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 $239.67. Assume the daily rate follows a normal probability distribution with a standard deviation of $22.93.

Complete parts a through d below. (round each answer to 4 decimal places)

a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $259?

b. What is the probability that a randomly selected luxury hotel's daily rate will be more than $268?

c. What is the probability that a randomly selected luxury hotel's daily rate will be between $236 and $256?

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?

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    259      
u = mean =    239.67      
          
s = standard deviation =    22.93      
          
Thus,          
          
z = (x - u) / s =    0.843000436      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.843000436   ) =    0.8003859 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    268      
u = mean =    239.67      
          
s = standard deviation =    22.93      
          
Thus,          
          
z = (x - u) / s =    1.235499346      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.235499346   ) =    0.10832236 [ANSWER]

***********************

c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    236      
x2 = upper bound =    256      
u = mean =    239.67      
          
s = standard deviation =    22.93      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.160052333      
z2 = upper z score = (x2 - u) / s =    0.712167466      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.436419925      
P(z < z2) =    0.76181946      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.325399535   [ANSWER]

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d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    239.67      
z = the critical z score =    1.281551566      
s = standard deviation =    22.93      
          
Then          
          
x = critical value =    $269.0559774   [ANSWER]  
  

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