To calculate and compare the z-scores for players A and B, you would need to use the means and standard deviations of the number of home runs scored in each of their respective seasons. Here's the approach to solve this:

1. Locate the Mean and Standard Deviation: Use the provided mean and standard deviation values (which should be accessible via the "Click the icon" instruction on your quiz interface). Let's denote:

   • For Player A's season:
     • Mean = $\mu_A$
     • Standard Deviation = $\sigma_A$
   • For Player B's season:
     • Mean = $\mu_B$
     • Standard Deviation = $\sigma_B$

2. Calculate the Z-scores:

   • The z-score formula is: $z = \frac{X - \mu}{\sigma}$
   • For Player A: $z_A = \frac{63 - \mu_A}{\sigma_A}$
   • For Player B: $z_B = \frac{58 - \mu_B}{\sigma_B}$

3. Compare the Z-scores: Once you have calculated $z_A$ and $z_B$, compare them to determine which is higher. The higher z-score indicates the more impressive performance relative to the season's average performance.

4. Determine the Result: If $z_A > z_B$, Player A's performance was more impressive. If $z_B > z_A$, then Player B's performance was more impressive.

Would you be able to access the mean and standard deviation values, or would you like any further assistance with this calculation?

Additional Questions:

1. What is the general significance of a z-score in statistics?
2. How does changing the standard deviation affect the z-score for a given value?
3. Why might using z-scores be more meaningful than comparing raw scores in this context?
4. Can z-scores be negative, and what would a negative z-score indicate in this scenario?
5. How might you use z-scores in other fields outside of sports?

Tip:

Remember, z-scores are a great way to understand how unusual or typical a specific value is within a distribution, especially when comparing different datasets or distributions.