Question

The answers are in red. Please do explain all parts

According to a recent survey, the average daily rate for a luxury hotel is $234.19. Assume the daily rate follows a normal probability distribution with a standard deviation of $22.42 Complete parts a through d below a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $248? 0.7310 (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 $262? 0.1074 (Round to four decimal places as needed.) c. What is the probability that a randomly selected luxury hotel's daily rate will be between $233 and $253? 0.3204 (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 $ 262.92. (Round to the nearest cent as needed.)

Answer 1

Let X follows N(/u= 234.19,0 = 22.42) (a) X -и 248- 234.19 P(X<248 P 22.42 (excelfunction( NORMSDIST (2) = P(z < 0.62) 0.7324 (b) P(X 262) 1-P(X < 262) 262 234.19 --PX- =1- P 22.42 -1-P(z 1.24) excelfunction NORMSDIST (2) = 1-0.8925 0.1075 (c) 233- 234.19 X-и 253-234.19 P(233 < X < 253) = P 22.42 22.42 Excel function, - P(-0.05 <z < 0,84) (= NORMSDIST(2) 0.7995 0.4801 0.3194

d) Using excel - 234.19 +1.282(22.42) Zo,90NORMSINV(0.90) 1.282 = 262.9224 262.92