A physical fitness association is including the mile run in its secondary-school fitness test
for boys. The time for this event for boys in secondary school is normally distributed
with a mean of 450 seconds and a standard deviation of 40 seconds. If the association
wants to designate the fastest 10% as excellent, what time should the association set for
this criterion?
Z = (score-mean)/SD
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (.10) in the smaller portion and its Z score. Insert data into above equation and solve for score.
To determine the time criterion for designating the top 10% as excellent, we need to find the corresponding z-score.
1. Start by sketching the normal distribution curve representing the mile run times for boys in secondary school. The mean is 450 seconds, and the standard deviation is 40 seconds.
2. We know that the fastest 10% should be designated as excellent. This means we are looking for the z-score that corresponds to the cutoff for the top 10%.
3. Look up the z-score associated with the top 10% in a standard normal distribution table or by using a statistical calculator. The z-score for the top 10% is approximately 1.28.
4. Once you have the z-score, you can use the z-score formula to find the corresponding time value. The formula is:
z = (x - mean) / standard deviation
Rearrange the formula to solve for x:
x = (z * standard deviation) + mean
Plug in the values:
x = (1.28 * 40) + 450
x ≈ 51.2 + 450
x ≈ 501.2 seconds
Therefore, the association should set a time criterion of approximately 501.2 seconds for the mile run to designate the top 10% as excellent.