
$160-165$	$10$
$165-170$	$8$
$170-175$	$9$
$175-180$	$5$

Class	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$	$250-300$	$300-350$
Frequency	$10$	$15$	$30$	$20$	$15$	$8$	$2$

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$3$	$x$	$7$	$7$	$y$

If $Z+M=34$ and $M+\bar{x}=40,$ then $M=\ldots \ldots \ldots . . .$

Similar Questions

Find the mean,median and mode of the following frequency distribution:

Class $0-50$ $50-100$ $100-150$ $150-200$ $200-250$ $250-300$ $300-350$

Frequency $10$ $15$ $30$ $20$ $15$ $8$ $2$

The mean of the following data is $55.5$ and the total frequency is $20$. Find the missing frequencies $x$ and $y$.
Class $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $3$ $x$ $7$ $7$ $y$

If $3 \bar{x} = 2 M = 60,$ then $Z = \dots$

If mode $= 25$ and mean $= 19,$ then median $= \dots$

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