Basic Z-Score & Standard Normal Question

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Homework Help Overview

The discussion revolves around calculating probabilities using the standard normal distribution, specifically focusing on Z-scores and their corresponding cumulative probabilities.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate probabilities using the normal cumulative distribution function (nmcdf) for given Z-scores. Questions arise regarding the correctness of the calculations and the answers provided in the textbook.

Discussion Status

Some participants are questioning the accuracy of the problems as stated, particularly regarding the interpretation of the Z-score ranges. There is a notable lack of consensus on the validity of the textbook answers, with some suggesting potential errors in the book.

Contextual Notes

Participants express uncertainty about the correctness of the problems and the answers provided, indicating a need for clarity on the setup of the Z-score scenarios.

kuahji
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If Z is a random variable having the standard normal distribution, find
P(Z[tex]\geq[/tex]-.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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