Given Information:

\(\displaystyle\mu={133.5}\) and \(\displaystyle\sigma={5.2}\)

\(\displaystyle{n}={230}\)

To find the probability that a sample of size \(\displaystyle{n}={230}\) is randomly selected with a mean between 133.6 and 134.1:

Based on the concept of Central limit theorem, if a random sample of size 'n' is drawn from a population with mean \(\displaystyle\mu\) and standard deviation \(\displaystyle\sigma\), the sampling distribution of sample mean \(\displaystyle\overline{{{x}}}\) is approximately normally distributed with mean \(\displaystyle\mu_{{\overline{{{x}}}}}=\mu\) and standard deviation \(\displaystyle\sigma_{{\overline{{{x}}}}}={\frac{{\sigma}}{{\sqrt{{{n}}}}}}\).

Mean of sample mean = \(\displaystyle\mu_{{\overline{{{x}}}}}=\mu={133.5}\)

Standard deviation \(\displaystyle\sigma_{{\overline{{{x}}}}}={\frac{{\sigma}}{{\sqrt{{{n}}}}}}={\frac{{{5.2}}}{{\sqrt{{{230}}}}}}={0.343}\)

Required probability can be obtained as follows:

\(\displaystyle{P}{\left({133.6}{<}\overline{{{x}}}{<}{134.1}\right)}={P}{\left(\overline{{{x}}}{<}{134.1}\right)}-{P}{\left(\overline{{{x}}}{<}{133.6}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{\overline{{{x}}}-\mu}}{{\sigma}}}{<}{\frac{{{134.1}-{133.5}}}{{{0.343}}}}\right)}-{P}{\left({\frac{{\overline{{{x}}}-\mu}}{{\sigma}}}{<}{\frac{{{133.6}-{133.5}}}{{{0.343}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{1.75}\right)}-{P}{\left({Z}{<}{0.29}\right)}\)</span>

\(\displaystyle={0.95994}−{0.61409}={0.34585}{\left[{u}{\sin{{g}}}{s}{\tan{{d}}}{a}{r}{d}{\left\|{a}\right\|}{l}{t}{a}{b}\le\right]}\)

Therefore, probability that a sample of size \(\displaystyle{n}={230}\) is randomly selected with a mean between 133.6 and 134.1 is 0.3459

\(\displaystyle\mu={133.5}\) and \(\displaystyle\sigma={5.2}\)

\(\displaystyle{n}={230}\)

To find the probability that a sample of size \(\displaystyle{n}={230}\) is randomly selected with a mean between 133.6 and 134.1:

Based on the concept of Central limit theorem, if a random sample of size 'n' is drawn from a population with mean \(\displaystyle\mu\) and standard deviation \(\displaystyle\sigma\), the sampling distribution of sample mean \(\displaystyle\overline{{{x}}}\) is approximately normally distributed with mean \(\displaystyle\mu_{{\overline{{{x}}}}}=\mu\) and standard deviation \(\displaystyle\sigma_{{\overline{{{x}}}}}={\frac{{\sigma}}{{\sqrt{{{n}}}}}}\).

Mean of sample mean = \(\displaystyle\mu_{{\overline{{{x}}}}}=\mu={133.5}\)

Standard deviation \(\displaystyle\sigma_{{\overline{{{x}}}}}={\frac{{\sigma}}{{\sqrt{{{n}}}}}}={\frac{{{5.2}}}{{\sqrt{{{230}}}}}}={0.343}\)

Required probability can be obtained as follows:

\(\displaystyle{P}{\left({133.6}{<}\overline{{{x}}}{<}{134.1}\right)}={P}{\left(\overline{{{x}}}{<}{134.1}\right)}-{P}{\left(\overline{{{x}}}{<}{133.6}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{\overline{{{x}}}-\mu}}{{\sigma}}}{<}{\frac{{{134.1}-{133.5}}}{{{0.343}}}}\right)}-{P}{\left({\frac{{\overline{{{x}}}-\mu}}{{\sigma}}}{<}{\frac{{{133.6}-{133.5}}}{{{0.343}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{1.75}\right)}-{P}{\left({Z}{<}{0.29}\right)}\)</span>

\(\displaystyle={0.95994}−{0.61409}={0.34585}{\left[{u}{\sin{{g}}}{s}{\tan{{d}}}{a}{r}{d}{\left\|{a}\right\|}{l}{t}{a}{b}\le\right]}\)

Therefore, probability that a sample of size \(\displaystyle{n}={230}\) is randomly selected with a mean between 133.6 and 134.1 is 0.3459