91Ó°ÊÓ

Given in the question that, the distance of fly balls hit to the outfield is normally distributed with a mean of $250$feet and a standard deviation of $50$feet. We randomly sample $49$fly balls.

Consider $\stackrel{¯}{X}$ as the average distance in feet for $49$fly balls, then

localid="1649344986876" $\stackrel{¯}{X}~N\left(250,\frac{50}{\sqrt{49}}\right)$

According to the information,

$\stackrel{¯}{X}~N\left(250,\frac{50}{\sqrt{49}}\right)=N(250,7.143)$

Let's find the probability:

$P(\stackrel{¯}{X}<240)=P\left(\frac{\stackrel{¯}{X}−250}{7.143}<\frac{240−250}{7.143}\right)$

$=P\left(\frac{\stackrel{¯}{x}−250}{7.143}<−1.4\right)$

$=\mathrm{ϕ}(−1.4)$

$=1−0.9192$

$=0.0808$

The distance of fly balls hit to the outfield is normally distributed with a mean of $250$feet and a standard deviation of $50$feet. We randomly sample $49$fly balls.

Let's find the value of $z$from

$P(\mathrm{Î\pounds }x<z)=0.8$

$P\left(\frac{\mathrm{Î\pounds }x−240}{7.143}<\frac{z−240}{7.143}\right)=0.8$

If a variable $z$has a normal distribution, then

$P(Z<0.84)=0.8$

Now we have

$\frac{z−240}{7.143}=0.84$

$z-240=6$

$z=246$