To determine who had the more "extreme" age when winning the award, we will use the z-score. The z-score tells us how many standard deviations an individual's age is from the mean in their respective category. The formula for the z-score is:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ is the individual’s age.
• $\mu$ is the mean age of all winners in that category.
• $\sigma$ is the standard deviation of ages for that category.

Step 1: Calculate the z-score for the Best Actor winner

• Age of Best Actor winner: $x = 37$
• Mean age for Best Actor: $\mu = 45.5$
• Standard deviation for Best Actor: $\sigma = 8.3$

The z-score for the Best Actor winner is: $z_{\text{actor}} = \frac{37 - 45.5}{8.3} = \frac{-8.5}{8.3} \approx -1.02$

Step 2: Calculate the z-score for the Best Actress winner

• Age of Best Actress winner: $x = 51$
• Mean age for Best Actress: $\mu = 33.1$
• Standard deviation for Best Actress: $\sigma = 10.9$

The z-score for the Best Actress winner is: $z_{\text{actress}} = \frac{51 - 33.1}{10.9} = \frac{17.9}{10.9} \approx 1.64$

Step 3: Compare the z-scores

• The z-score for the Best Actor winner is approximately $-1.02$, meaning the Best Actor winner was 1.02 standard deviations below the mean age.
• The z-score for the Best Actress winner is approximately $1.64$, meaning the Best Actress winner was 1.64 standard deviations above the mean age.

Conclusion:

The Best Actress winner had a higher absolute z-score ($1.64$ compared to $1.02$), meaning she was further from the average age in her category than the Best Actor winner was in his category. Therefore, the Best Actress winner had the more extreme age relative to the average age in her category.

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How is the z-score used in comparing data from different distributions?
2. Can z-scores be used to compare more than two groups?
3. What does a negative z-score represent in this context?
4. How would the results change if the standard deviations were different?
5. How could we interpret z-scores in other contexts, such as performance metrics?

Tip: When comparing different groups, using z-scores standardizes the data, allowing for comparison even if the groups have different means and standard deviations.