To estimate the reasonable range of scores for the entire population of 5,000 students based on the sample of 125, we can use the formula for the confidence interval:
\[ CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean (average score)
- \(z\) is the z-score for the desired confidence level
- \(\sigma\) is the standard deviation of the sample
- \(n\) is the sample size
Given:
- \(\bar{x} = 18\)
- \(z = 1.96\) (for 95% confidence level)
- \(\sigma = 2.5\)
- \(n = 125\)
First, we calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{125}} = \frac{2.5}{11.1803} \approx 0.2236 \]
Now we calculate the margin of error (ME):
\[ ME = z \cdot SE = 1.96 \cdot 0.2236 \approx 0.438 \]
Next, we can construct the confidence interval:
\[ CI = \bar{x} \pm ME = 18 \pm 0.438 \]
Calculating the limits of the confidence interval:
- Lower limit = \(18 - 0.438 \approx 17.562\)
- Upper limit = \(18 + 0.438 \approx 18.438\)
Rounding to whole numbers, the reasonable range of scores for the 5,000 students could be estimated from around 17 to 18.
The options provided do not include a range that aligns with our calculation. However, based on the closest approximation and standard ranges in educational contexts, from the choices listed, the best represented range of scores might be "16 to 20", as it encompasses the estimated range we calculated.