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To solve for the \( z \)-value, we'll use the formula for the \( z \)-score in the context of the sampling distribution of the sample mean \( \bar{X} \):

\[
z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}
\]

where:
- \( \mu = 230 \) (population mean),
- \( \sigma = 27 \) (population standard deviation),
- \( \bar{X} = 231 \) (sample mean),
- \( n = 27 \) (sample size).

### Step-by-Step Solution:

1. **Calculate the Standard Error (SE)**:
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{27}{\sqrt{27}} = \frac{27}{5.196} \approx 5.19
   \]

2. **Compute the \( z \)-score**:
   \[
   z = \frac{231 - 230}{5.19} = \frac{1}{5.19} \approx 0.19
   \]

### Answer:
The correct \( z \)-value is approximately \( 0.19 \), which matches the option **0.19**.

Would you like further clarification on any step?

Here are some additional questions for further practice:
1. What would the \( z \)-score be if \( \bar{X} \) was 233?
2. How would the answer change if the sample size \( n \) was increased?
3. What is the effect of a larger population standard deviation \( \sigma \) on the \( z \)-score?
4. How would you interpret a \( z \)-score of 1.96 in the context of a sampling distribution?
5. What assumptions are we making about the distribution of \( X \)?

**Tip:** When finding \( z \)-scores for sample means, always use the standard error instead of the population standard deviation.

Let X be a normally distributed random variable. Consider the sampling distribution of X̄. Find the z-value for X̄ when μ = 230, σ = 27, X̄ = 231, and n = 27.

Learn how to calculate the z-value for a sample mean in a normally distributed random variable. This example uses a population mean (μ) of 230, a population standard deviation (σ) of 27, a sample mean (X̄) of 231, and a sample size (n) of 27.

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