Step 1

According to the provided information,

\(\displaystyle\mu={29.3}\)

\(\displaystyle\sigma={65.1}\)

\(\displaystyle{n}={142}\)

\(\displaystyle{P}{\left({27.7}{<}{X}{<}{35.3}\right)}={P}{\left({\frac{{{27.7}-{29.3}}}{{\frac{{65.1}}{\sqrt{{{142}}}}}}}{<}{\frac{{{X}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{35.3}-{29.3}}}{{\frac{{65.1}}{\sqrt{{{142}}}}}}}\right)}\)

\(\displaystyle={P}{\left(-{0.29}{<}{Z}{<}{1.1}\right)}{\left({Z}={\frac{{{X}-\mu}}{{\sigma}}}\right)}\)

\(\displaystyle={0.4784}\) (From standard normal table)

Therefore, the probability that a sample of size \(\displaystyle{n}={142}\) is randomly selected with a mean between 27.7 and 35.3.

\(\displaystyle{P}{\left({27.7}{<}\overline{{{X}}}{<}{35.3}\right)}={0.4784}\)