# Let x be a binomial random variable with n=20 and p=42/100. Calculate the mean, variance and standard deviation of the random variable x.

emancipezN 2021-08-04
Let x be a binomial random variable with $$\displaystyle{n}={20}$$ and $$\displaystyle{p}=\frac{{42}}{{100}}.$$
a) Calculate the mean, variance and standard deviation of the random variable x.
b) Use the results of part a to calculate the intervals $$\displaystyle\mu\pm\sigma,\ \mu\pm{2}\sigma,$$ and $$\displaystyle\mu\pm{3}\sigma$$. Find the probability that an observation will fall into each of these intervals.
c) Are the results of part b consistent with Tchebysheff’s Theorem? With the Empirical Rule? Why or why not?

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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.