Step 1

Solution:

Let X be the value.

From the given information, X follows normal distribution with mean \(\displaystyle\mu={120.6}\) and a standard deviation \(\displaystyle\sigma={48.5}\). The sample size is 105.

Step 2

The probability that a single randomly selected value is greater than 114.9 is

\(\displaystyle{P}{\left({X}{>}{114.9}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{114.9}-{120.6}}}{{{48.5}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{0.118}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{0.118}\right)}\)</span>

\(\displaystyle={1}-{0.4530}={0.5470}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{0.118},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)

Solution:

Let X be the value.

From the given information, X follows normal distribution with mean \(\displaystyle\mu={120.6}\) and a standard deviation \(\displaystyle\sigma={48.5}\). The sample size is 105.

Step 2

The probability that a single randomly selected value is greater than 114.9 is

\(\displaystyle{P}{\left({X}{>}{114.9}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{>}{\frac{{{114.9}-{120.6}}}{{{48.5}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{0.118}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}{<}-{0.118}\right)}\)</span>

\(\displaystyle={1}-{0.4530}={0.5470}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{0.118},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)