According to a recent? survey, the average daily rate for a luxury hotel is ?$23

Question

According to a recent? survey, the average daily rate for a luxury hotel is ?$235.44. Assume the daily rate follows a normal probability distribution with a standard deviation of ?$22.11.

Complete parts a through d below.

a. What is the probability that a randomly selected luxury? hotel's daily rate will be less than $ 255? ?(Round to four decimal places as? needed.)

b. What is the probability that a randomly selected luxury? hotel's daily rate will be more than $ 266 ??(Round to four decimal places as? needed.)

c. What is the probability that a randomly selected luxury? hotel's daily rate will be between $ 245 and $ 265??(Round to four decimal places as? needed.)

d. The managers of a local luxury hotel would like to set the? hotel's average daily rate at the 85th ?percentile, which is the rate below which 85?% of? hotels' rates are set. What rate should they choose for their? hotel? The managers should choose a daily rate of ?$___.(Round to the nearest cent as? needed.)

Explanation / Answer

Solution : Given that mean ? = 235.44 , standard deviation ? = 22.11

a. P(x < 255) = P((x - ?)/? < (255 - 235.44)/22.11)

= P(Z < 0.8847)

= 0.8106

b. P(x > 266) = P((x - ?)/? > (266 - 235.44)/22.11)

= P(Z > 1.3822)

= 0.0838

c. P(245 < x < 265) = P((245 - 235.44)/22.11 < Z < (265 - 235.44)/22.11)

= P(0.4324 < Z < 1.3370)

= 0.2435

d. 85th percentile of Z-score = 1.036

=> 85th percentile of x = ? + Z*?

= 235.44 + 1.036*22.11

= 258.35 (nearest cent)

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