Questions: According to a recent survey, the average daily rate for a luxury hotel is 234.51. Assume the daily rate follows a normal probability distribution with a standard deviation of 21.93. 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 ? (Round to four decimal places as needed.)

According to a recent survey, the average daily rate for a luxury hotel is 234.51. Assume the daily rate follows a normal probability distribution with a standard deviation of 21.93. 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 ?
(Round to four decimal places as needed.)

Transcript text: According to a recent survey, the average daily rate for a luxury hotel is $\$ 234.51$. Assume the daily rate follows a normal probability distribution with a standard deviation of $\$ 21.93$. Complete parts a through d below. a. What is the probability that a randomly selected luxury hotel's daily rate will be less than $\$ 248$ ? $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Convert $X$ to a standard normal variable using the Z-score formula

Given $X = 248$, $\mu = 234.51$, and $\sigma = 21.93$, we calculate the Z-score as follows: \[Z = \frac{X - \mu}{\sigma} = \frac{248 - 234.51}{21.93} = 0.615\]

Step 2: Use the Z-score to find the probabilities

The probability of being less than $X$ is 0.731. The probability of being greater than $X$ is 0.269.