Consider,

\(\displaystyle{P}{\left({59.5}{<}\overline{{{X}}}{<}{98.6}\right)}={P}{\left({\frac{{{59.5}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{\overline{{{X}}}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{98.6}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{{59.5}-\mu}}{{{\left({\frac{{{32.2}}}{{\sqrt{{{15}}}}}}\right)}}}}{<}{z}{<}{\frac{{{98.6}-{77}}}{{{\left({\frac{{{32.2}}}{{\sqrt{{{15}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.10488}{<}{z}{<}{2.59803}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{2.59803}\right)}-{P}{\left({z}{<}-{2.10488}\right)}\)</span>

\(\displaystyle={0.995312}-{0.017651}={0.9777}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{F}{r}{o}{m}\ {t}{h}{e}\ {E}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{2.10488},{0},{1},{T}{R}{U}{E}\right)}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left({2.59803},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)

Thus, the probability that a sample of size \(\displaystyle{n}={15}\) is randomly selected with a mean between 59.5 and 98.6 is 0.9777.