Suppose you have a normally distributed population with a mean of 47.76, standard deviation of 2.3.  Consider the sampling distribution with sample size of 36.  There is an interval (Lower, Upper) centered at 47.76 that contains 83% of the sampling distribution.  Find the Upper bound of this interval.

Round your answers to 4 decimal places.

To solve this problem, we will use the properties of the sampling distribution of the sample mean.

Given:
- Population mean \(\mu = 47.76\)
- Population standard deviation \(\sigma = 2.3\)
- Sample size \(n = 36\)
- The interval contains 83% of the sampling distribution, and we are asked to find the **Upper bound**.

### Steps:

1. **Standard Error (SE) of the Sample Mean**:  
   The standard error is calculated as:
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{2.3}{\sqrt{36}} = \frac{2.3}{6} = 0.3833
   \]

2. **Find the z-value corresponding to 83%**:
   Since we are looking for a symmetric interval centered at the mean, we can split the 83% in half, leaving 41.5% on each side. So, we are looking for a z-value such that 41.5% of the normal distribution is below it.

   From a standard normal table, the z-value corresponding to an area of 0.415 (one side of the distribution) is approximately:
   \[
   z = 1.4051
   \]

3. **Find the Upper bound**:
   The upper bound is given by:
   \[
   \text{Upper bound} = \mu + z \times SE = 47.76 + 1.4051 \times 0.3833
   \]
   \[
   \text{Upper bound} = 47.76 + 0.5386 = 48.2986
   \]

Thus, the **Upper bound** of the interval is approximately **48.2986**.

### Final Answer:  
The Upper bound is \( \boxed{48.2986} \).

Would you like further details or have any questions?

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Here are 5 related questions for deeper understanding:

1. What is the role of standard error in constructing confidence intervals?
2. How would the interval change if the sample size increased?
3. What z-value corresponds to a 90% confidence level for the sampling distribution?
4. Can we apply the Central Limit Theorem to non-normal distributions? Under what conditions?
5. How would the interval change if the standard deviation were larger?

**Tip**: Increasing the sample size reduces the standard error, making your interval narrower.

Suppose you have a normally distributed population with a mean of 47.76, standard deviation of 2.3. Consider the sampling distribution with sample size of 36. There is an interval (Lower, Upper) centered at 47.76 that contains 83% of the sampling distribution. Find the Upper bound of this interval.

Discover how to calculate the upper bound of an interval containing 83% of a sampling distribution, given a normally distributed population with a mean of 47.76, a standard deviation of 2.3, and a sample size of 36. Learn about standard error, z-scores, and confidence intervals, suitable for college-level statistics students.

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