Given that,

\(\displaystyle\mu={181}\),

\(\displaystyle\sigma={41.8}\)

\(\displaystyle{n}={144}\)

The probability that a single randomly selected value is between 176.5 and 183.1 is,

\(\displaystyle{P}{\left({176.5}{<}{X}{<}{183.1}\right)}={P}{\left({\frac{{{176.5}-{181}}}{{{41.8}}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{183.1}-{181}}}{{{41.8}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{0.108}{<}{z}{<}{0.050}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{0.050}\right)}-{P}{\left({z}{<}-{0.108}\right)}\)</span>

\(\displaystyle=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({0.050}\right)}\right)}\backslash-{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left(-{0.108}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left({U}{\sin{{g}}}\ {E}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n},\backslash{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({z}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}\right.}\)

\(\displaystyle={0.5199}-{0.4570}={0.0629}\)

Therefore, the required probability is, 0.0629.

\(\displaystyle\mu={181}\),

\(\displaystyle\sigma={41.8}\)

\(\displaystyle{n}={144}\)

The probability that a single randomly selected value is between 176.5 and 183.1 is,

\(\displaystyle{P}{\left({176.5}{<}{X}{<}{183.1}\right)}={P}{\left({\frac{{{176.5}-{181}}}{{{41.8}}}}{<}{\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{183.1}-{181}}}{{{41.8}}}}\right)}\)</span>

\(\displaystyle={P}{\left(-{0.108}{<}{z}{<}{0.050}\right)}\)</span>

\(\displaystyle={P}{\left({z}{<}{0.050}\right)}-{P}{\left({z}{<}-{0.108}\right)}\)</span>

\(\displaystyle=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({0.050}\right)}\right)}\backslash-{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left(-{0.108}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}{\left({b}{e}{g}\in{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}{\left({U}{\sin{{g}}}\ {E}{x}{c}{e}{l}{f}{u}{n}{c}{t}{i}{o}{n},\backslash{\left(={N}{O}{R}{M}{D}{I}{S}{T}{\left({z}\right)}\right)}{e}{n}{d}{\left\lbrace{m}{a}{t}{r}{i}{x}\right\rbrace}\right)}\right.}\)

\(\displaystyle={0.5199}-{0.4570}={0.0629}\)

Therefore, the required probability is, 0.0629.