91Ó°ÊÓ

Given in the question that, the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.

$X$is a random variable with a mean of $250$and a standard deviation of $50$that represents the distance of fly balls hit to the outfield in baseball. If a random variable $X$has a normal distribution with a mean of $\mathrm{μ}$and a standard deviation of s, it is denoted as $X~N(\mathrm{μ},s)$. As a result, the random variable $X$in this situation is denoted as:$X~N(250,50)$

Given that the distance $\left(X\right)$ of fly balls hit to the outfield is such that $X~N(250,50)$, we must calculate the probability that a randomly selected fly ball went less than 220 feet.

According to the information, we know that $X~N(250,50)$

Here,

$$\begin{gathered} \mathrm{μ}=250 \\ s=50 \end{gathered}$$

Let's take $z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

The chance that a randomly chosen fly ball will go less than 220 feet is calculated as follows:

$$\begin{gathered} P(x<220)=P\left(\frac{x-\mathrm{μ}}{\mathrm{σ}}<\frac{220-250}{50}\right) \\ =P(z<-0.6) \end{gathered}$$

The shaded zone in the graph below depicts the needed probability. As a result, it can be written:

$$\begin{gathered} P(z<-0.6)=0.5-P(-0.6≤z≤0) \\ =0.5 \end{gathered}$$

Due to symmetry:

$-P(0≤z≤0.6)=0.5$

From normal table: $-0.2258=0.2743$

Given that the distance $\left(X\right)$ of fly balls hit to the outfield is such that $X~N(250,50)$, the 80th percentile of the distribution of fly balls.

The needed probability is calculated as follows:

$P(x<k)=P\left(\frac{x-\mathrm{μ}}{\mathrm{σ}}<\frac{k-250}{50}\right)=P\left(z<z_1\right)=0.8$

We can deduce the following from the normal distribution tables:

$z_1=0.8416$

In the equation, $x=\mathrm{μ}+z\mathrm{σ}$replace $\mathrm{x}$with $\mathrm{k}$and ${\mathrm{z}}_1$, we get

$k=250+0.8416*50=292.08$

The following graph depicts the corresponding region: