# A Normal Distribution Problem

• Dec 12th 2010, 05:24 PM
starkWonder
A Normal Distribution Problem
Okay, I'm not sure wether this is Basic or Advanced, but I am having trouble processing how to find the area in a standard normal distribuution.

Here is the textbook's basic example; the mean is zero, the standard deviation is one, and the z-score is .75. I understand that the equation is Z=(X-mean)divided by standard deviation, and that the back of the book says the answer is .7734, but I don't understand what I must put in the calculator to make this data into that answer.

Also, what if I knew the area, but not the z-score? What would I do then?

EDIT: Oh DUH, forgot. The area before the z-score.
• Dec 12th 2010, 05:48 PM
awkward
Quote:

Originally Posted by starkWonder
Okay, I'm not sure wether this is Basic or Advanced, but I am having trouble processing how to find the area in a standard normal distribuution.

Here is the textbook's basic example; the mean is zero, the standard deviation is one, and the z-score is .75. I understand that the equation is Z=(X-mean)divided by standard deviation, and that the back of the book says the answer is .7734, but I don't understand what I must put in the calculator to make this data into that answer.

Also, what if I knew the area, but not the z-score? What would I do then?

Look up .75 in a table of the cumulative normal distribution-- there is probably one in your book. If you knew the area but not Z, you would reverse this process.

It's possible that your calculator may have a function that will do this for you, but most simple calculators do not.
• Dec 12th 2010, 05:56 PM
starkWonder
The table was in the back cover, and it indeed says the intersection between .7 and .05 is .7734. Now I just feel foolish for not thinking of that. Thank you for the shockingly fast late-night reply, awkward.

...Incidentally, would a mean besides zero or a standard deviation besides one require a totally different process? I'm figuring yes, but drawing a blank as to what the equation would look like.
• Dec 13th 2010, 02:04 PM
awkward
If you have something other than a mean of zero and a standard deviation of 1, you convert it to an equivalent "Z-score" by the formula you quoted earlier, Z = (X - mean) / (standard deviation). Z has a normal distribution with mean zero and standard deviation 1, so you can use your table.