Question

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.

Answer 1

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)