Given Information:

\(\displaystyle\mu={129.7}\)

\(\displaystyle\sigma={7.7}\)

\(\displaystyle{n}={10}\)

Step 2

The probability that a sample of size \(\displaystyle{n}={10}\) is randomly selected with a mean less than 130.9.

\(\displaystyle{z}={\frac{{{x}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}={\frac{{{130.9}-{129.7}}}{{{\frac{{{7.7}}}{{\sqrt{{{n}}}}}}}}}={0.4928}\)

By referring to the z distribution, the p-value at \(\displaystyle{z}={0.492}\) is 0.311

Therefore, The probability that a sample of size \(\displaystyle{n}={10}\) is randomly selected with a mean less than 130.9 is 0.311