Question 20, 7.2.33-T HW Score: $26.61 \%, 8.25$ of 31 points Save Assume the random variable $X$ is normally distributed, with mean $\mu=53$ and standard deviation $\sigma=9$. Find the 15 th percentile. The 15 th percentile is $\square$. (Round to two decimal places as needed.)

Step 1 ：We are given a normally distributed random variable $X$ with mean $\mu = 53$ and standard deviation $\sigma = 9$. We are asked to find the 15th percentile of this distribution.

Step 2 ：The percentile of a value in a distribution is the percentage of values in the distribution that are less than that value. Therefore, the 15th percentile is the value such that 15% of the values are less than it.

Step 3 ：We can find this value using the Z-score formula, which is $\frac{X - \mu}{\sigma}$, where $X$ is the value we're looking for, $\mu$ is the mean, and $\sigma$ is the standard deviation. However, in this case, we need to find $X$, so we rearrange the formula to $X = Z \cdot \sigma + \mu$.

Step 4 ：The Z-score for the 15th percentile can be found from a standard normal distribution table. The Z-score for the 15th percentile is approximately -1.036.

Step 5 ：Substituting the Z-score, the mean, and the standard deviation into the formula, we get $X = -1.036 \cdot 9 + 53$, which simplifies to $X = 43.67$.

Step 6 ：Thus, the 15th percentile of the given normal distribution is $\boxed{43.67}$.