# Math Help - Statistics - College Level - Again!

1. ## Statistics - College Level - Again!

If I have found Z scores that are higher than the table, how do I go about finding the percentage?

Thanks!

Does this need further clarification? The prof said this regarding the probs: For any Z value over 4 you can assume that the area of the larger portion is 100% and that of the smaller portion is 0%.

I still can't get my mind around how to begin.

2. Originally Posted by LisaSTLMO
If I have found Z scores that are higher than the table, how do I go about finding the percentage?

Thanks!

Does this need further clarification? The prof said this regarding the probs: For any Z value over 4 you can assume that the area of the larger portion is 100% and that of the smaller portion is 0%.

I still can't get my mind around how to begin.

If you have an 83, 83+, 84, or 84+ you can do this pretty easily...

From the home screen go to the distribution menu. [2nd] [VARS]

Highlighted right away is [1] normalcdf(

Hit [ENTER].

Now you should see normalcdf( on your screen

Since you don't have an infinity button (like the 89), type in a really large negative number (use -1 E 9) where E is retrieved by hitting [2nd] [,].

Hit [,].

Then hit [ENTER]

Let's say my z value was 5.3

Then my percentage would be normalcdf(-1E9,5.3) $\approx 0.999999942$.

Rounding to 2 decimal places, this would give us 100% or 1.

As z gets larger and larger, we assume that the percentage is 100% or 1.

Hope this makes sense.

3. Originally Posted by Chris L T521

If you have an 83, 83+, 84, or 84+ you can do this pretty easily...

From the home screen go to the distribution menu. [2nd] [VARS]

Highlighted right away is [1] normalcdf(

Hit [ENTER].

Now you should see normalcdf( on your screen

Since you don't have an infinity button (like the 89), type in a really large negative number (use -1 E 9) where E is retrieved by hitting [2nd] [,].

Hit [,].

Then hit [ENTER]

Let's say my z value was 5.3

Then my percentage would be normalcdf(-1E9,5.3) $\approx 0.999999942$.

Rounding to 2 decimal places, this would give us 100% or 1.

As z gets larger and larger, we assume that the percentage is 100% or 1.

Hope this makes sense.
Another reason for treating anything beyond z~=3 as having cumulative prob of 1 is that generally the assumptions underlying treating the distribution as normal are no longer valid this far into the tails of the distribution, so you are looking at spurious precission out their.

RonL