a) The \(\displaystyle{99}\%\) confidence interval for the average number of kilometres an automobile is driven is obtained below:

The value of mean is 23,500 kilometres, population standard deviation is 3,900 and sample size \(\displaystyle{\left({n}\right)}={100}\)

Critical value:

From the standard normal distribution table, for the \(\displaystyle{99}\%\) confidence level the critical values is \(\displaystyle{\left({z}^{{\cdot}}\right)}{2.58}\).

The confidence interval formula for the population mean is,

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

Substitute mean \(\displaystyle={23},{500}\), standard deviation \(\displaystyle{\left(\sigma\right)}\) is 3,900 and sample size \(\displaystyle{\left({n}\right)}={100}\).

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

\(\displaystyle={23},{500}\pm{2.58}{\frac{{{3.900}}}{{\sqrt{{{100}}}}}}\)

\(\displaystyle={23},{500}\pm{2.58}{\left({390}\right)}\)

\(\displaystyle={23},{500}\pm{1006.2}\)

\(\displaystyle={\left({23},{500}-{1006.2},\ {23},{500}+{1006.2}\right)}\)

\(\displaystyle={\left({22},{493.8},\ {24},{506.2}\right)}\)

Thus, the \(\displaystyle{99}\%\) confidence interval for the average number of kilometres an automobile is between \(\displaystyle{22},{493.8}\) and \(\displaystyle{24},{506.2}\).

b) The required formula is obtained below:

\(\displaystyle{M}{E}={z}_{{{\frac{{\alpha}}{{{2}}}}}}{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}\)

Substitute 3,900 for \(\displaystyle\sigma\), 100 for n and 2.58 for \(\displaystyle{Z}_{{{\frac{{\alpha}}{{{2}}}}}}\).

\(\displaystyle{M}{E}={2.58}{\left({\frac{{{3},{900}}}{{\sqrt{{{100}}}}}}\right)}\)

\(\displaystyle={2.58}{\left({\frac{{{3},{900}}}{{{10}}}}\right)}\)

\(\displaystyle={2.58}{\left({390}\right)}\)

\(\displaystyle={1006.2}\)

With \(\displaystyle{99}\%\) confidence that the possible size of our error will not exceed the 1006.2 kilometres.