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.

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.