I'm in an introductory stats class, I have a question regarding the standard deviation. So normally, when I am asked to find the probability or percentage of something, I use normalcdf.
Can somebody explain to me when we use normalcdf (under what circumstances), and also when using normalcdf, when do we divide the standard deviation by square root of n?
-- Thanks
There are many different normal distributions depending on the choice of the population mean $\mu$ and the population standard deviation $\sigma.$ It is not possible to put tables for all of them in the back of a book.
However, all normal distributions have fundamentally the same shape (shifted left or right by the choice of $\mu$ and stretched or squeezed by the choice of $\sigma.)$ So by putting only the table for the standard normal distribution (with $\mu = 0$ and $\sigma = 1)$ in the back of the book, it becomes feasible to solve probability problems for the great variety of normal distributions.
Suppose there is a college entrance exam with with normally distributed scores and we know that $\mu = 400$ and $\sigma = 50.$ If State U admits only students with scores above 430, then what proportion of students taking the standardized test can be admitted? We have $X \sim \mathsf{Norm}(\mu = 400, \sigma = 50)$ and we want to find $P(X > 430).$ According to the distribution of the test, we have the figure below. The total area under any density curve is $!.$ We want the area under the normal density curve to the right of the vertical red line. Just by eye, it looks as if the answer may be something like 25% or 30%.
We can change this into a problem about a standard normal random variable $Z$ by using the standardization process: $Z = \frac{X - 400}{50}.$ This is the method:
$$P(X > 430) = P\left(\frac{X-400}{50} > \frac{430-400}{50}\right) = P(Z > 30/50 = 0.6).$$
Notice that the events inside the parentheses in each member of this equation are the same; we used 'legal' arithmetic to get from one to the next.
Now, you can look at the CDF of the standard normal distribution in your book and determine that the area under a standard normal curve to the left of $0.6$ is $0.7237,$ so the area to the right of $0.6$ must be $1 - 0.7237 = 0.2743.$ (I'm sorry I can't say exactly how to get the $0.7237$ because different books have slightly different kinds of tables; for instance, you might have to get it as $.5000 + .2237 = .7237.)$
So the answer is $P(X > 430) = 0.2743.$
The figure below, is essentially the same as the one above, but this time the numbers on the horizontal axis are 'z-scores' matching the standard normal distribution.