Question # A population of values has a normal distribution with \mu=239.5 and \sigma=32.7.

Random variables
ANSWERED A population of values has a normal distribution with $$\displaystyle\mu={239.5}$$ and $$\displaystyle\sigma={32.7}$$. You intend to draw a random sample of size $$\displaystyle{n}={139}$$.
Find the probability that a sample of size $$n=139$$ is randomly selected with a mean greater than 235.9.
$$\displaystyle{P}{\left({M}{>}{235.9}\right)}=$$? 2020-10-26

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