10) the heights of nfl football players are approximately normally distributed with a mean of 71.5 inches…

10) the heights of nfl football players are approximately normally distributed with a mean of 71.5 inches and a standard deviation of 2.3 inches. brock purdy is 6 feet 1 inches (74 inches) tall. what percentile does that place him at? (a) 43rd (b) 50th (c) 74th (d) 86th (e) 94th

10) the heights of nfl football players are approximately normally distributed with a mean of 71.5 inches and a standard deviation of 2.3 inches. brock purdy is 6 feet 1 inches (74 inches) tall. what percentile does that place him at? (a) 43rd (b) 50th (c) 74th (d) 86th (e) 94th

Answer

Explanation:

Step1: Calculate z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 74$ (Brock Purdy's height), $\mu=71.5$ (mean height) and $\sigma = 2.3$ (standard deviation). So, $z=\frac{74 - 71.5}{2.3}=\frac{2.5}{2.3}\approx1.09$.

Step2: Find the percentile

We use the standard normal distribution table (z - table). Looking up the value of $z = 1.09$ in the z - table, the corresponding cumulative probability is approximately $0.8621$. This means Brock Purdy is at the approximately 86th percentile.

Answer:

D. 86th