According to a recent survey, the average daily rate for a luxury hotel is $239.

Question

According to a recent survey, the average daily rate for a luxury hotel is $239.79. Assume the daily rate follows a normal probability distribution with a standard deviation of S21.97. Complete parts a through d below. a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $256? 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 $265? (Round to four decimal places as needed.) c. What is the probability that a randomly selected luxury hotel's daily rate will be between S238 and $258? L/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 75th percentile, which is the rate below which 75% 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

mean is 239.79 and s is 21.97

z is given as (x-mean)/s

a) P(x<256)=P(z<(256-239.79)/21.97)=P(z<0.74), from normal distribution table we get 0.7704

b) P(x>265)=P(z>(265-239.79)/21.97)=P(z>1.15) or 1-P(z<1.15), from normal table we get 1-0.8749=0.1251

c) P(238<x<258)=P((238-239.79)/21.97<z<(258-239.79)/21.97)=P(-0.08<z<0.83) or P(z<0.83)-(1-P(z<0.08))

from normal distribution table we get 0.7967-(1-0.5319)=0.3286

d) for 75th percentile, the z value is 0.68 from normal distribution table so answer is mean+z*s= 239.79+0.68*21.97=254.7296

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