Questions: Question 5 1 pts Suppose you have a normally distributed population with a mean of 48.2, standard deviation of 3.23. Consider the sampling distribution with sample size of 25. There is an interval (Lower, Upper) centered at 48.2 that contains 82% of the sampling distribution. Find the Upper bound of this interval. Round your answers to 4 decimal places.

We have a normally distributed population with the following parameters:

• Mean (\( \mu \)): \( 48.2 \)
• Standard Deviation (\( \sigma \)): \( 3.23 \)
• Sample Size (\( n \)): \( 25 \)
• Confidence Level: \( 82\% \)

The standard error of the mean (\( SE \)) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3.23}{\sqrt{25}} = 0.6460 \]

To find the z-scores that correspond to the middle \( 82\% \) of the distribution, we calculate the significance level: \[ \alpha = 1 - 0.82 = 0.18 \] Since this is a two-tailed test, we divide by \( 2 \): \[ \alpha/2 = 0.09 \] Using a z-table or calculator, we find the z-score that corresponds to \( 1 - 0.09 = 0.91 \): \[ z \approx 1.6954 \]

The confidence interval is given by: \[ \bar{x} \pm z \cdot SE \] Substituting the values: \[ 48.2 \pm 1.6954 \cdot 0.6460 \] Calculating the bounds: \[ \text{Lower Bound} = 48.2 - 1.6954 \cdot 0.6460 \approx 47.1048 \] \[ \text{Upper Bound} = 48.2 + 1.6954 \cdot 0.6460 \approx 49.2952 \]

The confidence interval is: \[ (47.1048, 49.2952) \] Thus, the upper bound of the interval is: \[ \boxed{49.2952} \]