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 $203 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $56. 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 $139 per night (to 4 decimals)?

c. What is the probability that a hotel room costs between $199 and $300 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.

Solutions

Expert Solution

Given,

= 203 , = 56

We convert this to standard normal as

P(X < x) = P(Z < ( x - ) / )

a)

P(X >= 227) = P(Z > ( 227 - 203) / 56)

= P(Z > 0.43)

= 0.3336

b)

P(X < 139) = P(Z < ( 139 - 203) / 56)

= P(Z < -1.14)

= 0.1271

c)

P(199 < X < 300) = P(X < 300) - P(X < 199)

= P(Z < ( 300 - 203) / 56) - P(Z < ( 199 - 203) / 56)

= P(Z < 1.73) - P(Z < -0.07)

= 0.9582 - 0.4721

= 0.4861

d)

We have to calculate x such that P(X > x) = 0.20

That is P(X < x) = 0.80

P(Z < ( x - ) / ) = 0.80

From Z table z-score for the probability of 0.80 is 0.8416

(x - ) / ) = 0.8416

( x - 203 ) / 56 = 0.8416

Solve for x

x = 250.13

= 251 (Rounded up to next dollar)


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