To solve this problem, let's go through each part step by step.

Part 1: Transform the age to a z-score.

The formula for calculating a z-score is:

$z = \frac{(X - \mu)}{\sigma}$

Where:

• $X$ is the value (age of the winner)
• $\mu$ is the mean
• $\sigma$ is the standard deviation

Given:

• $X = 36$
• $\mu = 28.6$
• $\sigma = 3.4$

Now, substituting the values into the formula:

$z = \frac{(36 - 28.6)}{3.4} = \frac{7.4}{3.4} \approx 2.18$

So the z-score is approximately 2.18.

Part 2: Interpret the results.

The z-score of 2.18 means that the age of 36 years is 2.18 standard deviations above the mean age of the winners.

Part 3: Determine whether the age is unusual.

In general, a z-score greater than 2 or less than -2 is considered unusual in a bell-shaped distribution. Since 2.18 is greater than 2, the age of 36 years can be considered unusual.

Summary of Answers:

• (a) The z-score is 2.18.
• (b) An age of 36 is 2.18 standard deviations above the mean.
• (c) The age is unusual because it is more than 2 standard deviations from the mean.

Would you like to dive deeper into any of these parts, or do you have any other questions? Here are some related questions to consider:

1. What does it mean for a distribution to be bell-shaped?
2. How can z-scores be used in other contexts?
3. What is the significance of the mean and standard deviation in a dataset?
4. How do you interpret negative z-scores?
5. Can z-scores help identify outliers in a dataset?

Tip: Always consider the context of the data when interpreting z-scores, as the implications can vary based on the field of study.