Respuesta :
Answer:
[tex]P(50000<X<60000)=P(\frac{50000-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{60000-\mu}{\sigma})=P(\frac{50000-54000}{12000}<Z<\frac{60000-54000}{12000})=P(-0.33<z<0.5)[/tex]
And we can find this probability with this difference:
[tex]P(-0.33<z<0.5)=P(z<0.5)-P(z<-0.33)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.33<z<0.5)=P(z<0.5)-P(z<-0.33)=0.691-0.371=0.320[/tex]
Explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(54000,12000)[/tex]
Where [tex]\mu=54000[/tex] and [tex]\sigma=12000[/tex]
We are interested on this probability
[tex]P(50000<X<60000)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(50000<X<60000)=P(\frac{50000-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{60000-\mu}{\sigma})=P(\frac{50000-54000}{12000}<Z<\frac{60000-54000}{12000})=P(-0.33<z<0.5)[/tex]
And we can find this probability with this difference:
[tex]P(-0.33<z<0.5)=P(z<0.5)-P(z<-0.33)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.33<z<0.5)=P(z<0.5)-P(z<-0.33)=0.691-0.371=0.320[/tex]
