Answer:
To find the minimum save percentage for recruits to be considered the top 20% of goalies, we'll need to use the z-score formula and the standard normal distribution table (z-table).
The z-score formula is:
\[z = \frac{(x - \mu)}{\sigma}\]
Where:
- \(z\) is the z-score.
- \(x\) is the value we want to find (in this case, the minimum save percentage).
- \(\mu\) is the mean save percentage (given as 91%).
- \(\sigma\) is the standard deviation (given as 10).
To find the z-score that corresponds to the top 20% of the distribution, we'll use the z-table or a calculator with a cumulative standard normal distribution function. For the top 20%, we're looking for the z-score that leaves 80% below it.
Using a z-table or calculator, you can find that the z-score for the top 20% is approximately 0.84 (rounded to two decimal places).
Now, we can rearrange the z-score formula to solve for \(x\):
\[x = \mu + (z \cdot \sigma)\]
Substitute the values:
\[x = 91 + (0.84 \cdot 10)\]
Calculating this:
\[x = 91 + 8.4\]
So, the minimum save percentage for recruits to be considered the top 20% of goalies is approximately 99.40%. Rounded to two decimal places, this would be "99.40%."