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 $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

Answer 1

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)