To find the reasonable range of scores for the 5,000 students based on the average score and the standard deviation from the sample of 125 students, we can use the confidence interval formula for the population mean. The formula for the confidence interval is:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean (18 in this case),
- \(z\) is the z-score corresponding to the desired confidence level (1.96 for 95% confidence),
- \(\sigma\) is the standard deviation (2.5 in this case),
- \(n\) is the sample size (125 in this case).
Let's calculate the standard error first:
\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{125}} \approx \frac{2.5}{11.18} \approx 0.224 \]
Now we can compute the margin of error:
\[ \text{Margin of Error} = z \times \text{Standard Error} = 1.96 \times 0.224 \approx 0.439 \]
Now we can construct the confidence interval around the sample mean:
\[ \text{Lower Limit} = \bar{x} - \text{Margin of Error} = 18 - 0.439 \approx 17.561 \] \[ \text{Upper Limit} = \bar{x} + \text{Margin of Error} = 18 + 0.439 \approx 18.439 \]
Thus, the 95% confidence interval for the average score of the 5,000 students is approximately (17.56, 18.44).
From the options given:
- 7 to 43
- 4875 to 5,125
- 16 to 20
The reasonable range of scores, based on the calculated confidence interval and the context provided, would be 16 to 20, as it encompasses the average score range we calculated.