If text{Mean} = x(3 times text{Median} - text | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The empirical relationship between Mean,Median,and Mode is given by the formula: $	ext{Mode} = 3 	imes 	ext{Median} - 2 	imes 	ext{Mean}$.Rea

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(C) The empirical relationship between Mean,Median,and Mode is given by the formula: $	ext{Mode} = 3 	imes 	ext{Median} - 2 	imes 	ext{Mean}$.Rearranging the terms to solve for $	ext{Mean}$:$2 	imes 	ext{Mean} = 3 	imes 	ext{Median} - 	ext{Mode}$.$	ext{Mean} = \frac{1}{2}(3 	imes 	ext{Median} - 	ext{Mode})$.Comparing this with the given equation $	ext{Mean} = x(3 	imes 	ext{Median} - 	ext{Mode})$,we get $x = 1/2$.

Class	$1 - 10$	$11 - 20$	$21 - 30$	$31 - 40$	$41 - 50$
$f_i$	$5$	$7$	$8$	$6$	$4$

If $\text{Mean} = x(3 \times \text{Median} - \text{Mode})$,then what is the value of $x$?

Similar Questions

In a moderately asymmetrical distribution,the mode and mean are $7$ and $4$ respectively. The median is:

Find the mode of the following frequency distribution.

Class $1 - 10$ $11 - 20$ $21 - 30$ $31 - 40$ $41 - 50$

$f_i$ $5$ $7$ $8$ $6$ $4$

The sum of the absolute deviations is minimum when taken with respect to:

The median is independent of the change of which of the following?

Karl Pearson's coefficient of skewness of a distribution is $0.32$. Its standard deviation $(S.D.)$ is $6.5$ and the mean is $39.6$. The median of the distribution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module