The probability that mean value is between 11.5 and 14 is 0.8884, which is obtained below:

\(\displaystyle{P}{\left({x}{1}{<}\overline{{{X}}}{<}{x}{2}\right)}={P}{\left({\frac{{{x}{1}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{z}{<}{\frac{{{x}{2}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)</span>

\(\displaystyle{P}{\left({11.5}{<}\overline{{{X}}}{<}{14}\right)}={P}{\left({\frac{{{11.5}-{13.2}}}{{{\frac{{{5}}}{{\sqrt{{{60}}}}}}}}}{<}{z}{<}{\frac{{{14}-{13.2}}}{{{\frac{{{5}}}{{\sqrt{{{60}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.636}{<}{z}{<}{1.240}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{1.240}\right)}-{P}{\left({z}{<}-{2.636}\right)}{\left[{U}{s}{e}\ {s}{\tan{{d}}}{a}{r}{d}\ {\left\|{a}\right\|}{l}\ {t}{a}{b}\le\right]}\)</span>

=0.89251-0.00415=0.8884.

\(\displaystyle{P}{\left({x}{1}{<}\overline{{{X}}}{<}{x}{2}\right)}={P}{\left({\frac{{{x}{1}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{z}{<}{\frac{{{x}{2}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)</span>

\(\displaystyle{P}{\left({11.5}{<}\overline{{{X}}}{<}{14}\right)}={P}{\left({\frac{{{11.5}-{13.2}}}{{{\frac{{{5}}}{{\sqrt{{{60}}}}}}}}}{<}{z}{<}{\frac{{{14}-{13.2}}}{{{\frac{{{5}}}{{\sqrt{{{60}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.636}{<}{z}{<}{1.240}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{1.240}\right)}-{P}{\left({z}{<}-{2.636}\right)}{\left[{U}{s}{e}\ {s}{\tan{{d}}}{a}{r}{d}\ {\left\|{a}\right\|}{l}\ {t}{a}{b}\le\right]}\)</span>

=0.89251-0.00415=0.8884.