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 90thpercentile, 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
