To determine the probability that the sample mean for a sample of size 30 will be more than 56, we need more context, specifically the population mean ($\mu$) and the population standard deviation ($\sigma$). With these values, we can proceed with the following steps:

Step-by-Step Solution

1. Determine the Z-Score: We use the Z-score formula for a sample mean: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$ where:

   • $\bar{X} = 56$ (the sample mean we're interested in),
   • $\mu$ is the population mean,
   • $\sigma$ is the population standard deviation,
   • $n = 30$ (sample size).

2. Calculate the Z-Score (with values for $\mu$ and $\sigma$ provided).

3. Use the Z-Table: Once the Z-score is calculated, we use the provided Z-table to find the corresponding probability. This table provides the cumulative probability up to the Z-score. For the probability that the sample mean is greater than 56, we would find $P(Z > \text{calculated Z-score})$, which is $1 - P(Z \leq \text{Z-score})$.

Please provide the values of the population mean and standard deviation if available. Alternatively, I can guide you through with assumed values for practice.

Would you like to proceed with this or have further questions?

Follow-up Questions:

1. What are the population mean and standard deviation for this problem?
2. Would you like to understand the derivation of the Z-score formula for sample means?
3. Do you need a refresher on how to interpret Z-scores and probabilities?
4. Are you familiar with using the Z-table, or would you like a walkthrough?
5. Would you like to know how sample size affects the Z-score calculation?

Tip:

For probability questions involving sample means, knowing the standard error ($\sigma / \sqrt{n}$) is essential, as it adjusts the spread of the sampling distribution based on sample size.