The times for completing one circuit of a bicycle course are normally distributed with a mean of 81.7 minutes and a standard deviation of 8.6 minutes. An association wants to sponsor a race and provide prizes for the top (that is, fastest) 15% of riders. Where should they expect to set the cutoff time to earn a prize?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.15) that correspond to a Z score. Insert Z score and other values into the above equation and solve for the score.
To determine the cutoff time for earning a prize, we first need to find the z-score associated with the top 15% of riders. The z-score helps us measure how many standard deviations a particular value is from the mean in a normal distribution.
Step 1: Calculate the z-score
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = cutoff time (unknown)
μ = mean time = 81.7 minutes
σ = standard deviation = 8.6 minutes
To find the z-score for the top 15%, we need to find the corresponding area under the normal curve using a z-table or a statistical calculator. Since we are interested in the top 15%, we subtract this value from 1 to find the area in the tail.
Step 2: Find the z-score for the top 15%
The area in the tail is 1 - 0.15 = 0.85.
Using a standard normal distribution table, we look for the z-score that corresponds to an area of 0.85. The closest value is around 1.036.
Step 3: Substitute the values back into the z-score formula to solve for x (cutoff time).
1.036 = (x - 81.7) / 8.6
Rearranging the formula:
x - 81.7 = 1.036 * 8.6
x - 81.7 = 8.87696
x ≈ 90.57696
Therefore, the association should expect to set the cutoff time around 90.6 minutes to earn a prize.