To find the 95% confidence interval for the mean reading scores of all fifth-graders, we can use the formula for the confidence interval:
\[ \text{CI} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
where:
- \(\bar{x}\) is the sample mean (82),
- \(z\) is the z-score corresponding to the desired confidence level (for 95%, \(z \approx 1.96\)),
- \(\sigma\) is the population standard deviation (15),
- \(n\) is the sample size (35).
Now we calculate the standard error (SE):
\[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{35}} \approx \frac{15}{5.916} \approx 2.53 \]
Next, we calculate the margin of error (ME):
\[ \text{ME} = z \times \text{SE} \approx 1.96 \times 2.53 \approx 4.96 \]
Now, we can calculate the confidence interval:
\[ \text{CI} = 82 \pm 4.96 \]
This gives us:
\[ \text{Lower limit} = 82 - 4.96 \approx 77.04 \ \text{Upper limit} = 82 + 4.96 \approx 86.96 \]
Rounding to the nearest whole number, we obtain:
\[ \text{Lower limit} \approx 77 \ \text{Upper limit} \approx 87 \]
Thus, the 95% confidence interval for the mean reading scores of all fifth-graders is:
\[ \boxed{(77,87)} \]