In a recent awards​ ceremony, the age of the winner for the Best Actor award was 34 and the age of the winner for the Best Actress award was 54. For all recipients of Best​ Actor, the mean age is 43.4 years and the standard deviation is 7.6 years. For all recipients of Best​ Actress, the mean age is 37.9 years and the standard deviation is 10.9 years.​ (All ages are determined at the time of the awards​ ceremony.) Relative to the award​ category, who had the more extreme age when winning the​ award, the winner of Best Actor or the winner of Best​ Actress? Explain.

To determine who had the more extreme age relative to their respective award categories, we can use **z-scores**. A z-score tells us how many standard deviations a value is from the mean. The formula for calculating a z-score is:

\[
z = \frac{{X - \mu}}{{\sigma}}
\]

Where:
- \( X \) is the individual score (age in this case),
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.

### 1. Calculate the z-score for the Best Actor winner:
Given:
- Winner's age \( X = 34 \),
- Mean age for Best Actor \( \mu = 43.4 \),
- Standard deviation for Best Actor \( \sigma = 7.6 \).

\[
z_{\text{actor}} = \frac{{34 - 43.4}}{{7.6}} = \frac{{-9.4}}{{7.6}} \approx -1.24
\]

### 2. Calculate the z-score for the Best Actress winner:
Given:
- Winner's age \( X = 54 \),
- Mean age for Best Actress \( \mu = 37.9 \),
- Standard deviation for Best Actress \( \sigma = 10.9 \).

\[
z_{\text{actress}} = \frac{{54 - 37.9}}{{10.9}} = \frac{{16.1}}{{10.9}} \approx 1

Learn how to calculate and compare z-scores to determine which winner, Best Actor (age 34) or Best Actress (age 54), had the more extreme age at the awards ceremony. This problem uses the z-score formula and covers concepts in statistics, suitable for students in Grades 10-12.

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