Suppose 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.

a. If X = distance in feet for a fly ball, then X ~ _____(_____,_____)

b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.

c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.

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 $\mu$and a standard deviation of s, it is denoted as $X~N(\mu ,s)$. 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,

$$\begin{gathered} \mu =250 \\ s=50 \end{gathered}$$

Let's take $z=\frac{x-\mu }{\sigma }$

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

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

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

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

Due to symmetry:

$-P(0\le z\le 0.6)=0.5$

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:

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

We can deduce the following from the normal distribution tables:

$z_1=0.8416$

In the equation, $x=\mu +z\sigma$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: