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

[karush]

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 =

 

[romsek]

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

 

[karush]

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

ok I can see that at 1.88 goes to .97 on table
or using P to z calculator but still what is $\Phi^{-1}$

 

[karush]

Overleaf PDF statistics with MHB replies
Im hoping to get this up to 100+ pages
Mahalo much for so much valuable input

 

[romsek]

$$\Phi^{-1}(p)$$ is the inverse of $$\Phi(z)$$

If you are given a probability $$p, \Phi^{-1}(p)$$ returns the associated z-score of $$p$$

 

[HOI]

Since [math]\Phi(1.88)= 0.97[/math], [math]\Phi^{-1}(0.97)= 1.88[/math]