Given n=12 and p=0.6.

Then,

Mean = np

\(\displaystyle={13}\times{0.6}\)

=7.8

Standard deviation \(\displaystyle=\sqrt{{{n}{p}{\left({1}-{p}\right)}}}\)

\(\displaystyle=\sqrt{{{7.8}{\left({1}-{0.6}\right)}}}\)

\(\displaystyle=\sqrt{{{3.12}}}\)

=1.7663

Step 2

So,

\(\displaystyle{P}{\left({X}\ge{11}\right)}={P}{\left(\frac{{{X}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\ge\frac{{{11}-{m}{e}{a}{n}}}{{{s}{\tan{{d}}}{a}{r}{d}{d}{e}{v}{i}{a}{t}{i}{o}{n}}}\right)}\)

\(\displaystyle={P}{\left({Z}\ge\frac{{{11}-{7.8}}}{{{1.7663}}}\right)}\)

\(\displaystyle={P}{\left({Z}\ge{1.81}\right)}\) From the right tailed z - table.

=0.4649

Therefore, p (at least 11)=0.4649