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

Answer:

$P(210$

And we can find this probability with this difference

$P(0.114$

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

$P(0.114$

Step-by-step 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

$P(210$

The z score formula is given by:

$z=\frac{x-\mu}{\sigma}$

Using the formula we got:

$P(210$

And we can find this probability with this difference

$P(0.114$

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

$P(0.114$