Assume a binomial probability distribution has p=.60\ and\ n=200. a. what

Wotzdorfg 2021-09-17 Answered
Assume a binomial probability distribution has \(\displaystyle{p}={.60}\ {\quad\text{and}\quad}\ {n}={200}\).
a. what are the mean and standard deviation?
b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain.
c. what is the probability of 100 to 110 successes?

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Expert Answer

mhalmantus
Answered 2021-09-18 Author has 19233 answers

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)
\(\displaystyle={0.4985}-{0.4147}\)
\(\displaystyle={0.0838}\)
\(\displaystyle\therefore{P}{\left({100}{<}{x}{<}{110}\right)}={0.0838}\)

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