Step 1

From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={99.6}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={35.1}\)

\(\displaystyle{X}\sim{N}{\left({99.6},{35.1}\right)}\)

Sample size \(\displaystyle{\left({n}\right)}={84}\)

Step 2

The required probability that a single randomly selected value is between 98.5 and 100.7 can be obtained as:

\(\displaystyle{P}{\left({98.5}{<}{X}{<}{100.7}\right)}={P}{\left({\frac{{{98.5}-{99.6}}}{{{35.1}}}}{<}{\frac{{{x}-\mu}}{{\sigma}}}{<}{\frac{{{100.7}-{99.6}}}{{{35.1}}}}\right)}\)

\(\displaystyle={P}{\left(-{0.031}{<}{Z}{<}{0.031}\right)}\)

\(\displaystyle={P}{\left({Z}{<}{0.031}\right)}-{P}{\left({Z}{<}-{0.031}\right)}\)

\(\displaystyle={P}{\left({Z}{<}{0.031}\right)}-{\left[{1}-{P}{\left({z}{<}{0.031}\right)}\right]}\)

\(\displaystyle={2}{P}{\left({Z}{<}{0.031}\right)}-{1}\)

\(\displaystyle={2}{\left({0.5124}\right)}-{1}={0.0248}\) (Using standard normal table)

Thus, the required probability is 0.0248.