Question

New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night.† Assume that room rates are normally distributed with a standard deviation of $55. (a) What is the probability that a hotel room costs $235 or more per night? (Round your answer to four decimal places.) (b) What is the probability that a hotel room costs less than $120 per night? (Round your answer to four decimal places.) (c) What is the probability that a hotel room costs between $210 and $300 per night? (Round your answer to four decimal places.) (d) What is the cost in dollars of the 10% most expensive hotel rooms in New York City? (Round your answer to the nearest cent.)

Answer 1

please see detailed answers below

Explanation:

we can work these out with z scores and use of a z-table.

formula is z = (X - υ) / σ, where X is test statistic, υ is the mean and σ is the standard deviation.

a) z = (X - υ) / σ

= (235 - 204) / 55 = 0.5636.

now go to a z-table. find +0.5 along left column. now find 0.06 on top row. look where these two meet on the table. number is 0.71226. this is area to the left of z = 0.5636. since we want to find probability of at least $235, we need area to the right.

*total area under a normal curve always = 1.

so, area to the right is 1 - 0.71226 = 0.2877 = p(at least $235).

b) z = (X - υ) / σ

= (120 - 204) / 55 = -1.527.

we find just like in part a). area for this z-score is 0.6301, to the left.

p(< $120) = 0.6301.

c) for $300:

z = (X - υ) / σ

= (300 - 204) / 55 = 1.745.

area to left is 0.95950.

for $210:

z = (X - υ) / σ

= (210 - 204) / 55 = 0.109.

area to left = 0.54380.

p($210 < Z < $300) = p($300) - p($210)

= 0.95950 - 0.54380

= 0.4157.

d) top 10% means we need z area of 0.9.

z-score for that is 1.285.

z = (X - υ) / σ

1.285 = (X - 204) / 55

X - 204 = 1.285(55) = 70.675

X = 70.675 + 204

= 274.675

so cost of 10% most expensive is $274.68 (to nearest cent).