A population of values has a normal distribution with μ=239.5 and σ=32.7. You intend to...

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.
${\displaystyle X\sim N(239.5,32.7)}$
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:
${\displaystyle P\left(M>235.9\right)=P\left(\frac{M-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{235.9-239.5}{\frac{32.7}{\sqrt{139}}}\right)}$
${\displaystyle =P(Z\succ 1.298)}$
${\displaystyle =1-P(Z<-1.298)}$
${\displaystyle =1-0.0971=0.9029}$ (Using standard normal table)
Hence, the required probability is 0.9029.