Can somebody explain how to figure this out... or could help me with this question?

Question

How many statdard deviation above and below teh mean do the quartiles of ANY Normal distribution lie? (Use the standard Normal distribution to answer this question).

Answer 1

The standard normal distribution has a probability density function of...
P(x) = (1 / sqrt(2*pi)) * e^(-x^2 / 2)
which has a mean of 0 and a standard deviation of 1.

To find the quartiles, you need to (numerically) integrate this function from
x = -infinity
to
x = x_quarter
where x_quarter is the position such that exactly 1/4 of the probability distribution lies to the left, and 3/4 lies to the right. This means you're looking for the following result:
integral[P(x), x, -infinity, a] = 0.25

The simplest way is to just guess values for x_quarter and keep refining your guess until you get as close as you like. This integral can't be solve analytically, so you have to use a calculator or computer to integrate it numerically.

The result I obtain after guessing a few times is
a = -0.67449

So, integrating from -infinity to -0.67449 covers 1/4 of the distribution. Thus x = -0.67449 is the first quartile. The second quartile is just the mean, or x = 0. The third quartile is symmetrical, so it's x = 0.67559.

Answer 2

Just go to your friendly z-tables and look at the standard deviation corresponding to the 25th percentile of the distribution.