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 single randomly selected value is greater than 235.9 can be obtained as:

\(\displaystyle{P}{\left({X}{>}{235.9}\right)}={P}{\left({\frac{{{x}-\mu}}{{\sigma}}}{>}{\frac{{{235.9}-{239.5}}}{{{32.7}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{0.110}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{0.110}\right)}\)

\(\displaystyle={1}-{0.4562}={0.5438}\) (Using standard normal table)

Thus, the required probability is 0.5438.