Respuesta :
Answer:
[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]
And we can find this probability with this difference
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]
And we can find the difference with the normal standard distirbution or excel:
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]
Step-by-step explanation:
Let X the random variable that represent the hotel room cost of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(202,70)[/tex]
Where [tex]\mu=202[/tex] and [tex]\sigma=70[/tex]
We are interested on this probability
[tex]P(210<X<290)[/tex]
The z score formula is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using the formula we got:
[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]
And we can find this probability with this difference
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]
And we can find the difference with the normal standard distirbution or excel:
[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]
