Step 1

Given,

Mean \(= 49\)

Standard deviation \(= 79.5\)

Sample size \(= 84\)

Step 2

Consider,

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

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

\(\displaystyle={P}{\left({Z}{>}{0.294}\right)}\)

\(\displaystyle={1}-{P}{\left({Z}\leq{0.294}\right)}\)

\(\displaystyle={1}-{0.6156}={0.3844}\) (From the standard normal table)

The probability that a single randomly selected value is greater than 72.4 is, 0.3844.