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

Answer:

$z=0.842<\frac{a-204}{55}$

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.  

Step-by-step 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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842$  

$P(z<\frac{a-\mu}{\sigma})=0.842$

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

$z=0.842<\frac{a-204}{55}$

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.