# Solved (Free): 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

#### ByDr. Raju Chaudhari

Apr 12, 2021

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.
The probability that a fly ball travels less than 220 feet is:

a. 0.3000
b. 0.2743
c. 0.2358
d. None of these

Sketch the graph:

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

a. 80% of the fly balls will travel more than 292 feet.
b. 80% of the fly balls will travel less than 292 feet.
c. 80% of the fly balls will travel 292 feet.
d. 80% of the fly balls are not fly balls at all, they are bee balls.

#### Solution

a) Let $X$ denote the distance in feet for a fly ball. Given that $\mu = 250$ and $\sigma = 50$.

$X\sim N(250, 50^2)$.

b) The probability that this ball traveled fewer than 220 feet is

 \begin{aligned} P(X< 220) &=P\bigg(\frac{X-\mu}{\sigma} < \frac{220-250}{50}\bigg)\\ &=P\bigg(Z< -0.6\bigg)\\ &= P(Z< -0.6)\\ &=0.2743 \end{aligned}

Let the 80th percentile of the distribution of fly balls be $a$.

 \begin{aligned} & P(X< a) =0.8\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}< \frac{a-250}{50}\big)=0.8\\ &\Rightarrow P(Z< \frac{a-250}{50}\big)=0.8\\ &\Rightarrow \frac{a-250}{50}= 0.84\\ & \qquad \text{(From normal statistical table)}\\ &\Rightarrow a = 250 + 0.84* 50\\ &\Rightarrow a = 292 \end{aligned}

Correct answer is (b), i.e., 80% of the fly balls will travel less than 292 feet.