Given information-

Population mean,\(\displaystyle\mu={221.6}\)

Population standard deviation, \(\displaystyle\sigma={44.1}\)

Sample size, \(\displaystyle{n}={42}\)

Let, X be the randomly selected value is approximately normally distributed.

\(\displaystyle{X}\sim{N}{\left({221.6},{44.1}\right)}\)

The probability that a sample of size n=42 is randomly selected with a mean between 221.6 and 229.1-

\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({\frac{{{221.6}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{42}}}}}}}{<}{\frac{{{X}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{229.1}-{221.6}}}{{\frac{{44.1}}{\sqrt{{{42}}}}}}}\right)}\)</span>

\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({0}{<}{Z}{<}{1.102}\right)}\)</span>

\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={P}{\left({Z}{<}{1.102}\right)}-{P}{\left({Z}{<}{0}\right)}\)</span>

\(\displaystyle{P}{\left({221.6}{<}{X}{<}{229.1}\right)}={0.8648}-{0.5}={0.3648}\)</span>

(From excel using formula = NORM.S. DIST (0.102,TRUE))

(From excel using formula = NORM.S. DIST (0,TRUE))

Hence, the probability that a sample of size \(\displaystyle{n}={42}\) is randomly selected with a mean between 221.6 and 229.1 is 0.3648.