According to a recent survey, the average daily rate for a luxury hotel is $236.

Question

According to a recent survey, the average daily rate for a luxury hotel is $236.68. Assume the daily rate follows a normal probability distribution with a standard deviation of $21.92. Complete parts a through d below. a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $251? (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 $264? (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 80th percentile, which is the rate below which 80% 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

given that mean = 236.68

sd = 21.92 , please keep the z tables handy for this

a)

P(X<251) , so we use the z table as

Z = (X-mean)/sd

(251- 236.68 )/21.92 = 0.6532

P(Z<0.6532) , from the z tables

We conclude that:

P ( Z<0.6532 )=0.7422

b)

P(X>264) , so we use the z table as

Z = (X-mean)/sd

(264- 236.68 )/21.92 = 1.246

P ( Z>1.246 )=1P ( Z<1.246 )=10.8944=0.1056

c)

P(X>245) , so we use the z table as

Z = (X-mean)/sd

(245- 236.68 )/21.92 = 0.3795

P(X<265) , so we use the z table as

Z = (X-mean)/sd

(265- 236.68 )/21.92 = 1.29

To find the probability of P (0.3795<Z<1.29), we use the following formula:

P (0.3795<Z<1.29 )=P ( Z<1.29 )P (Z<0.3795 )

We see that P ( Z<1.29 )=0.9015

We see that  P ( Z<0.3795 )=0.648.

At the end we have:

P (0.3795<Z<1.29 )=0.2535

d)

here we are given that p= 0.80 so we use

Z = (X - Mean)/sd

so we calculate z score using the probability as 0.80

0.8416*21.92 + 236.68 = X

X = 255.12

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