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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.

A normal distribution has a mean of 101 and a standard deviation of 4. Find the z-score for a data value of 113.

Calculate the z-score for a data value of 113 from a normal distribution with a mean of 101 and a standard deviation of 4. Learn how to apply the z-score formula for understanding deviations from the mean. Suitable for high school statistics students.

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