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

[karush]

$\tiny{7.68}$
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

 

[romsek]

\(\displaystyle \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

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

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 \(\displaystyle \Phi^{-1}(p)\) as they give you probabilities.

For example (a)

\(\displaystyle P(z < z^*) = 0.25\\
\text{we look up in the table $0.25$ (actually I just use software) and find that it corresponds to a "z-score", as it's commonly known, of $-0.67449$}\\
\dfrac{z^* - \mu}{\sigma} = -0.67449\\~\\
z^* = \mu - 0.67449\sigma = 10.651\)

 

[karush]

mahalo
ill try the rest

\(\displaystyle \text{let $\Phi(z)$ be the CDF of the standard normal. That function they make tables of. You'll need some form of that.}\\
\text{Then $\Phi\left(\dfrac{z - \mu}{\sigma}\right)$ is the CDF of a normal distribution centered at $\mu$ with standard deviation $\sigma$}\\
\text{So $P(z < z^*) = \Phi\left(\dfrac{z^*-\mu}{\sigma}\right)$}\)
Similarly
\(\displaystyle P(z > z^*) = 1-\Phi\left(\dfrac{z^*-\mu}{\sigma}\right)\)
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 \(\displaystyle \Phi^{-1}(p)\) as they give you probabilities.
For example (a)
\(\displaystyle P(z < z^*) = 0.25\\
\text{we look up in the table $0.25$ (actually I just use software) and find that it corresponds to a "z-score", as it's commonly known, of $-0.67449$}\\
\dfrac{z^* - \mu}{\sigma} = -0.67449\\
z^* = \mu - 0.67449\sigma = 10.651\)

mahalo
ill try the rest

 

[karush]

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

 

[romsek]

use calculate z from p