The problem in the image is asking to find a 95% confidence interval for the average score of all 1159 seniors at Central High School based on the data from a sample.

Given Data:

• Sample size (n): 50
• Sample mean ($\bar{x}$): 81%
• Sample standard deviation (s): 11%
• Confidence level: 95%

Formula for the Confidence Interval:

The formula for a confidence interval for the mean is:

$\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ = sample mean
• $t_{\alpha/2}$ = t-value for the desired confidence level (depends on degrees of freedom $df = n - 1$)
• $s$ = sample standard deviation
• $n$ = sample size

Steps:

1. Find the $t_{\alpha/2}$ value: For a 95% confidence level and $df = 50 - 1 = 49$, the t-value can be found using a t-distribution table or calculator.
2. Calculate the margin of error: $\text{Margin of Error} = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$
3. Calculate the confidence interval: Add and subtract the margin of error from the sample mean.

Let's calculate the interval.The 95% confidence interval for the average score of all 1159 seniors at Central High School is approximately:

$(77.87\%, 84.13\%)$

This means we are 95% confident that the true average score of all seniors falls within this interval.

Would you like more details or have any questions? Here are some related questions you might consider:

1. How does the sample size affect the width of the confidence interval?
2. What assumptions are made when using a t-distribution for this calculation?
3. How would the confidence interval change if we increased the confidence level to 99%?
4. What is the impact of the sample standard deviation on the confidence interval?
5. How would you interpret this confidence interval in a practical context?

Tip: When interpreting confidence intervals, remember they provide a range of plausible values for the population parameter, not the certainty of including every individual score within that range.