-7.6.69 Determine the value z^* that...

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SUMMARY

The discussion focuses on determining the critical values \( z^* \) that separate specific percentiles of the standard normal distribution using the cumulative distribution function (CDF) \( \Phi(z) \). The values identified are \( z^* = 1.88 \) for the largest 3%, \( z^* = 2.33 \) for the largest 1%, \( z^* = 1.75 \) for the smallest 4%, and \( z^* = 1.28 \) for the smallest 10%. The inverse function \( \Phi^{-1}(p) \) is essential for calculating these values, as it returns the z-score corresponding to a given probability \( p \).

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karush
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Determine the value $z^*$ that...
a. Separates the largest $3\%$ of all z values from the others
$=1.88$
b. Separates the largest $1\%$ of all z values from the others
$=2.33$
c. Separates the smallest $4\%$ of all z values from the others
$=1.75$
d. Separates the smallest $10\%$ of all z values from the others
$=1.28$

OK just can't seem to find an example of how these are stepped thru
the book answer follows the =
 
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once again $$\Phi(z)$$ is the CDF of the standard normal distribution
you'll need some way to compute the inverse of this function to complete this problem.

a)$$ \Phi(z^*) = 0.97\\ z^* = \Phi^{-1}(0.97)= 1.88079$$

b) $$\Phi(z^*) = 0.99$$

c) $$\Phi(z^*) = 0.04$$

d)$$ \Phi(z^*) = 0.1$$
 
romsek said:
once again $$\Phi(z)$$ is the CDF of the standard normal distribution
you'll need some way to compute the inverse of this function to complete this problem.

a)$$ \Phi(z^*) = 0.97\\ z^* = \Phi^{-1}(0.97)= 1.88079$$

b) $$\Phi(z^*) = 0.99$$

c) $$\Phi(z^*) = 0.04$$

d)$$ \Phi(z^*) = 0.1$$

mahalo I was unaware of the use of that symbol

Screenshot 2021-09-02 11.25.39 AM.png


ok I can see that at 1.88 goes to .97 on table
or using P to z calculator but still what is $\Phi^{-1}$
 
Last edited:
$$\Phi^{-1}(p)$$ is the inverse of $$\Phi(z)$$

If you are given a probability $$p, \Phi^{-1}(p)$$ returns the associated z-score of $$p$$
 
Since [math]\Phi(1.88)= 0.97[/math], [math]\Phi^{-1}(0.97)= 1.88[/math]
 

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