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 $53. Use Table 1 in Appendix B.

a. What is the probability that a hotel room costs $224 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 $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.

Answer 1

Solution :

Given that ,

mean = = 205

standard deviation = = 53

P(x 224) = 1 - P(x < 224)

= 1 - P((x - ) / (224 - 205) / 53)

= 1 -  P(z 0.36)  

= 1 - 0.6406   

= 0.3594

P(x 224) = 0.3594

Probability = 0.3594

b)

P(x < 139) = P((x - ) / < ( 139 - 205) / 53) = P(z < -1.25)

Using standard normal table,

P(x < 139) = 0.1056

Probability = 0.1056

(c)

P(201 < x < 299) = P((201 - 205 / 53) < (x - ) / < (299 - 205) / 53) )

P(201 < x < 299) =  P(-0.08 < z < 1.77)

P(201 < x < 299) = P(z < 1.77) - P(z < -0.08)

P(201 < x < 299) = 0.9616 - 0.4681 = 0.4935

Probability = 0.4935

(d)

P(Z > z) = 20%

1 - P(Z < z) = 0.20

P(Z < z) = 1 - 0.20 = 0.80

P(Z < 0.84) = 0.80

z = 0.84

Using z-score formula,

x = z * +

x = 0.84 * 53 + 205 = 249.52 = 250

Cosr = $250