Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55.a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)?c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)?d. What is the cost of the 20% most expensive hotel rooms in New York City? Round up to the next dollar.

Answer 1

a) We can find the z score for the value of 225 and we got:

`z = (225-204)/(55)=0.382`

And we can find this probability with the complement rule:

`P(z>0.382)=1-P(z<0.382)=1-0.6488=0.3512`

b)
`z = (140-204)/(55)=-1.164`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-1.164)=0.1223`

c)
`P(200<X<300)=P((200-\mu)/(\sigma)<(X-\mu)/(\sigma)<(300-\mu)/(\sigma))=P((200-204)/(55)<Z<(300-204)/(55))=P(-0.072<z<1.74)`

And we can find this probability with this difference:

`P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)=0.9595-0.4710=0,4885`

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

`P(X>a)=0.2` (a)

`P(X<a)=0.8` (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 0.8 of the area on the left and 0.2 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))=0.8`

`P(z<(a-\mu)/(\sigma))=0.8`

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`

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".

Part a

Let X the random variable that represent the costs 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`

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)`

We can find the z score for the value of 225 and we got:

`z = (225-204)/(55)=0.382`

And we can find this probability with the complement rule:

`P(z>0.382)=1-P(z<0.382)=1-0.6488=0.3512`

Part b

`P(X<140)`

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:

`z = (140-204)/(55)=-1.164`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-1.164)=0.1223`

Part c

`P(200<X<300)`

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(200<X<300)=P((200-\mu)/(\sigma)<(X-\mu)/(\sigma)<(300-\mu)/(\sigma))=P((200-204)/(55)<Z<(300-204)/(55))=P(-0.072<z<1.74)`

And we can find this probability with this difference:

`P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-0.072<z<1.74)=P(z<1.74)-P(z<-0.072)=0.9595-0.4710=0,4885`

Part d

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

`P(X>a)=0.2` (a)

`P(X<a)=0.8` (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 0.8 of the area on the left and 0.2 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))=0.8`

`P(z<(a-\mu)/(\sigma))=0.8`

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`