Step 1

From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={197}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={68}\)

Let X be a random variable which represents the value.

\(\displaystyle{X}\sim{N}{\left({197},{68}\right)}\)

Step 2

Sample size \(\displaystyle{\left({n}\right)}={181}\)

The required probability that a sample of size \(\displaystyle{n}={181}\) is randomly selected with a mean between 198.5 and 205.6 \(\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}\)</span> can be obtained as:

\(\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}={P}{\left({\frac{{{198.5}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{205.6}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left({0.297}{<}{Z}{<}{1.701}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{1.701}\right)}-{P}{\left({Z}{<}{0.297}\right)}\)</span>

\(\displaystyle={0.9555}-{0.6168}={0.3387}\) (Using standard normal table)

Thus, the required probability is 0.3387.

From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={197}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={68}\)

Let X be a random variable which represents the value.

\(\displaystyle{X}\sim{N}{\left({197},{68}\right)}\)

Step 2

Sample size \(\displaystyle{\left({n}\right)}={181}\)

The required probability that a sample of size \(\displaystyle{n}={181}\) is randomly selected with a mean between 198.5 and 205.6 \(\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}\)</span> can be obtained as:

\(\displaystyle{P}{\left({198.5}{<}{M}{<}{205.6}\right)}={P}{\left({\frac{{{198.5}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{M}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{181}}}}}}}}}{<}{\frac{{{205.6}-{197}}}{{{\frac{{{68}}}{{\sqrt{{{181}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left({0.297}{<}{Z}{<}{1.701}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{1.701}\right)}-{P}{\left({Z}{<}{0.297}\right)}\)</span>

\(\displaystyle={0.9555}-{0.6168}={0.3387}\) (Using standard normal table)

Thus, the required probability is 0.3387.