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)?

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:

Where and

And let represent the sample mean, the distribution for the sample mean is given by:

We are interested on this probability

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

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

And we can find this probability on this way:

the graph on the bottom right

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09