Respuesta :
Answer:
[tex]P(67.5<X<70.8)=P(\frac{67.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70.8-\mu}{\sigma})=P(\frac{67.5-69.4}{2.32}<Z<\frac{70.8-69.4}{2.32})=P(-0.819<z<0.603)[/tex]
[tex]P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)=0.7267-0.2064=0.5203[/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 heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(69.4,2.32)[/tex]
Where [tex]\mu=69.4[/tex] and [tex]\sigma=2.32[/tex]
We are interested on this probability
[tex]P(67.5<X<70.8)[/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(67.5<X<70.8)=P(\frac{67.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70.8-\mu}{\sigma})=P(\frac{67.5-69.4}{2.32}<Z<\frac{70.8-69.4}{2.32})=P(-0.819<z<0.603)[/tex]
And we can find this probability with this difference:
[tex]P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.819<z<0.603)=P(z<0.603)-P(z<-0.819)=0.7267-0.2064=0.5203[/tex]
