Thread: A Normal Distribution Problem

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 distribution. I am posting it in both forums, just in case.

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.

Ok, so you're looking for (the area before the z score). You can use your calculator or use a z-table (which should be in the appendices of most statistics books; people prefer to use the calculator because its quicker). If you're using the TI-83/84 calculator, all you need to punch in is: normalcdf(-1E99,.75). It should return a value of about .7734.

In general, there are two more arguments that can be put in, but since you're dealing with the standard normal distribution, those two arguments are not necessary (they are already pre-programmed to be 0 and 1, the mean and standard deviation respectively). So in this case, normalcdf(-1E99,.75,0,1) will give the same value as normalcdf(-1E99,.75).

If you use a table, look up the z value 0.75 by looking for the 0.7 row and the 0.05 column, then find their intersection. In most books, this will give you the value of area to the left of the z value. Sometimes, they give the area to the right of the z value; you would then need to subtract that value from 1 to get the desired answer.

If you need to go the other way, you can use the inverse norm feature on the calculator. So for instance on the TI-83/84, invnorm(.7734) should output the z value .75 (approximately).

I hope this clarifies things.