To calculate the confidence interval for the average score of the 5,000 students based on the sample of 125 students, we can use the formula for the confidence interval:
\[ \text{Confidence Interval} = \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \(\bar{x}\) = sample mean (average score) = 18
- \(z\) = z-score for 95% confidence level = 1.96
- \(\sigma\) = standard deviation = 2.5
- \(n\) = sample size = 125
First, calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{125}} = \frac{2.5}{11.1803} \approx 0.224 \]
Now, we calculate the margin of error (ME):
\[ ME = z \times SE = 1.96 \times 0.224 \approx 0.439 \]
Now, we can determine the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 18 - 0.439 \approx 17.561 \] \[ \text{Upper limit} = \bar{x} + ME = 18 + 0.439 \approx 18.439 \]
Thus, the 95% confidence interval for the average score of the 5,000 students is approximately (17.56, 18.44).
Given the options provided:
- 4,875 to 5,125
- 7 to 43
- 2
- 16 to 20
The reasonable range of scores for the 5,000 students, based on the calculations, is 16 to 20.