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

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Conditions Value
Click For Summary
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.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{7.68}$
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:
Mathematics news on Phys.org
$$\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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
12K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K