Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is per night (USA Today, April , ). Assume that room rates are normally distributed with a standard deviation of . a. What is the probability that a hotel room costs or more per night (to 4 decimals)?

Answer 1

Answer:

$P(X$

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:

$X \sim N(204,55)$  

Where $\mu=204$ and $\sigma=55$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

We are interested on this probability

$P(X>225)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>225)=P(\frac{X-\mu}{\sigma}>\frac{225-\mu}{\sigma})$

$=P(Z>\frac{225-204}{55})=P(Z>0.382)$

And we can find this probability on this way:

$P(Z>0.382)=1-P(Z$