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?

Answer 1

Answer:

a) 0.8000

b) 0.1080

c) 0.3260

d) $269.1

Step-by-step explanation:

Mean = xbar = $239.67

standard deviation = $22.93

a) The probability that a randomly selected luxury​ hotel's daily rate will be less than $259 = P(x < 259)

We need to standardize the $259 in z-score.

The standardized z-score is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (259 - 239.67)/22.93 = 0.843

To determine the probability that a randomly selected luxury​ hotel's daily rate will be less than $259

P(x < 259) = P(z < 0.843)

We'll use data from the normal probability table for these probabilities

P(x < 259) = P(z < 0.843) = 1 - P(z ≥ 0.843) = 1 - P(z ≤ - 0.843) = 1 - 0.2 = 0.8000

b) The probability that a randomly selected luxury​ hotel's daily rate will be more than $268 = P(x > 268)

We need to standardize the $268 in z-score.

z = (x - xbar)/σ = (268 - 239.67)/22.93 = 1.235

To determine the probability that a randomly selected luxury​ hotel's daily rate will be more than $268

P(x > 268) = P(z > 1.235)

We'll use data from the normal probability table for these probabilities

P(x > 268) = P(z > 1.235) = 1 - P(z ≤ 1.235) = 1 - 0.892 = 0.1080

c) The probability that a randomly selected luxury​ hotel's daily rate will be between $236 and $256 = P(236 < x < 256)

We need to standardize the $236 and $256 in z-score.

z = (x - xbar)/σ = (256 - 239.67)/22.93 = 0.712

z = (x - xbar)/σ = (236 - 239.67)/22.93 = - 0.16

To determine the probability that a randomly selected luxury​ hotel's daily rate will be between $236 and $256

P(236 < x < 256) = P(-0.16 < z < 0.712)

We'll use data from the normal probability table for these probabilities

P(236 < x < 256) = P(-0.16 < z < 0.712) = P(z < 0.712) - P(z < -0.16) = [1 - P(z ≥ 0.712)] - [1 - P(z ≥ -0.16)] = [1 - P(z ≤ -0.712)] - [1 - P(z ≤ 0.16)] = (1 - 0.238) - (1 - 0.564) = 0.762 - 0.436 = 0.3260

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?

We need to obtain the z' value that corresponds to P(z ≤ z') = 0.90

From the normal distribution table,

z' = 1.282

z' = (x - xbar)/σ

1.282 = (x - 239.67)/22.93

x = (1.282 × 22.93) + 239.67 = $269.1