According to a recent survey, the average daily rate for a luxury hotel is exist

Question

According to a recent survey, the average daily rate for a luxury hotel is exist234.31. Assume the daily rate follows a normal probability distribution with a standard deviation of exist22.56. Complete parts a through d below. a. What is the probability that a randomly selected luxury hotel's daily rate will be less than exist256? 8315 (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 exist264? 0934 (Round to four decimal places as needed.) c. What is the probability that a randomly selected luxury hotel's daily rate will be between exist230 and exist250? 3333 (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 90th percentile, which is the rate below which 90% of hotel's rates are set. What rate should they choose for their hotel? The managers should choose a daily rate of exist (Round to the nearest cent as needed.)

Explanation / Answer

From z table,

P(z < 1.28) = 0.90

Hence,

Daily rate for 90th percentile

= 234.31 + 1.28(22.56)

= 263.19

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