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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.

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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.

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