-7.68 Determine the value z* to satisfy the following conditions

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Discussion Overview

The discussion revolves around determining the value of \( z^* \) for a normal distribution with given parameters, specifically focusing on various probabilities associated with the standard normal distribution. Participants aim to solve multiple conditions related to cumulative probabilities.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant introduces the problem and specifies the parameters of the normal distribution, indicating a desire to tackle the conditions sequentially.
  • Another participant explains the use of the cumulative distribution function (CDF) of the standard normal distribution, denoted as \( \Phi(z) \), and how it relates to the probabilities given in the problem.
  • The same participant provides an example calculation for one of the conditions, showing how to find \( z^* \) using the inverse function \( \Phi^{-1}(p) \) and the relationship between \( z^* \), \( \mu \), and \( \sigma \).
  • A subsequent post attempts to calculate \( z^* \) for a different condition but indicates a potential error in the calculation.
  • Another participant suggests calculating \( z \) from the given probabilities without providing specific details.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations, as one participant acknowledges a mistake in their approach, and others have not yet provided complete solutions for all conditions.

Contextual Notes

Some calculations appear to depend on the correct application of the inverse CDF and the relationship between the parameters of the normal distribution. There are indications of unresolved mathematical steps and potential errors in calculations.

karush
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$\tiny{7.68}$
Normal curve $\quad\mu=12\quad\sigma = 2$
Let z denote a variable that a standard normal distribution.
Determintie the value $z*$ to satisfy the following conditions:
a. $P(z<z*)=.025$
b. $P(z<z*)=.01$
c. $P(z<z*)=.05$
d. $P(z<z*)=.02$
e. $P(z>z*)=.01$
f. $P(z>z*)\textbf{ or }(z<-z*)=.20$

ok I want to do these one at a time especially a,d and f

example on page 411 of text
SSCtw.png
 
Last edited:
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$$\text{let $\Phi(z)$ be the CDF of the standard normal. That function they make tables of. You'll need some form of that.}\\~\\

\text{Then $\Phi\left(\dfrac{z - \mu}{\sigma}\right)$ is the CDF of a normal distribution centered at $\mu$ with standard deviation $\sigma$}\\~\\

\text{So $P(z < z^*) = \Phi\left(\dfrac{z^*-\mu}{\sigma}\right)$}$$

Similarly

$$P(z > z^*) = 1-\Phi\left(\dfrac{z^*-\mu}{\sigma}\right)$$

Finally for (f) the probability of two non-intersecting intervals is the sum of the individual interval probabilities.

You'll actually have to use the inverse function $$\Phi^{-1}(p)$$ as they give you probabilities.

For example (a)

$$P(z < z^*) = 0.25\\
\text{we look up in the table $0.25$ (actually I just use software) and find that it corresponds to a "z-score", as it's commonly known, of $-0.67449$}\\
\dfrac{z^* - \mu}{\sigma} = -0.67449\\~\\
z^* = \mu - 0.67449\sigma = 10.651$$
 
Last edited:
mahalo
ill try the rest
romsek said:
$$\text{let $\Phi(z)$ be the CDF of the standard normal. That function they make tables of. You'll need some form of that.}\\
\text{Then $\Phi\left(\dfrac{z - \mu}{\sigma}\right)$ is the CDF of a normal distribution centered at $\mu$ with standard deviation $\sigma$}\\
\text{So $P(z < z^*) = \Phi\left(\dfrac{z^*-\mu}{\sigma}\right)$}$$
Similarly
$$P(z > z^*) = 1-\Phi\left(\dfrac{z^*-\mu}{\sigma}\right)$$
Finally for (f) the probability of two non-intersecting intervals is the sum of the individual interval probabilities.
You'll actually have to use the inverse function $$\Phi^{-1}(p)$$ as they give you probabilities.
For example (a)
$$P(z < z^*) = 0.25\\
\text{we look up in the table $0.25$ (actually I just use software) and find that it corresponds to a "z-score", as it's commonly known, of $-0.67449$}\\
\dfrac{z^* - \mu}{\sigma} = -0.67449\\
z^* = \mu - 0.67449\sigma = 10.651$$

mahalo
ill try the rest
 
Last edited:
Determintie the value z∗z∗ to satisfy the following conditions

[c] $P(z<z*)=.05$
$\quad\dfrac{z^* - \mu}{\sigma} =-5.975 \textbf{ so } z^*=(12)-5.975(2)=.05$

did something wrong...

z score calc
 
use calculate z from p
 

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