Karl Pearson's coefficient of skewness of a d | vedclass

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

Question

(B) The Karl Pearson's coefficient of skewness $(S_k)$ is given by the formula: $S_k = \frac{	ext{Mean} - 	ext{Mode}}{\sigma}$.Given: $S_k = 0.32$,M

Vedclass · Accepted Answer

$38.81$

Vedclass · Answer

(B) The Karl Pearson's coefficient of skewness $(S_k)$ is given by the formula: $S_k = \frac{	ext{Mean} - 	ext{Mode}}{\sigma}$.Given: $S_k = 0.32$,Mean $(M)$ = $39.6$,and $\sigma = 6.5$.Substituting the values: $0.32 = \frac{39.6 - M_o}{6.5}$.$39.6 - M_o = 0.32 	imes 6.5 = 2.08$.$M_o = 39.6 - 2.08 = 37.52$.We know the empirical relationship: $	ext{Mode} = 3 	imes 	ext{Median} - 2 	imes 	ext{Mean}$.$37.52 = 3 	imes 	ext{Median} - 2 	imes 39.6$.$37.52 = 3 	imes 	ext{Median} - 79.2$.$3 	imes 	ext{Median} = 37.52 + 79.2 = 116.72$.$	ext{Median} = \frac{116.72}{3} \approx 38.9067$.Rounding to the nearest provided option,the correct value is $38.81$.

Class	$0-6$	$6-12$	$12-18$	$18-24$	$24-30$
Frequency	$a$	$b$	$12$	$9$	$5$

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:

Similar Questions

For a normal distribution,if the mean is $M$,mode is $M_0$,and median is $M_d$,then:

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

Consider the following frequency distribution:

Class $0-6$ $6-12$ $12-18$ $18-24$ $24-30$

Frequency $a$ $b$ $12$ $9$ $5$

If $\text{mean} = \frac{309}{22}$ and $\text{median} = 14$,then the value of $(a-b)^{2}$ is equal to $.....$

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

If in a frequency distribution,the mean and median are $21$ and $22$ respectively,then its mode is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module