According to a recent survey, the average daily rate for a luxury hotel is $234.

Question

According to a recent survey, the average daily rate for a luxury hotel is $234.28

Assume the daily rate follows a normal probability distribution with a standard deviation of $21.25. 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?

(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 $ 263?

(Round to four decimal places as needed.)

c. What is the probability that a randomly selected luxury hotel's daily rate will be between $231 and $ 251?

(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 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 234.28 and s is 21.25. z is given as (x-mean)/s

a) P(x<249)=P(z<(249-234.28)/21.25)=P(z<0.69), from normal distribution table we get 0.7549

b) P(x>263)=P(z>(263-234.28)/21.25)=P(z>1.35) or 1-P(z<1.35). from normal table we get 1-0.9115=0.0885

c) P(231<x<251)=P((231-234.28)/21.25<z<(251-234.28)/21.25)=P(-0.15<z<0.79) or P(z<0.79)-P(z<-0.15) or P(z<0.79)-(1-P(z<0.15)) =0.7852-(1-0.5596)=0.3448

d) for 90% , the z value is 1.28 thus value is mean+s*z=234.28+1.28*21.25=261.48

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