Step 1

From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={20}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={4}\)

\(\displaystyle{X}\sim{N}{\left({20},{4}\right)}\)

Step 2

The required proportion of the population which is less than 18 can be obtained as:

\(\displaystyle{P}{\left({X}{<}{18}\right)}={P}{\left({\frac{{{x}-\mu}}{{\sigma}}}{<}{\frac{{{18}-{20}}}{{{4}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}-{0.5}\right)}={0.3085}\)</span> (Using standard normal table)

Thus, the required proportion is 0.3085.

From the provided information,

Mean \(\displaystyle{\left(\mu\right)}={20}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}={4}\)

\(\displaystyle{X}\sim{N}{\left({20},{4}\right)}\)

Step 2

The required proportion of the population which is less than 18 can be obtained as:

\(\displaystyle{P}{\left({X}{<}{18}\right)}={P}{\left({\frac{{{x}-\mu}}{{\sigma}}}{<}{\frac{{{18}-{20}}}{{{4}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}-{0.5}\right)}={0.3085}\)</span> (Using standard normal table)

Thus, the required proportion is 0.3085.