Step 1

Given,

Mean = 49

Standard deviation = 79.5

Sample size = 84

Step 2

Consider,

\(\displaystyle{P}{\left({M}{>}{72.4}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-{49}}}{{\frac{{79.5}}{\sqrt{{{84}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{2.70}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\leq{2.70}\right)}\)

\(\displaystyle={1}-{0.9965}={0.0035}\) (Using standard normal table)

The probability that a sample of size \(\displaystyle{n}={84}\) is randomly selected with a mean greater than 72.4 is, 0.0035.

Given,

Mean = 49

Standard deviation = 79.5

Sample size = 84

Step 2

Consider,

\(\displaystyle{P}{\left({M}{>}{72.4}\right)}={P}{\left({\frac{{{M}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{\frac{{{72.4}-{49}}}{{\frac{{79.5}}{\sqrt{{{84}}}}}}}\right)}\)

\(\displaystyle={P}{\left({Z}{>}{2.70}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\leq{2.70}\right)}\)

\(\displaystyle={1}-{0.9965}={0.0035}\) (Using standard normal table)

The probability that a sample of size \(\displaystyle{n}={84}\) is randomly selected with a mean greater than 72.4 is, 0.0035.