91Ó°ÊÓ

$X$is a random variable in baseball that represents the distance travelled by fly balls hit to the outfield.

Mean $250$and standard deviation of$50$.

If a random variable $X$has a mean $\mathrm{μ}$and a standard deviation of s, we know that it is normally distributed., then it is denotes as $X~N(\mathrm{μ},s)$.

The random variable role="math" localid="1653569453344" $X$is denoted as:

Let $X$be a random variable in baseball that represents the distance travelled by fly balls hit to the outfield.

$\mathrm{μ}=250$,

$\mathrm{σ}=50$.

The probability that a randomly chosen fly ball travels less than $220$feet is calculated as follows:

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

$=P(z<-0.6)$

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

$P(z<-0.6)=0.5-P(-0.6≤z≤0)$

$=0.5$

$-P(0≤z≤0.6)$(due to symmetry)

$=0.5$$-0.2258$(from normal tables)

$=0.2743$

Given data:

Let $X$be a random variable in baseball that represents the distance travelled by fly balls hit to the outfield.

$\mathrm{μ}=250$,

$\mathrm{σ}=50$

The required 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:

$z_1=0.8416$

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

$k=250+0.8416*50$

$=292.08$

The following graph represents the corresponding region: