Question

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 267 feet and a standard deviation of 39 feet. Let X be the distance in feet for a fly ball.c. Find the 70th percentile for the distribution of distance of fly balls. Round to 2 decimal places.

Answer 1

To find the 70th percentile we look for the z value that gives the following probability:

$P(Z\leq z)=0.70$

Looking at the standard normal distribution table we notice that this probability happens when:

$z=0.5244$

Then we need the z score to be 0.5244 .Now that we know this we use its definition to find the value of x that gives this z score:

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ \frac{x-267}{39}=0.5244 \\ x-267=20.4516 \\ x=267+20.4516 \\ x=287.45 \end{gathered}$

Therefore, the 70th percentile is 287.45