Question

New York City is known for it's tourist attractions and high priced real estate. The mean hotel room rate is $202 per night. Assume that the room rates are normally distributed with a standard deviation of $70.What is the probability that a hotel room costs between $210 and $290?

Answer 1

`P(210<X<290)=P((210-\mu)/(\sigma)<(X-\mu)/(\sigma)<(290-\mu)/(\sigma))=P((210-202)/(70)<Z<(290-202)/(70))=P(0.114<z<1.26)`

And we can find this probability with this difference

`P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)`

And we can find the difference with the normal standard distirbution or excel:

`P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351`

Explanation:

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

`X \sim N(202,70)`

Where
`\mu=202` and
`\sigma=70`

We are interested on this probability

`P(210<X<290)`

The z score formula is given by:

`z=(x-\mu)/(\sigma)`

Using the formula we got:

`P(210<X<290)=P((210-\mu)/(\sigma)<(X-\mu)/(\sigma)<(290-\mu)/(\sigma))=P((210-202)/(70)<Z<(290-202)/(70))=P(0.114<z<1.26)`

And we can find this probability with this difference

`P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)`

And we can find the difference with the normal standard distirbution or excel:

`P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351`