What is the relationship between mean,median,and mode for a moderately skewed distribution?
- AMode = Median - $2$ Mean
- BMode = $2$ Median - Mean
- CMode = $2$ Median - $3$ Mean
- DMode = $3$ Median - $2$ Mean
Explore More
Similar Questions
The mode of the following data set is $0, 1, 6, 7, 2, 3, 7, 6, 6, 2, 6, 0, 5, 6, 0$.
Easy
View SolutionIn any discrete series (when all values are not same),the relationship between $M.D.$ about mean and $S.D.$ is:
Easy
View SolutionKarl 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:
Difficult
View SolutionThe mode of the following frequency distribution is:
Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$ $25-30$ Frequency $(f_i)$ $3$ $4$ $7$ $11$ $2$ $5$
| Class | $0-5$ | $5-10$ | $10-15$ | $15-20$ | $20-25$ | $25-30$ |
| Frequency $(f_i)$ | $3$ | $4$ | $7$ | $11$ | $2$ | $5$ |
Difficult
View SolutionConsider 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 $.....$
| 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 $.....$
DifficultJEE MAIN 2021
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo