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

`P(X<225)=0.351`

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,(\sigma)/(√(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=(x-\mu)/(\sigma)`

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

`P(X>225)=P((X-\mu)/(\sigma)>(225-\mu)/(\sigma))`

`=P(Z>(225-204)/(55))=P(Z>0.382)`

And we can find this probability on this way:

`P(Z>0.382)=1-P(Z<0.382)=0.351`