Question

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. We randomly sample $49$ fly balls.

a. If $\overline{X}=$ average distance in feet for $49$ fly balls, then $\text{Missing open brace for subscript}$

b. What is the probability that the $49$ balls traveled an average of less than $240$ feet? Sketch the graph. Scale the horizontal axis for $\overline{X}$ . Shade the region corresponding to the probability. Find the probability.

c. Find the $80th$ percentile of the distribution of the average of $49$ fly balls.

Short answer

Part a. $\overline{X}=N(250,\frac{50}{\sqrt{49}})$

Part b. The probability that $49$ balls travelled an average of less than $240$ is $0.0808$.

Part c.The \(80^th\) percentile of the distribution of the average of $49$ fly balls is $256.01$ feet.

Step-by-step solution

Part a. Step 1. Given information

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. $\overline{X}=$ Average distance in feet for $49$ balls.

Part a. Step 2. Calculation

Here the sample size n is $49$.

$\overline{X}=N({\mu }_x,\frac{{\sigma }_x}{\sqrt{n}}$

${\mu }_x=250$

${\sigma }_x=50$

$n=49$

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

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

Part b. Step 1. Calculation

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

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 $0.0808$.

Part c. Step 1. Calculation

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

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

$P(\overline{x}<k)=0.80$

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.