The average middle-distance runner at a local high school runs the mile in 4.5 minutes, with a standard deviation of 0.3 minute. Find the percentage of a runners that will run the mile in less than 4 minutes?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To find the percentage of runners who will run the mile in less than 4 minutes, we will first need to convert the given information into a standardized score using the z-score formula:
z = (x - μ) / σ,
where "x" is the value we want to convert, "μ" is the mean, and "σ" is the standard deviation.
In this case, we want to find the percentage of runners who run the mile in less than 4 minutes, so our "x" value will be 4 minutes.
Calculating the z-score:
z = (4 - 4.5) / 0.3
z = -0.5 / 0.3
z ≈ -1.67
Now, we need to find the corresponding area under the normal distribution curve for a z-score of -1.67. We can use a standard normal distribution table (also known as the Z-table) or a statistical calculator to find this area.
Using a Z-table, we search for the row corresponding to -1.6 and the column corresponding to 0.07 (0.67 - 0.60). The intersection gives us the value 0.4522.
However, since we have a negative z-score, we need to find the area of the left tail of the distribution, which is 1 minus the value we found.
So, the area to the left of the z-score -1.67 is approximately 1 - 0.4522 = 0.5478.
To find the percentage of runners, we multiply the result by 100:
Percentage = 0.5478 * 100 ≈ 54.78%.
Therefore, approximately 54.78% of the runners are expected to run the mile in less than 4 minutes.