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

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:

Difficult
View Solution

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:

Difficult
View Solution

In a frequency distribution,if the mean and median are $21$ and $22$ respectively,then what is the approximate mode?

The mode of the following frequency distribution is:
Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$ $25-30$
Frequency $(f_i)$ $3$ $4$ $7$ $11$ $2$ $5$

Difficult
View Solution

In a series,if $4$ appears $3$ times,$5$ appears $4$ times,$6$ appears $5$ times,$7$ appears $8$ times,$8$ appears $7$ times,and $9$ appears $6$ times,then the mode of the numbers is = .......

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo