Basic Z-Score & Standard Normal Question

[kuahji]

If Z is a random variable having the standard normal distribution, find
P(Z\geq-.79)

So I punch it into my calculator nmcdf(-.79,infinity,0,1) & I get the answer .7852

The book however has the answer .2148 (essentially 1-ans).

Which is what I don't understand...

Have I forgotten something really simple, or do you think there is an error in the book. Most of the time its me :(.

 

[kuahji]

Also, P(-1.90>z>.44)

The only thing I could think of was nmcdf(-infinity,-1.90,0,1)+nmcdf(.44,infinity,0,1) which gives .3587 but the book gives the answer .6143.

 

[Hurkyl]

Are you sure you've written the problems correctly? P(-1.90>z>.44) is obviously zero, no matter how z is distributed.

 

[kuahji]

Are you sure you've written the problems correctly? P(-1.90>z>.44) is obviously zero, no matter how z is distributed.

100% sure. I think its an error in the book... I don't really see how it can be possible. If you flip the signs around it makes sense though.