The \(\displaystyle{95}\%\) CI is given by

\(\displaystyle{C}{I}={p}\pm{z}_{{{0.05}}}\sqrt{{{\frac{{{p}{q}}}{{{n}}}}}}\)

\(\displaystyle{C}{I}={0.3}\pm{1.96}\sqrt{{{\frac{{{0.3}\times{0.7}}}{{{100}}}}}}\)

\(\displaystyle{C}{I}={\left[{0.21},\ {0.39}\right]}\)

Option (e) is correct.

\(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.39}\right)}={0.95}\)