0
$\begingroup$

Many of the articles discuss about the relationship between mean and standard deviation of numbers above zero stating that mean should be greater than standard deviation

But if the numbers or the data has negative integers with scores ranging from -10 to +10, is it necessary that mean should be greater than standard deviation.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ many of which articles? $\endgroup$
    – Glen_b
    Commented Jan 19, 2013 at 11:17
  • $\begingroup$ I suspect this is a sort of poor-man's Normality test: the the data are all positive and the standard deviation is larger than the mean then the data obviously are not close to Normal. $\endgroup$
    – Jonathan Christensen
    Commented Jan 19, 2013 at 15:06
  • $\begingroup$ Even after checking for normality the mean is less than SD.i mentioned that the raw scores range from -10 to +10. $\endgroup$
    – user19828
    Commented Jan 21, 2013 at 3:15
  • $\begingroup$ in this website also there is one query about the relationship between mean and SD and it is mentioned that mean is expected to be greater than SD $\endgroup$
    – user19828
    Commented Jan 21, 2013 at 3:21

1 Answer 1

Reset to default
2
$\begingroup$

The mean can be smaller than the standard deviation in your example, and even when all of the numbers in the dataset are positive. For example, the set {1, 1, 1, 1, 1, 1, 100} has a mean of just over 15, but the SD is about 37.4.

In the case where all of the numbers are negative, the mean will certainly be smaller than the SD, since the SD will always be >= 0 (since it's the square root of the variance), and the mean always negative.

In short, the relationship between the mean and the standard deviation depends on the specific dataset--there's no requirement that the mean be larger than the SD.

Improve this answer
$\endgroup$
1
  • $\begingroup$ But in the example of {1,1,1,1,1,1,100} the data is not normally distributed if it is normally distributed?? $\endgroup$
    – user19828
    Commented Jan 21, 2013 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.