Question

Problem 2 Suppose that the distance of fty balls hit to the outfield (in baseball) is normally distributed with a mean of 225 feet and a standard deviation of 54 feet. We randomly sample 47 fly balls Let X- average distance in feet for 47 fly balls Enter numbers as integers or fractions in p/q form, or as decimals accurate to nearest 0.01 a. (.20) X ~(pick one) b. (20) Use the mean and standard deviation ofX to determine the z value for X 240 c (20) lilustrate PX > 240) as an area by adjusting the slider along the horizontal axis to control the z value. 0.00 0.300 0.2 0.1

Answer 1

Solution:

a. $\boldsymbol{\bar{X}\sim Normal (225,7.88)}$

Explanation:

$\mu_{\bar{X}}=\mu=225$

$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{54}{\sqrt{47}}=7.88$

b. Use the mean and standard deviation of $\bar{X}$ to determine the z value for $\bar{X}=240$

Answer:

$z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$

  $=\frac{240-225}{7.88}$

  $\boldsymbol{=1.90}$

$\boldsymbol{\therefore z=1.90}$

d. What is the probability that the 47 balls traveled an average of greater than 240 feet?

Answer: We have to find:

$P(z>1.90)$

Using the standard normal table we have:

$P(z>1.90)=0.03$

Therefore, the probability that the 47 balls traveled an average of greater than 240 feet is 0.03

e. Find the 80th percentile of the distribution of the average of 47 fly balls. That is, find x so that $P(\bar{X}<x)=0.8$

Answer: The z value corresponding to 80th percentile is:

$z(0.8)=0.84$

Therefore using the z score formula we have:

$z=\frac{x-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$

$0.84=\frac{x-225}{7.88}$

$0.84 \times 7.88=x-225$

$6.6192=x-225$

$x=225&plus;6.6192$

$\boldsymbol{x=231.62}$