Respuesta :
Answer:
[tex]P(8<X<12)=P(\frac{8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{12-\mu}{\sigma})=P(\frac{8-10}{3.6}<Z<\frac{12-10}{3.6})=P(-0.56<z<0.56)[/tex]
And we can find this probability with this difference:
[tex]P(-0.56<z<0.56)=P(z<0.56)-P(z<-0.56)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.56<z<0.56)=P(z<0.56)-P(z<-0.56)=0.7123-0.2877=0.4246[/tex]
Step-by-step 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 waiting time of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(10,3.6)[/tex]
Where [tex]\mu=10[/tex] and [tex]\sigma=3.6[/tex]
We are interested on this probability
[tex]P(8<X<12)[/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(8<X<12)=P(\frac{8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{12-\mu}{\sigma})=P(\frac{8-10}{3.6}<Z<\frac{12-10}{3.6})=P(-0.56<z<0.56)[/tex]
And we can find this probability with this difference:
[tex]P(-0.56<z<0.56)=P(z<0.56)-P(z<-0.56)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.56<z<0.56)=P(z<0.56)-P(z<-0.56)=0.7123-0.2877=0.4246[/tex]
