Question

suppose the mean 40 yard dash time of a perfessional football player is 4.41 seconds with a standard deviation of 0.15 seconds. assume normal distribution.what is the percent of who run 40-yard dash in between 4.11and 4.41 seconds

Answer 1

The percent of runners between these two times is 47.72.

We find the z-scores associated with each end of this interval using the formula
z=(X-μ)/σ

For the lower end,
z=(4.11-4.41)/0.15 = -0.3/0.15 = -2

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.0228.

For the upper end:
z=(4.41-4.41)/0.15 = 0/0.15 = 0

Using a z-table we see that the area to the left of, less than, this score is 0.5000.

We want the area between these times, so we subtract:
0.5000-0.0228 = 0.4772

This corresponds with 47.72%.