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.