To solve the problem, we need to calculate the z-score using the formula provided:

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

Where:

• $x$ is the data value (113),
• $\mu$ is the mean (101),
• $\sigma$ is the standard deviation (4).

Now, substituting the values:

$Z = \frac{113 - 101}{4} = \frac{12}{4} = 3.0000$

So, the z-score is 3.0000 when rounded to four decimal places.

Would you like further details or clarifications? Here are some additional questions related to this problem:

1. What does a z-score represent in the context of a normal distribution?
2. How would the z-score change if the data value was less than the mean?
3. What is the probability associated with a z-score of 3 in a standard normal distribution?
4. How can the z-score formula be modified for other distributions?
5. Why is it important to standardize data using the z-score?

Tip: The z-score indicates how many standard deviations a data point is from the mean in a standard normal distribution.