Basic Z-Score & Standard Normal Question

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The discussion centers on calculating probabilities using the standard normal distribution, specifically P(Z ≥ -0.79) and P(-1.90 < Z < 0.44). The user employed the nmcdf function, yielding P(Z ≥ -0.79) = 0.7852, while the book states the answer as 0.2148, indicating a misunderstanding of the cumulative distribution function. For the second probability, the user calculated P(-1.90 < Z < 0.44) as 0.3587, while the book claims it to be 0.6143, leading to further confusion. The consensus is that the book likely contains errors in these problems.

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

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