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A global population study of professional hockey goalies finds a population save percentage of 91 ± 10 (μ ± σ^²). Using this information, what would the minimum save percentage be for recruits to be considered the top 20% of goalies? (Hint: the equation for z-scores is z = (x−μ)​/σ ; please round your answer to two decimal places, e.g., "12.34\%" or "12.34") Minimum save percentage =

User Hasienda
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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%."

User Celeritas
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