The mean is 198.8, standard deviation is 69.2, and sample size is 147.

The probability that a sample of size \(\displaystyle{n}={147}\) is randomly selected with a mean between 184 and 205.1 is,

\(\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({\frac{{{184}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}{<}{\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{205.1}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{-{14.8}}}{{{5.70775}}}}{<}{z}{<}{\frac{{{6.3}}}{{{5.7075}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.593}{<}{z}{<}{1.104}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}\)</span>

The probability of z less than 1.104 can be obtained using the excel formula “=NORM.S.DIST(1.104,TRUE)”. The probability value is 0.8652.

The probability of z less than –2.593 can be obtained using the excel formula “=NORM.S.DIST(–2.593,TRUE)”. The probability value is 0.0048.

The required probability value is,

\(\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}\)</span>

\(\displaystyle={0.8652}−{0.0048}={0.8604}\)

Thus, the probability that a sample of size \(\displaystyle{n}={147}\) is randomly selected with a mean between 184 and 205.1 is 0.8604.

The probability that a sample of size \(\displaystyle{n}={147}\) is randomly selected with a mean between 184 and 205.1 is,

\(\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({\frac{{{184}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}{<}{\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{205.1}-{198.8}}}{{{\left({\frac{{{69.2}}}{{\sqrt{{{147}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{-{14.8}}}{{{5.70775}}}}{<}{z}{<}{\frac{{{6.3}}}{{{5.7075}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.593}{<}{z}{<}{1.104}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}\)</span>

The probability of z less than 1.104 can be obtained using the excel formula “=NORM.S.DIST(1.104,TRUE)”. The probability value is 0.8652.

The probability of z less than –2.593 can be obtained using the excel formula “=NORM.S.DIST(–2.593,TRUE)”. The probability value is 0.0048.

The required probability value is,

\(\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}={P}{\left({z}{<}{1.104}\right)}-{P}{\left({z}{<}-{2.593}\right)}\)</span>

\(\displaystyle={0.8652}−{0.0048}={0.8604}\)

Thus, the probability that a sample of size \(\displaystyle{n}={147}\) is randomly selected with a mean between 184 and 205.1 is 0.8604.