Question

According to a recent? survey, the average daily rate for a luxury hotel is ?$234.02. Assume the daily rate follows a normal probability distribution with a standard deviation of ?$22.34. Complete parts a through d below.

a. What is the probability that a randomly selected luxury? hotel's daily rate will be less than $249?
b. What is the probability that a randomly selected luxury? hotel's daily rate will be more than $265?
c. What is the probability that a randomly selected luxury? hotel's daily rate will be between $228 and $248?
d. The managers of a local luxury hotel would like to set the? hotel's average daily rate at the 75th ?percentile, which is the rate below which 75?% of? hotels' rates are set. What rate should they choose for their? hotel?

Answer 1

Given,

= 234.02, = 22.34

We convert this to standard normal as

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

a)

P( X < 249) = P( Z < 249 - 234.0 / 22.34)

= P( Z < 0.6705)

= 0.7487

b)

P( X > 265) = P( Z > 265 - 234.02 / 22.34)

= P( Z > 1.3868)

= 1 - P( Z < 1.3868)

= 1 - 0.9172

= 0.0828

c)

P( 228 < X < 248) = P( X < 248) - P( X < 228)

= P( Z < 248 - 234.02 / 22.34) - P( Z < 228 - 234.02 / 22.34)

= P( Z < 0.6258) - P( Z < -0.2695)

= P( Z < 0.6258) - ( 1 - P( Z < 0.2695) )

= 0.7343 - ( 1 - 0.6062)

= 0.3405

d)

We have to calculate x such that

P( X < x) = 0.75

That is

P( Z < x - / ) = 0.75

From Z table, z-score for the probability of 0.75 is 0.6745

x - / = 0.6745

Put the values of and in above equation and solve for x.

x - 234.02 / 22.34 = 0.6745

x = 249.088

x = $249.09