I am just wondering what a standard deviation means when the distribution is non-normal.

I was looking around and was referenced to read Chebyshev's Inequality which states:

$$P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^{2}}$$

My question is related to a grade that I got in a class. The TA stated that the mean was 85.6% with a standard deviation of 12%.

So by using Chebyshev, I think that means that the probability that a grade from the sample is greater than 1.2$\sigma$ from the mean is less than or equal to 69.44%? (I used 1.2$\sigma$ because it is impossible to get over 100%.)

So essentially, there is less than or equal than 69.44% probability that my grade is between $85.6 \pm 14.4$, or greater than a 30.56% chance that my grade is above 100% (impossible) or below 71.2%. Thus, I can conclude that 30.56% of the students got below 71.2%.

Am I interpreting this correct? Is there anything else I can use the standard deviation for when the distribution is not normal? I am pretty sure that the grade distribution is not normal but I do not know what it actually looks like. Any clarification would be greatly appreciated.


Well variance is clear enough, right? It's the average of squared distances from the mean: $$ \sigma^2 = \frac1n \sum_{i=1}^n (X_i - \mu)^2 $$ And then standard deviation is just the square root of variance.

None of this is related to the distribution being normal or not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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