Step 1

Given:

\(\displaystyle\mu={192.6}\),

\(\displaystyle\sigma={34.4}\),

\(\displaystyle{n}={173}\).

The Z-score follows standard normal distribution.

Step 2

\(\displaystyle{P}{\left[{X}{<}{186.1}\right]}={P}{\left[{\frac{{{X}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{186.1}-{192.6}}}{{{\frac{{{34.4}}}{{\sqrt{{{173}}}}}}}}}\right]}\)

\(\displaystyle={P}{\left[{Z}{<}-{2.49}\right]}={0.0066}\) (Use standard normal table)

The probability that a sample of size \(\displaystyle{n}={173}\) is randomly selected with a mean less than 186.1 is 0.0066.