The provided information are:

Sample size \(\displaystyle{\left({n}\right)}={30}\)

Sample mean \(\displaystyle{\left(\overline{{{x}}}\right)}={2800}\)

Population standard deviation \(\displaystyle{\left(\sigma\right)}={130}\)

Confidence level \(\displaystyle={99}\%\)

Step 2

a) Using the standard normal table, the z-critical value at \(\displaystyle{99}\%\) confidence level is 2.58.

The confidence interval is:

\(\displaystyle{C}{I}=\overline{{{x}}}\pm{z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={2800}\pm{2.58}\times{\frac{{{130}}}{{\sqrt{{{30}}}}}}\)

\(\displaystyle={\left({2738.9},\ {2861.1}\right)}\)

Thus, the \(\displaystyle{99}\%\) confidence interval is \(\displaystyle{\left({2738.9},\ {2861.1}\right)}\)

Step 3

b) If many confidence interval of same size \(\displaystyle={\left({30}\right)}\) is taken and confidence interval is calculated for each of the sample, then about \(\displaystyle{99}\%\) of them will contain the true population parameter.

Thus, the correct option is (A).