Assume the random variable $X$ is normally distributed with mean $\mu=50$ and standard deviation $\sigma=7$. Find the 89th percentile. The 89th percentile is (Round to two decimal places as needed.)

Step 1 ：Given that the random variable $X$ is normally distributed with a mean ($\mu$) of 50 and a standard deviation ($\sigma$) of 7, we are to find the 89th percentile.

Step 2 ：The percentile of a distribution is the value below which a certain percent of the observations fall. For example, the 89th percentile is the value below which 89% of the observations may be found.

Step 3 ：In a normal distribution, percentiles can be found using the z-score formula, which is: \(Z = \frac{X - \mu}{\sigma}\).

Step 4 ：In this case, we are given the percentile (89th), the mean (50), and the standard deviation (7), and we need to find the corresponding value from the data (X). So, we need to rearrange the z-score formula to solve for X: \(X = Z\sigma + \mu\).

Step 5 ：We can find the z-score corresponding to the 89th percentile using a z-table or a statistical calculator. The z-score for the 89th percentile is approximately 1.23.

Step 6 ：Substitute the z-score, the given mean, and the standard deviation into the formula to find the corresponding value from the data: \(X = 1.23*7 + 50\).

Step 7 ：Calculate the value to get approximately 58.59.

Step 8 ：Final Answer: The 89th percentile of a normally distributed random variable with mean 50 and standard deviation 7 is approximately \(\boxed{58.59}\).