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

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To determine the value of z* for various probabilities under a normal distribution with a mean of 12 and a standard deviation of 2, the cumulative distribution function (CDF) of the standard normal distribution, denoted as Φ(z), is utilized. The relationship between z* and the probabilities is established through the equations P(z < z*) = Φ((z* - μ)/σ) and P(z > z*) = 1 - Φ((z* - μ)/σ). For specific probabilities, the inverse function Φ⁻¹(p) is applied to find the corresponding z-scores. An example calculation shows that for P(z < z*) = 0.025, the resulting z* is approximately 10.651. The discussion emphasizes the importance of using statistical tables or software for accurate z-score determination.
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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
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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)

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