The time it takes to run a road race is normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. What values cuts off the slowest 12% of the racers?
http://davidmlane.com/hyperstat/z_table.html
But, you need to learn how to do this with the tables, as painful as those are to use.
To find the cutoff value that separates the slowest 12% of the racers, you need to work with the concept of z-scores and the standard normal distribution.
Step 1: Determine the z-score corresponding to the desired percentile.
The percentile is given as 12%. Since we want to find the cutoff value for the slowest racers, we need to look at the left tail of the normal distribution. To find the z-score that corresponds to the 12th percentile, we subtract 12% from 100%, resulting in 88%.
Step 2: Find the z-score using a standard normal distribution table or calculator.
Using a standard normal distribution table (sometimes called a z-table), you can find the z-score that corresponds to 88%. The closest z-score to 88% is approximately -1.18.
Step 3: Use the z-score formula to find the cutoff value.
The z-score formula is: z = (X - μ) / σ
Where:
- z represents the z-score
- X represents the cutoff value
- μ represents the mean of the distribution
- σ represents the standard deviation of the distribution
Since we want to find the cutoff value, we rearrange the formula to solve for X:
X = z * σ + μ
Step 4: Plug in values to find the cutoff value.
Given:
μ (mean) = 190 minutes
σ (standard deviation) = 21 minutes
z = -1.18 (from Step 2)
Using the formula, we substitute the values:
X = -1.18 * 21 + 190
X ≈ 166.22
Therefore, the cutoff time that separates the slowest 12% of the racers is approximately 166.22 minutes.