Answer:
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.
Explanation:
We are given the following information in the question:
Mean, μ = 50
Standard Deviation, σ = 8
We are given that the distribution of number of rushing yards per game is a bell shaped distribution that is a normal distribution.
Formula:
![z_(score) = \displaystyle(x-\mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/high-school/pad6rntb722qswc0kw4hmbstruityvpgp4.png)
P(running back has 64 or fewer rushing yards)
![P( x \leq 64) = P( z \leq \displaystyle(64 - 50)/(8)) = P(z \leq 1.75)](https://img.qammunity.org/2021/formulas/mathematics/college/y1m9zt72k9jcvqy2evqj93tv7a12uplduw.png)
Calculation the value from standard normal z table, we have,
![P(x \leq 64) = 0.9599](https://img.qammunity.org/2021/formulas/mathematics/college/ng3jbjicah6ys7nw39kp3q79pmj50irkme.png)
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.