If Z = 36.8 and M = 33.6,then bar{x} = ldots | vedclass

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(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:$Z = 3M - 2\bar{x}$Given values are $Z = 3

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:$Z = 3M - 2\bar{x}$Given values are $Z = 36.8$ and $M = 33.6$.Substituting these values into the formula:$36.8 = 3(33.6) - 2\bar{x}$Calculate $3 	imes 33.6$:$36.8 = 100.8 - 2\bar{x}$Rearrange the equation to solve for $\bar{x}$:$2\bar{x} = 100.8 - 36.8$$2\bar{x} = 64$Divide by $2$:$\bar{x} = \frac{64}{2} = 32$Thus,the value of $\bar{x}$ is $32$.

Class	$50-70$	$70-90$	$90-110$	$110-130$	$130-150$	$150-170$
Frequency	$10$	$18$	$7$	$6$	$5$	$4$

Class	$0-4$	$4-8$	$8-12$	$12-16$	$16-20$	$20-24$	$24-28$	$28-32$	$32-36$
Frequency	$6$	$8$	$17$	$23$	$16$	$15$	$f$	$4$	$3$

Class	Frequency $(f)$
$110-120$	$6$
$120-130$	$25$
$130-140$	$48$
$140-150$	$72$
$150-160$	$116$
$160-170$	$60$
$170-180$	$38$
$180-190$	$22$
$190-200$	$3$

If $Z = 36.8$ and $M = 33.6$,then $\bar{x} = \ldots \ldots \ldots \ldots .$

Similar Questions

Find the mean of the following data by all the three methods:

Class $50-70$ $70-90$ $90-110$ $110-130$ $130-150$ $150-170$

Frequency $10$ $18$ $7$ $6$ $5$ $4$

If $Z - M = 4$,then $M - \bar{x} = \dots$

The mean of the following data is $16.$ Find the missing frequency $f.$

Class $0-4$ $4-8$ $8-12$ $12-16$ $16-20$ $20-24$ $24-28$ $28-32$ $32-36$

Frequency $6$ $8$ $17$ $23$ $16$ $15$ $f$ $4$ $3$

Find the median of the following frequency distribution:

Class Frequency $(f)$

$110-120$ $6$

$120-130$ $25$

$130-140$ $48$

$140-150$ $72$

$150-160$ $116$

$160-170$ $60$

$170-180$ $38$

$180-190$ $22$

$190-200$ $3$

$\frac{Z-M}{M-\bar{x}}=\ldots \ldots \ldots \ldots$

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