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

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

What is the relationship between mean,median,and mode for a moderately skewed distribution?

Let $a_1, a_2, \dots, a_{101}$ be a group of real numbers such that $a_i > a_{i+1}$ for all values of $i$ and the mean square deviation of the group is minimum about the number $a_{51}$. Then the mode of the group will be:

Find the mode of the following frequency distribution.

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

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

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

The mode of the following data set is $0, 1, 6, 7, 2, 3, 7, 6, 6, 2, 6, 0, 5, 6, 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module