Question

In: Statistics and Probability

New York City is the most expensive city in the United States for lodging. The mean...

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 $227 or more per night (to 4 decimals)?

b. What is the probability that a hotel room costs less than $143 per night (to 4 decimals)?

c. What is the probability that a hotel room costs between $201 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.
$ or - Select your answer -more less

Solutions

Expert Solution

Given that mean = 205 and standard deviation =55

(A) To find:- Probability that a hotel room costs 227 or more

P(X>227) = normalcdf(lower, upper, mean, standard deviation)

setting lower = 227, upper =9999, mean = 205 and standard deviation =55

=normalcdf(227,9999,205,55)

= 0.3446

(B) To find:- Probability that a hotel room costs less than 143

P(X<143) = normalcdf(lower, upper, mean, standard deviation)

setting lower = -9999, upper =143, mean = 205 and standard deviation =55

=normalcdf(-9999,143,205,55)

= 0.1298

(C) To find:- Probability that a hotel room costs between 201 and 299

P(201<X<299) = normalcdf(lower, upper, mean, standard deviation)

setting lower = 201, upper =299, mean = 205 and standard deviation =55

=normalcdf(201,299,205,55)

= 0.4853

(D) top 20% means percentile value corresponding to bottom 80%

using z and percentile table, we get z value = 0.842 for top 20% expensive hotel rooms

let x be the required cost

using the formula

x = mean +(z*standard deviation)

= 205 +(0.842*55)

= 205+46.31

= 252 (rounded to nest dollar)


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