bit a

we are suppose to do first three sub parts

Given, a binomial probability distribution has

\(\displaystyle{p}={0.60}\ {\quad\text{and}\quad}\ {n}={200}\)

a) Mean \(\displaystyle{\left(\mu\right)}={n}{p}\)

\(\displaystyle\mu={200}\cdot{0.60}\)

\(\displaystyle\mu={120}\)

Standard deviation \(\displaystyle{\left(\sigma\right)}=\sqrt{{{n}{P}{\left({1}-{P}\right)}}}\)

\(\displaystyle\sigma=\sqrt{{{200}\cdot{0.60}\cdot{\left({1}-{0.60}\right)}}}\)

\(\displaystyle\sigma={6.9282}\)

bit b

b) Here \(\displaystyle{n}{P}={200}\cdot{0.60}\)

\(\displaystyle{n}{P}={120}{>}{5}\)

\(\displaystyle{n}{\left({1}-{P}\right)}={200}\cdot{\left({1}-{0.60}\right)}\)

\(\displaystyle={200}\cdot{0.40}\)

\(\displaystyle{n}{\left({1}-{P}\right)}={80}{>}{5}\)

both of which are more than 5

Hence normal distribution can be a good aproximation

bit c

c) Required probability is \(\displaystyle{P}{\left({100}{<}{x}{<}{110}\right)}\)

\(\displaystyle={P}{\left({100}-{0.5}{<}{x}{<}{110}+{0.5}\right)}{\left(\because\right.}\) continiity correction)

\(\displaystyle={P}{\left({99.5}{<}{x}{<}{110.5}\right)}\)

\(\displaystyle={P}{\left({\frac{{{99.5}-\mu}}{{\sigma}}}{<}{\frac{{{x}-\mu}}{{\sigma}}}{<}{\frac{{{110.5}-\mu}}{{\sigma}}}\right.}\)

\(\displaystyle={P}{\left({\frac{{{99.5}-{120}}}{{{6.9282}}}}{<}{7}{<}{\frac{{{110.5}-{120}}}{{{6.9282}}}}\right)}\)

\(\displaystyle={P}{\left(-{2.96}{<}{7}{<}-{1.37}\right)}\)

\(\displaystyle={P}{\left({0}{<}{7}{<}{2.96}\right)}-{P}{\left({0}{<}{7}{<}{1.37}\right)}{\left(\because\right.}\) by symmetry)

\(\displaystyle={0.4985}-{0.4147}\)

\(\displaystyle={0.0838}\)

\(\displaystyle\therefore{P}{\left({100}{<}{x}{<}{110}\right)}={0.0838}\)