Standard Deviation Conceptual [intro. Stats]

[END]

Hello, PF!

[My question pertains to a non-rigorous, undergraduate introductory Probability and Statistics course. I'm no math major, so please correct me if I've mishandled any terms or concepts as I try to express myself. I'm always eager to learn!]

* * *​

In a discussion of the standard deviation of a sample in relation to the 68-95-99.7 rule, the following "conceptual" example was given—or rather, made up on the spot—by our professor:

Assume $\bar{x}=50 \%$ and $s=20 \%$ for test scores (in units of percent correct), and assume that the sample represents the normal distribution (symmetrical and bell-shaped) of a test where no test score range below $0 \%$ and none above $100 \%$ (sorry, fellas, no extra credit).

It occurred to me that any score beyond $2.5$ standard deviations would be a score of more than $100 \%$ or less than $0 \%$. According to the three-sigma rule, this would still only encompass approximately $98.7\%$ of the scores meaning that approximately $1.3\%$ of the scores fall outside this possible range.

My question:

Is the above example even possible given the "parameters" (limits?—I can't find the right word) ${0 \%}≤x_i≤{100 \%}$?

And

Extrapolating this question to the overall concept, can any standard deviation $s$ of a normal distribution ever exceed the possible range of data points/values within that distribution?

My guess is that this was simple oversight and an error on the part of my professor.

Thank you!

 

[Stephen Tashi]

A random variable with a bounded range (such as 0 to 100) is not actually normally distributed. However, a normal distribution can often be used to get approximate answers to questions about such a random variable. Many problems in textbooks expect students to make such an approximation. I'd call this type of approximation a tradition, not an oversight.

 

[END]

Ah, bounded was the word I was looking for! Thank you, Stephen.

A random variable with a bounded range (such as 0 to 100) is not actually normally distributed.

Would the example then be considered a Truncated normal distribution ?

If this is the case, what would 2.5 or 3 standard deviations imply in relation to the three-sigma rule when the values simply cannot extend beyond the boundaries? Does the three-sigma/68-95-99.7 rule simply not apply to this (and other truncated distributions); i.e., this example would have a different set of probabilities in relation to the various standard deviations: $$Pr(\bar{x} - 2.5s ≤x≤ \bar{x} + 2.5s) = 1.00 \ ?$$



Thank you.

 

[Stephen Tashi]

Would the example then be considered a Truncated normal distribution ?

The example used the normal distribution as an approximation. If you want to make a different example, you could use a different distribution. A truncated normal distribution is but one example of what could be used.

i.e., this example would have a different set of probabilities in relation to the various standard deviations: $$Pr(\bar{x} - 2.5s ≤x≤ \bar{x} + 2.5s) = 1.00 \ ?$$

In general, a truncated normal distribution would have a different set of probability values for such an interval than a non-truncated normal distribution. (Keep in mind that a truncated normal distribution has a different "s" than the normal distribution that was truncated.)

 

[Hornbein]

There is no such thing as a normal distribution in the real world. It is a mathematical ideal. Quite often there are deviations in the tails. Often it is close enough for jazz.