Step 1

Given that

Let x be a binomial random variable with \(\displaystyle{n}={20}\) and \(\displaystyle{p}=\frac{{42}}{{100}}\)

The binomial probability distribution function is

\(\displaystyle{P}{\left[{X}={x}\right]}={P}{\left({x}\right)}\)

\(P[X=x]=\left(\begin{array}{c}a\\ b\end{array}\right)p^{x}q^{n-x};\ x=0,\ 1,\ \cdots\ ,\ n,\ 0<p<1,\ q=1-p\)

\(\displaystyle={0}.\ {o}{t}{h}{e}{r}{w}{i}{s}{e}\)

Mean of binomial distribution is np

\(\displaystyle{E}{\left({X}\right)}={n}\times\ {p}\)

\(\displaystyle{E}{\left({X}\right)}={20}\times{\frac{{{42}}}{{{100}}}}\)

\(\displaystyle{E}{\left({X}\right)}={8.4}\)

Variance of binomial distribution function is npq

\(\displaystyle{V}{a}{r}{\left({X}\right)}=\delta^{{{2}}}={n}{p}{q}\)

\(\displaystyle{V}{a}{r}{\left({X}\right)}={20}\times{\frac{{{42}}}{{{100}}}}\times{\left({1}-{\frac{{{42}}}{{{100}}}}\right)}\)

\(\displaystyle{V}{a}{r}{\left({X}\right)}={20}\times{0.42}\times{0.58}\)

\(\displaystyle{V}{a}{r}{\left({X}\right)}={4.872}\)

Standard deviation is \(\displaystyle{\left({s}{d}\right)}=\sqrt{{{V}{a}{r}{\left({X}\right)}}}\)

\(\displaystyle{s}{d}=\delta=\sqrt{{{V}{a}{r}{\left({X}\right)}}}\)

\(\displaystyle\delta=\sqrt{{{4.872}}}\)

\(\displaystyle\delta={2.2073}\)

Step 2

Use the results of part a to calculate the intervals \(\displaystyle\mu\pm\sigma,\ \mu\pm{2}\sigma,\) and \(\displaystyle\mu\pm{3}\sigma\).

The confidence interval for \(\displaystyle\mu\pm\sigma\)

\(\displaystyle{C}.{I}.={\left(\mu=\delta,\ \mu+\delta\right)}\)

\(\displaystyle{C}.{I}.={\left({8.4}-{2.2073},\ {8.4}+{2.2073}\right)}\)

\(\displaystyle{C}.{I}.={\left({6.1927},\ {10.6073}\right)}\)

The Confidence interval for \(\displaystyle\mu\pm{2}\sigma\)

\(\displaystyle{C}.{I}.={\left(\mu-{2}\delta,\ \mu+{2}\delta\right)}\)

\(\displaystyle{C}.{I}.={\left({8.4}-{2}\times{2.2073},\ {8.4}+{2}\times{2.2073}\right)}\)

\(\displaystyle{C}.{I}.={\left({3.9854},\ {12.8146}\right)}\)

The confidence interval for \(\displaystyle\mu\pm{3}\sigma\) is

\(\displaystyle{C}.{I}.={\left(\mu-{3}\delta,\ \mu+{3}\delta\right)}\)

\(\displaystyle{C}.{I}.={\left({8.4}-{3}\times{2.2073},\ {8.4}+{3}\times{2.2073}\right)}\)

\(\displaystyle{C}.{I}.={\left({1.7781},\ {15.0219}\right)}\)

Find the probability that an observation will fall into each of these intervals.

\(\begin{array}{|c|c|}\hline X & P(X\leq\ x) \\ \hline 0 & 0.000018559 \\ \hline 1 & 0.0002687 \\ \hline 2 & 0.001849 \\ \hline 3 & 0.004 \\ \hline 4 & 0.014 \\ \hline 5 & 0.036 \\ \hline 6 & 0.075 \\ \hline 7 & 0.122 \\ \hline 8 & 0.162 \\ \hline 9 & 0.177 \\ \hline 10 & 0.159 \\ \hline 11 & 0.119 \\ \hline 12 & 0.073 \\ \hline 13 & 0.037 \\ \hline 14 & 0.015 \\ \hline 15 & 0.005 \\ \hline 16 & 0.001 \\ \hline 17 & 0.000 \\ \hline 18 & 0.000 \\ \hline 19 & 0.000 \\ \hline 20 & 0.000 \\ \hline & \\ \hline Sum & 2.1261 \\ \hline \end{array}\)

\(\displaystyle{P}{\left({X}\leq\ {x}\right)}={\sum_{{{i}={1}}}^{{{20}}}}{P}{\left[{X}={x}\right]}\)

\(\displaystyle{P}{\left({X}\leq\ {x}\right)}={2.1261}\)

This is belong to the all the confidence interval that have written in above step.

From the chebysheff's theorem

\(\displaystyle{79.48}\%\) of the data value will lie within 2.2073 standard deviation from mean.