Stats: finding probability in normal distribution

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SUMMARY

The discussion focuses on finding the Z-score (Z0) such that P(Z > Z0) = 0.1234 in the context of normal distribution. The mean (Z) is identified as 0, and the closest value from the standard normal distribution table is 0.1217, corresponding to Z0 = -0.31 for the range -∞ ≤ Z ≤ Z0. However, the correct interpretation involves calculating P(Z < Z0) = 1 - P(Z > Z0), leading to Z0 = 1.155 based on the standard normal distribution table provided.

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Homework Statement



Find Z0 such that P(z > z0) = 0.1234

Homework Equations



The Attempt at a Solution



Z is the mean which is 0. So if Z0 is less than the mean it should be a negative number. Looking at the table 0.1234 does not show up but the closest is 0.1217 which is 0.31.
So Z0 is -0.31?
 
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It depends on if you're saying from -\infty \leq z \leq z0 or 0 \leq z \leq z0. The value you found is for the latter (+0.31). It sounds like intuitively you are thinking of the former case, but used the table for the latter case.

Here if you needed another table-- http://www.math.unb.ca/~knight/utility/NormTble.htm
 
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Z is not the mean, it is the standard normal random variable. You want to find P(Z > z0). So you want to find P(Z<z0) such that it equals to 1-P(Z>z0). 1-0.1234=0.8766, which corresponds to about z0=1.155 from the table David posted
 

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