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

Answer:

a) 0.3520 = 35.20% probability that a hotel room costs $225 or more per night.

b) 0.1038 = 10.38% probability that a hotel room costs less than $140 per night.

c) 0.4878 = 48.78% probability that a hotel room costs between $200 and $300 per night.

d) $251.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 204, \sigma = 55$

a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?

This probability is 1 subtracted by the pvalue of Z when X = 225. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{225 - 204}{55}$

$Z = 0.38$

$Z = 0.38$ has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

0.3520 = 35.20% probability that a hotel room costs $225 or more per night.

b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)?

This probability is the pvalue of Z when X = 140. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{140 - 204}{55}$

$Z = -1.16$

$Z = -1.16$ has a pvalue of 0.1038.

0.1038 = 10.38% probability that a hotel room costs less than $140 per night.

c. What is the probability that a hotel room costs between $200 and $300 per night (to 4 decimals)?

This probability is the pvalue of Z when X = 300 subtracted by the pvalue of Z when X = 200. So

X = 300

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{300 - 204}{55}$

$Z = 1.75$

$Z = 1.75$ has a pvalue of 0.9599.

X = 200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{200 - 204}{55}$

$Z = -0.07$

$Z = -0.07$ has a pvalue of 0.4721.

0.9599 - 0.4721 = 0.4878

0.4878 = 48.78% probability that a hotel room costs between $200 and $300 per night.

d. What is the cost of the 20% most expensive hotel rooms in New York City?

This is X when Z has a pvalue of 1-0.2 = 0.8. So X when Z = 0.84

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - 204}{55}$

$X - 204 = 0.84*55$

$X = 250.2$

Rounding to the next dollar.

$251.