obtain the probability that the sampling error made in estimatjng the popularion length of stay by the mean length of stay of a sample of 60 patients will be at most 2 days. The standard deviation of the sample distribution is 0.8521

To obtain the probability that the sampling error made in estimating the population length of stay by the mean length of stay of a sample of 60 patients will be at most 2 days, we need to calculate the probability using the **standard error** and the **Z-score** for the normal distribution.

### Key Information:
- Sample size (\(n\)) = 60
- Standard deviation (\(\sigma\)) = 0.8521
- Sampling error = 2 days

### Steps to calculate the probability:

1. **Calculate the standard error of the mean (SEM)**:
   The standard error of the mean is calculated by the formula:
   \[
   \text{Standard Error of the Mean (SEM)} = \frac{\sigma}{\sqrt{n}}
   \]
   where:
   - \(\sigma = 0.8521\) (standard deviation of the sample)
   - \(n = 60\) (sample size)

2. **Convert the sampling error into a Z-score**:
   The Z-score tells us how many standard deviations the error is from the mean. The formula for the Z-score is:
   \[
   Z = \frac{\text{sampling error}}{\text{SEM}}
   \]
   In this case, the sampling error is 2 days.

3. **Find the probability**:
   Once we have the Z-score, we use the standard normal distribution table to find the probability that the sampling error is within the given bounds.

Let’s calculate these steps.The Z-score is approximately \( 18.18 \).

### Step 3: Probability Calculation

Given that a Z-score of 18.18 is extremely high, this implies that the probability of the sampling error being at most 2 days is nearly 100%. In fact, for such a high Z-score, the probability is essentially 1 (or 100%).

This means that the chance of the sampling error being within 2 days is almost guaranteed in this scenario.

Would you like to go into more detail or have any further questions?

Here are five related questions that expand on this topic:
1. How would the probability change if the sampling error was reduced to 1 day?
2. What if the sample size increased to 100 patients, how would that affect the result?
3. How is the Z-score interpreted in terms of standard deviations?
4. Can you explain the importance of the central limit theorem in this context?
5. How does increasing the standard deviation affect the confidence in the estimate?

**Tip**: Larger sample sizes generally reduce the standard error, making estimates more precise.

Learn how to calculate the probability that the sampling error in estimating the population length of stay by the mean length of stay of a sample of 60 patients is at most 2 days. This solution covers the Z-score, standard error of the mean, and the application of the central limit theorem. Suitable for high school students in Grades 10-12.

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