Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night. Assume that room rates are normally distributed with a standard deviation of $55. What is the minimum cost that captures the 20% most expensive hotel rooms in New York City?

Answer 1

`z=0.842<(a-204)/(55)`

And if we solve for a we got

`a=204 +0.842*55=250.31`

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.

Explanation:

Previous concepts

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".

Solution to the problem

Let X the random variable that represent the room hotel rate of a population, and for this case we know the distribution for X is given by:

`X \sim N(204,55)`

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

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.20` (a)

`P(X<a)=0.80` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.842`

`P(z<(a-\mu)/(\sigma))=0.842`

But we know which value of z satisfy the previous equation so then we can do this:

`z=0.842<(a-204)/(55)`

And if we solve for a we got

`a=204 +0.842*55=250.31`

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.