From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={239.5}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={32.7}\)

Sample size \(\displaystyle{\left({n}\right)}={139}\)

Let X be a random variable which represents the value score.

\(\displaystyle{X}\sim{N}{\left({239.5},{32.7}\right)}\)

The required probability that a sample of size \(\displaystyle{n}={139}\) is randomly selected with a mean greater than 235.9 can be obtained as:

\(\displaystyle{P}{\left({M}{>}{235.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{>}{\frac{{{235.9}-{239.5}}}{{{\frac{{{32.7}}}{{\sqrt{{{139}}}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{1.298}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{1.298}\right)}\)</span>

\(\displaystyle={1}-{0.0971}={0.9029}\) (Using standard normal table)

Hence, the required probability is 0.9029.

Mean \(\displaystyle{\left(\mu\right)}={239.5}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={32.7}\)

Sample size \(\displaystyle{\left({n}\right)}={139}\)

Let X be a random variable which represents the value score.

\(\displaystyle{X}\sim{N}{\left({239.5},{32.7}\right)}\)

The required probability that a sample of size \(\displaystyle{n}={139}\) is randomly selected with a mean greater than 235.9 can be obtained as:

\(\displaystyle{P}{\left({M}{>}{235.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{>}{\frac{{{235.9}-{239.5}}}{{{\frac{{{32.7}}}{{\sqrt{{{139}}}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{1.298}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{1.298}\right)}\)</span>

\(\displaystyle={1}-{0.0971}={0.9029}\) (Using standard normal table)

Hence, the required probability is 0.9029.