The probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9 is,

\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{226.9}-{200}}}{{{\left({\frac{{{31.9}}}{{\sqrt{{{11}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{\frac{{{26.9}}}{{{9.6182}}}}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{2.797}\right)}\)</span>

The probability of z less than 2.797 can be obtained using the excel formula “=NORM.S.DIST(2.797,TRUE)”. The probability value is 0.9974.

The required probability value is,

\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({z}{<}{2.797}\right)}={0.9974}\)</span>

Thus, the probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9 is 0.9974.

\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({\frac{{{M}-\mu}}{{{\left({\frac{{\sigma}}{{\sqrt{{{n}}}}}}\right)}}}}{<}{\frac{{{226.9}-{200}}}{{{\left({\frac{{{31.9}}}{{\sqrt{{{11}}}}}}\right)}}}}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{\frac{{{26.9}}}{{{9.6182}}}}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{2.797}\right)}\)</span>

The probability of z less than 2.797 can be obtained using the excel formula “=NORM.S.DIST(2.797,TRUE)”. The probability value is 0.9974.

The required probability value is,

\(\displaystyle{P}{\left({M}{<}{226.9}\right)}={P}{\left({z}{<}{2.797}\right)}={0.9974}\)</span>

Thus, the probability that a sample of size \(\displaystyle{n}={11}\) is randomly selected with a mean less than 226.9 is 0.9974.