Let us denote X= population values

Given that X follows a normal distribution with mean \(\displaystyle=\mu={73.1}\) and standard deviation \(\displaystyle=\sigma={28.1}\)

Step 2

The probability that a single randomly selected value is greater than 69.7.

\(\displaystyle{P}{\left({X}{>}{69.7}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{{69.7}-{73.1}}}{{{28.1}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{0.120996441281}\right)}={0.5482}\)

Answer: 0.5482