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

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Discussion Overview

The discussion revolves around determining specific values of \( z^* \) that separate certain percentages of z values in the context of the standard normal distribution. Participants explore the use of the cumulative distribution function (CDF) and its inverse to find these values.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Post 1 presents a series of values for \( z^* \) that separate the largest and smallest percentages of z values, providing specific numerical answers.
  • Post 2 explains that \( \Phi(z) \) is the CDF of the standard normal distribution and indicates the need to compute the inverse function to solve the problem.
  • Post 3 reiterates the same points as Post 2, emphasizing the relationship between \( \Phi(z^*) \) and the specified probabilities.
  • Post 5 clarifies that \( \Phi^{-1}(p) \) is the inverse function that returns the z-score associated with a given probability \( p \).
  • Post 6 confirms the relationship between \( \Phi(1.88) \) and \( \Phi^{-1}(0.97) \), illustrating the use of the inverse function.

Areas of Agreement / Disagreement

Participants generally agree on the use of the CDF and its inverse to determine the values of \( z^* \), but there is some uncertainty regarding the notation and specific calculations involved.

Contextual Notes

Some participants express confusion about the notation \( \Phi^{-1} \) and its application, indicating a potential gap in understanding how to compute these values.

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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