Respuesta :
Answer:
[tex]P(z<\frac{a-\mu}{\frac{\sigma}{\sqrt{n}}})=0.1[/tex]
And replacing the value obtained we got:
[tex]z=-1.282<\frac{a-410000}{\frac{\sigma}{\sqrt{200}}}[/tex]
And if we solve for a we got
[tex]a=410000 -1.282*\frac{65000}{\sqrt{200}}=404107.68[/tex]
Step-by-step explanation:
Let X the random variable that represent the average home prices of a population, and for this case we know the distribution for X is given by:
Where [tex]\mu=410000[/tex] and [tex]\sigma=65000[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.90[/tex] (a)
[tex]P(X<a)=0.10[/tex] (b)
We want to find a value who accumulate 0.10 of the area on the left and 0.90 of the area on the right of the normal standard distributon and for this case it's z=-1.282
And using the distribution for the sample mean we can do this:
[tex]P(X<a)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{a-\mu}{\frac{\sigma}{\sqrt{n}}})=0.10[/tex]
[tex]P(z<\frac{a-\mu}{\frac{\sigma}{\sqrt{n}}})=0.1[/tex]
And replacing the value obtained we got:
[tex]z=-1.282<\frac{a-410000}{\frac{\sigma}{\sqrt{200}}}[/tex]
And if we solve for a we got
[tex]a=410000 -1.282*\frac{65000}{\sqrt{200}}=404107.68[/tex]
