NYC Probability

Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $205 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. Use Table 1 in Appendix B.

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 $299 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.

$(need answer) or - Select your answer -more or less

Answer 1

a. To find the probability that a hotel room costs $225 or more per night, we need to find the area to the right of $225 under the normal distribution curve.

First, we calculate the z-score:

z = (225 - 205) / 55 = 0.36

Using Table 1 in Appendix B, the area to the right of z = 0.36 is 0.3520. Therefore, the probability that a hotel room costs $225 or more per night is 0.3520 (to 4 decimals).

b. To find the probability that a hotel room costs less than $140 per night, we need to find the area to the left of $140 under the normal distribution curve.

First, we calculate the z-score:

z = (140 - 205) / 55 = -1.18

Using Table 1 in Appendix B, the area to the left of z = -1.18 is 0.1190. Therefore, the probability that a hotel room costs less than $140 per night is 0.1190 (to 4 decimals).

c. To find the probability that a hotel room costs between $200 and $299 per night, we need to find the area between $200 and $299 under the normal distribution curve.

First, we calculate the z-scores:

z1 = (200 - 205) / 55 = -0.09

z2 = (299 - 205) / 55 = 1.71

Using Table 1 in Appendix B, the area to the left of z = -0.09 is 0.4641, and the area to the left of z = 1.71 is 0.9564. Therefore, the area between z = -0.09 and z = 1.71 is:

0.9564 - 0.4641 = 0.4923

Therefore, the probability that a hotel room costs between $200 and $299 per night is 0.4923 (to 4 decimals).

d. To find the cost of the 20% most expensive hotel rooms in New York City, we need to find the z-score that corresponds to the 80th percentile. Using Table 1 in Appendix B, the z-score that corresponds to the 80th percentile is approximately 0.84.

Now, we can use the z-score formula to find the corresponding hotel room rate:

z = (x - 205) / 55

0.84 = (x - 205) / 55

x - 205 = 46.2

x = 251.2

Rounding up to the next dollar, the cost of the 20% most expensive hotel rooms in New York City is $252.