Questions: Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Find the 89th percentile. The 89th percentile is (Round to two decimal places as needed.)

Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Find the 89th percentile.

The 89th percentile is
(Round to two decimal places as needed.)

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

Solution

Solution Steps

Step 1: Determine the Z-Score for the 89th Percentile

To find the 89th percentile of a normally distributed random variable $ X $ with mean $ \mu = 50 $ and standard deviation $ \sigma = 7 $, we first calculate the z-score corresponding to the 89th percentile. The z-score is given by:

\[ z = 1.2265 \]

Step 2: Calculate the 89th Percentile Value

Using the z-score, we can find the actual value corresponding to the 89th percentile using the formula:

\[ x = \mu + z \cdot \sigma \]

Substituting the known values:

\[ x = 50 + 1.2265 \cdot 7 \]

Calculating this gives:

\[ x = 50 + 8.5855 = 58.5855 \]

Rounding to two decimal places, we find:

\[ x \approx 58.59 \]