Step 1

It is given that a population of values distributed normally with mean 26.8 and standard deviation 33.8. The sample size is 89.

Step 2

Calculate the probability that a sample of size \(\displaystyle{n}={89}\) is randomly selected with a mean between 17.1 and 25 is as follows:

\(\displaystyle{P}{\left({17.1}{<}{X}{<}{25}\right)}={P}{\left({\frac{{{17.1}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{X}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{25}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{{17.1}-{26.8}}}{{{\frac{{{33.8}}}{{\sqrt{{{89}}}}}}}}}{<}{Z}{<}{\frac{{{25}-{26.8}}}{{{\frac{{{33.8}}}{{\sqrt{{{89}}}}}}}}}\right)}\)</span>

\(\displaystyle{P}{\left(-{\frac{{{9.7}}}{{{3.58279}}}}{<}{Z}{<}{\frac{{{1.8}}}{{{3.58279}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.707}{<}{Z}{<}-{0.502}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}-{0.502}\right)}-{P}{\left({Z}{<}-{2.707}\right)}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {E}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}{s},\backslash{P}{\left({Z}{<}-{2.707}\right)}={0.0034}\backslash{P}{\left({Z}{<}-{0.502}\right)}={0.3078}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}\)</span>

\(\displaystyle={0.3078}-{0.0034}={0.3044}\)

Therefore, the value of \(\displaystyle{P}{\left({17.1}{<}{M}{<}{25}\right)}={0.3044}\)</span>.

It is given that a population of values distributed normally with mean 26.8 and standard deviation 33.8. The sample size is 89.

Step 2

Calculate the probability that a sample of size \(\displaystyle{n}={89}\) is randomly selected with a mean between 17.1 and 25 is as follows:

\(\displaystyle{P}{\left({17.1}{<}{X}{<}{25}\right)}={P}{\left({\frac{{{17.1}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{X}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{<}{\frac{{{25}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)</span>

\(\displaystyle={P}{\left({\frac{{{17.1}-{26.8}}}{{{\frac{{{33.8}}}{{\sqrt{{{89}}}}}}}}}{<}{Z}{<}{\frac{{{25}-{26.8}}}{{{\frac{{{33.8}}}{{\sqrt{{{89}}}}}}}}}\right)}\)</span>

\(\displaystyle{P}{\left(-{\frac{{{9.7}}}{{{3.58279}}}}{<}{Z}{<}{\frac{{{1.8}}}{{{3.58279}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{2.707}{<}{Z}{<}-{0.502}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}-{0.502}\right)}-{P}{\left({Z}{<}-{2.707}\right)}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {E}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}{s},\backslash{P}{\left({Z}{<}-{2.707}\right)}={0.0034}\backslash{P}{\left({Z}{<}-{0.502}\right)}={0.3078}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}\)</span>

\(\displaystyle={0.3078}-{0.0034}={0.3044}\)

Therefore, the value of \(\displaystyle{P}{\left({17.1}{<}{M}{<}{25}\right)}={0.3044}\)</span>.