Question

5,000 students in a district take a standardized test. 125 students’ scores are randomly selected. Of these 125 students, the average score was 18 out of 25. Given a standard deviation of 2.5 and a 95% level of confidence with a z-score of 1.96, what is a reasonable range of scores for the 5,000 students?(1 point)
Responses

4,875 to 5,125
4,875 to 5,125

7 to 43
7 to 43

2
2

16 to 20

Answer 1

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**.