To find a reasonable range of scores for the entire population of 5,000 students, we can use the formula for the confidence interval:
\[ \text{Confidence Interval} = \text{Sample Mean} \pm (z \times \frac{\text{Standard Deviation}}{\sqrt{n}}) \]
Where:
- Sample Mean = 18
- z = 1.96 (for 95% confidence)
- Standard Deviation = 2.5
- n = 125 (the number of students selected)
First, we calculate the standard error (SE):
\[ SE = \frac{\text{Standard Deviation}}{\sqrt{n}} = \frac{2.5}{\sqrt{125}} = \frac{2.5}{11.1803} \approx 0.224 \]
Next, we calculate the margin of error (ME):
\[ ME = z \times SE = 1.96 \times 0.224 \approx 0.439 \]
Now we can find the confidence interval:
\[ \text{Lower limit} = \text{Sample Mean} - ME = 18 - 0.439 \approx 17.561 \] \[ \text{Upper limit} = \text{Sample Mean} + ME = 18 + 0.439 \approx 18.439 \]
Thus, the confidence interval for the average score of the entire population of 5,000 students is approximately:
\[ (17.561, 18.439) \]
Rounding this to two decimal places, we get the range approximately between 17.56 and 18.44.
Looking at the available responses, the most reasonable answer that describes the average scores of the total students from the confidence interval we calculated is:
16 to 20
This range is broader than the confidence interval but captures the essence of the expected average score range for the entire student population.