A population of values has a normal distribution with \(\displaystyle\mu={116.5}\) and \(\displaystyle\sigma={63.7}\).

A random sample of size \(\displaystyle{\left({n}\right)}={244}\) is drawn.

We need to find the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean(M) between 104.7 and 112.8.

\(\displaystyle{M}\sim{N}{\left({116.5},{\frac{{{63.7}^{{{2}}}}}{{{n}}}}\right)}\) and \(\displaystyle{n}={244}\)

So, \(\displaystyle{Z}={\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}\sim{N}{\left({0},{1}\right)}\)

Thus, \(\displaystyle{P}{\left[{104.7}{<}{M}{<}{112.8}\right]}\)</span>

\(\displaystyle={P}{\left[{\frac{{\sqrt{{{n}}}{\left({104.7}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({112.8}-{116.5}\right)}}}{{{63.7}}}}\right]}\)</span>

\(\displaystyle={P}{\left[-{2.893}{<}{Z}{<}-{0.9073}\right]}\)</span>

\(\displaystyle={P}{\left[{Z}\leq-{0.9073}\right]}-{P}{\left[{Z}\leq-{2.8934}\right]}\)

\(\displaystyle={0.1821}-{0.0019}={0.1802}\)

So, the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean between 104.7 and 112.8 is 0.1802.

A random sample of size \(\displaystyle{\left({n}\right)}={244}\) is drawn.

We need to find the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean(M) between 104.7 and 112.8.

\(\displaystyle{M}\sim{N}{\left({116.5},{\frac{{{63.7}^{{{2}}}}}{{{n}}}}\right)}\) and \(\displaystyle{n}={244}\)

So, \(\displaystyle{Z}={\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}\sim{N}{\left({0},{1}\right)}\)

Thus, \(\displaystyle{P}{\left[{104.7}{<}{M}{<}{112.8}\right]}\)</span>

\(\displaystyle={P}{\left[{\frac{{\sqrt{{{n}}}{\left({104.7}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({M}-{116.5}\right)}}}{{{63.7}}}}{<}{\frac{{\sqrt{{{n}}}{\left({112.8}-{116.5}\right)}}}{{{63.7}}}}\right]}\)</span>

\(\displaystyle={P}{\left[-{2.893}{<}{Z}{<}-{0.9073}\right]}\)</span>

\(\displaystyle={P}{\left[{Z}\leq-{0.9073}\right]}-{P}{\left[{Z}\leq-{2.8934}\right]}\)

\(\displaystyle={0.1821}-{0.0019}={0.1802}\)

So, the probability that a sample of size \(\displaystyle{n}={244}\) is randomly selected with a mean between 104.7 and 112.8 is 0.1802.