For a normal distribution,if the mean is $M$,mode is $M_0$,and median is $M_d$,then:
- A$M > M_d > M_0$
- B$M < M_d < M_0$
- C$M = M_d \times M_0$
- D$M = M_d = M_0$
Explore More
Similar Questions
The sum of the absolute deviations is minimum when taken with respect to:
In a moderately asymmetrical distribution,the mode and mean are $7$ and $4$ respectively. The median is:
Easy
View SolutionFind the mode of the given frequency distribution.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $f_i$ $2$ $18$ $30$ $45$ $35$ $20$ $6$ $3$
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
| $f_i$ | $2$ | $18$ | $30$ | $45$ | $35$ | $20$ | $6$ | $3$ |
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 SolutionIf $\text{mean} = (3 \text{median} - \text{mode}) k$,then the value of $k$ is
Easy
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