Is there a possibility that variance is less than the standard deviation?
well, when can you have x^2 < x ?
If the standard deviation is less than 0 but not negative sir is it possible?
Example 0.02 is the standard deviation then the variance is 0.0004
they can never be negative, but as you say, if x < 1, then x^2 < x
Is it possible to have standard deviation less than 1?
of course. google your question and you can find several discussions of the topic.
No, it is not possible for the variance to be less than the standard deviation. The standard deviation is the square root of the variance, so by definition, the standard deviation will always be equal to or greater than the square root of the variance.
To understand why this is the case, let's explain how the variance and standard deviation are calculated. The variance is a measure of the spread or dispersion of a set of data points. It is calculated by taking the average of the squared differences between each data point and the mean. The formula for variance is:
Variance = (1 / n) * Σ (xᵢ - μ)²
where n is the number of data points, Σ represents the sum of the squared differences, xᵢ represents each data point, and μ is the mean of the data.
On the other hand, the standard deviation is the square root of the variance. It is calculated by taking the square root of the variance:
Standard Deviation = √(Variance)
Since the variance is expressed in squared units (e.g., squared meters, squared dollars), taking the square root gives us the standard deviation in the original units of the data.
Since squaring a number always results in a positive value, the sum of squared differences in the variance calculation is always positive. Taking the square root of the variance will then give us a positive standard deviation.
Therefore, the standard deviation can never be less than the variance. It will always be equal to or greater than the square root of the variance.