Questions: Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Find the 87th percentile. The 87th 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 87th percentile.

The 87th 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 87th percentile. The 87th percentile is $\square$ (Round to two decimal places as needed.)

Solution

Solution Steps

Step 1: Determine the Z-Score for the 87th Percentile

To find the 87th 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 87th percentile. This is done using the inverse cumulative distribution function (CDF) for the standard normal distribution:

\[ z_{0.87} \approx 1.1264 \]

Step 2: Convert the Z-Score to the Original Scale

Next, we convert the z-score back to the original scale using the formula:

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

Substituting the values we have:

\[ x = 50 + 1.1264 \cdot 7 \]

Calculating this gives:

\[ x \approx 50 + 7.8848 \approx 57.8848 \]

Rounding to two decimal places, we find:

\[ x \approx 57.88 \]