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.