91Ó°ÊÓ

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. \(\bar{X}=\) Average distance in feet for \(49\) balls.

Here the sample size n is \(49\).

\(\bar{X}=N(\mu_{x},\frac{\sigma_{x}}{\sqrt{n}}\)

\(\mu_{x}=250\)

\(\sigma_{x}=50\)

\(n=49\)

\(\bar{X}=N(250,\frac{50}{\sqrt{49}})\)

Hence, the distribution of \(\bar{X}\) is \(\bar{X}=N(250,\frac{50}{\sqrt{49}})\)

\(\bar{X}=N(250,\frac{50}{\sqrt{49}})\)

The probability that \(49\) balls travelled an average of less than \(240\) is

Hence, the probability that \(49\) balls travelled an average of less than \(240\) is \(0.0808\).

\(\bar{X}=N(250,\frac{50}{\sqrt{49}})\)

The \(80^th\) percentile of the distribution of the average of \(49\) fly balls is

\(P(\bar{x}<k)=0.80\)

Finding k,

\(invnorm(0.80,250,7.143)=256.01\)

Hence, the \(80^th\)percentile of the distribution of the average of \(49\) fly balls is \(256.01\) feet.