Question

New York City is one of the most expensive cities in the US for lodging. The mean hotel room rate is $244.00 per night; assume that room rates are normally distributed, with the standard deviation of $55.00 What is the probability that a hotel room costs between $250.00 and $285.00?

Answer 1

Answer:

22.96% probability that a hotel room costs between $250.00 and $285.00

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 244, \sigma = 55$

What is the probability that a hotel room costs between $250.00 and $285.00?

This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So

X = 285

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

$Z = \frac{285 - 244}{55}$

$Z = 0.75$

$Z = 0.75$ has a pvalue of 0.7734

X = 250

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

$Z = \frac{250 - 244}{55}$

$Z = 0.11$

$Z = 0.11$ has a pvalue of 0.5438

0.7734 - 0.5438 = 0.2296

22.96% probability that a hotel room costs between $250.00 and $285.00