Respuesta :
Answer:
[tex]P(X<225)=0.351[/tex]
Step-by-step explanation:
Assuming a mean of $204 per night and a deviation of $55.
a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?
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".
Let X the random variable that represent the cost per night at the hotel, and for this case we know the distribution for X is given by:
[tex]X \sim N(204,55)[/tex]
Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
We are interested on this probability
[tex]P(X>225)[/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(X>225)=P(\frac{X-\mu}{\sigma}>\frac{225-\mu}{\sigma})[/tex]
[tex]=P(Z>\frac{225-204}{55})=P(Z>0.382)[/tex]
And we can find this probability on this way:
[tex]P(Z>0.382)=1-P(Z<0.382)=0.351[/tex]
