To find the margin of error for the population mean, we'll use the formula for the margin of error [tex]\( ME \)[/tex]:
[tex]\[ ME = \frac{z \cdot s}{\sqrt{n}} \][/tex]
where:
- [tex]\( z \)[/tex] is the z-score corresponding to the desired confidence level,
- [tex]\( s \)[/tex] is the standard deviation, and
- [tex]\( n \)[/tex] is the sample size.
Given values:
- Z-score ([tex]\( z \)[/tex]) = 1.96 (for a 95% confidence level),
- Standard deviation ([tex]\( s \)[/tex]) = 2 days,
- Sample size ([tex]\( n \)[/tex]) = 35.
Let's go through the steps now:
1. Calculate the square root of the sample size ([tex]\( \sqrt{n} \)[/tex]):
[tex]\[ \sqrt{35} \approx 5.916 \][/tex]
2. Divide the standard deviation by the result from step 1:
[tex]\[ \frac{s}{\sqrt{n}} = \frac{2}{5.916} \approx 0.338 \][/tex]
3. Multiply the result by the z-score:
[tex]\[ ME = 1.96 \cdot 0.338 \approx 0.663 \][/tex]
Therefore, the margin of error for the population mean, at a 95% confidence level, is approximately 0.66.
So, the correct choice from the provided answers is:
[tex]\[ \boxed{0.66} \][/tex]
