Step 1

Solution:

Let X be the value.

From the given information, X follows normal distribution with mean \(\displaystyle\mu={182.5}\) and a standard deviation \(\displaystyle\sigma={49.4}\). The sample size is 15.

Step 2

The probability that a sample of size \(\displaystyle{n}={15}\) is randomly selected with a mean greater than 169.7 is

\(\displaystyle{P}{\left({M}{>}{169.7}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{169.7}-{182.5}}}{{\frac{{49.4}}{\sqrt{{{15}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{\sqrt{{{15}}}{\left(-{12.8}\right)}}}{{{49.4}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{1.003}\right)}\)

\(\displaystyle={1}-{0.1579}={0.8421}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{1.003},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)

Solution:

Let X be the value.

From the given information, X follows normal distribution with mean \(\displaystyle\mu={182.5}\) and a standard deviation \(\displaystyle\sigma={49.4}\). The sample size is 15.

Step 2

The probability that a sample of size \(\displaystyle{n}={15}\) is randomly selected with a mean greater than 169.7 is

\(\displaystyle{P}{\left({M}{>}{169.7}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{169.7}-{182.5}}}{{\frac{{49.4}}{\sqrt{{{15}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{\sqrt{{{15}}}{\left(-{12.8}\right)}}}{{{49.4}}}}\right)}\)

\(\displaystyle={P}{\left({Z}\succ{1.003}\right)}\)

\(\displaystyle={1}-{0.1579}={0.8421}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{U}{\sin{{g}}}\ {t}{h}{e}\ {e}{x}{c}{e}{l}\ {f}{u}{n}{c}{t}{i}{o}{n}\backslash={N}{O}{R}{M}.{D}{I}{S}{T}{\left(-{1.003},{0},{1},{T}{R}{U}{E}\right)}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)