Suppose xis a binomial random variable with n= 20 and

p= .7.

a. Find$\mathit{P}\left(x=14\right)$.

b. Find$\mathit{P}\left(x\le 12\right)$.

c. Find$\mathit{P}\left(x>12\right)$.

d. Find$\mathit{P}\left(9\le x\le 18\right)$.

e. Find$\mathit{P}\left(8<x<18\right)$.

f. Find$\mathit{\mu }$,${\mathit{\sigma }}^{\mathbf{2}}$, and$\mathit{\sigma }$.

g. What is the probability that xis in the interval$\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$?

X is a binomial random variable.

$$\begin{gathered} n=20 \\ p=0.7 \end{gathered}$$

The p.m.f of X is

$$\begin{gathered} f\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x};x=0,1,2,...,n \\ =\left(\begin{array}{c}20\\ x\end{array}\right){\left(0.7\right)}^x{\left(1-0.7\right)}^{20-x} \end{gathered}$$

Hence,$P\left(x=14\right)=0.19164$ .

b.

$$\begin{gathered} P\left(x\le 12\right)=1-P\left(x>12\right) \\ =1-\sum _{r=13}^{20}\left(\begin{array}{c}20\\ r\end{array}\right){\left(0.7\right)}^r{\left(1-0.7\right)}^{20-r} \\ =1-\left(0.16426+0.19164+0.17886+0.13042+0.0716+0.02785+0.00684+0.0008\right) \\ =1-0.77227 \\ =0.22773 \end{gathered}$$

Hence,$P\left(x\le 12\right)=0.22773$

c.

$$\begin{gathered} P\left(x>12\right)=\sum _{r=3}^{20}\left(\begin{array}{c}20\\ r\end{array}\right){\left(0.7\right)}^r{\left(1-0.7\right)}^{20-r} \\ =\left(0.16426+0.19164+0.17886+0.13042+0.0716+0.02785+0.00684+0.0008\right) \\ =0.77227 \end{gathered}$$

Hence,$P\left(x>12\right)=0.77227$

d.

$$\begin{gathered} P\left(9\le x\le 18\right)=\sum _{r=9}^{18}\left(\begin{array}{c}20\\ r\end{array}\right){\left(0.7\right)}^r{\left(1-0.7\right)}^{20-r} \\ \left(0.01201+0.03082+0.06537+0.1144+0.16426+0.19164+0.17886+0.13042+0.0716+0.02785\right) \\ =0.98723 \end{gathered}$$

Hence,$P\left(9\le x\le 18\right)=0.98723$

e.

$$\begin{gathered} P\left(8<x<18\right)=\sum _{r=9}^{17}\left(\begin{array}{c}20\\ r\end{array}\right){\left(0.7\right)}^r{\left(1-0.7\right)}^{20-r} \\ =\left(0.01201+0.03082+0.06537+0.1144+0.16426+0.19164+0.17886+0.13042+0.0716\right) \\ =0.95938 \end{gathered}$$

Hence,$P\left(8<x<18\right)=0.95938$

f.

$$\begin{gathered} \mu =np \\ =20\times 0.7 \\ =14 \\ {\sigma }^2=npq \\ =20\times 0.7\times 0.3 \\ =4.2 \\ \sigma =\sqrt{npq} \\ =\sqrt{4.2} \\ =2.0493 \end{gathered}$$

Hence,$\mu =14,{\sigma }^2=4.2and\sigma =2.0493$

g.

$$\begin{gathered} \mu +2\sigma =14+\left(2\times 2.0493\right) \\ =14+4.0986 \\ =18.0986 \\ \mu -2\sigma =14-\left(2\times 2.0493\right) \\ =14-4.0986 \\ =9.9014 \end{gathered}$$

The interval is$\left(9.9014,18.0986\right)$

$$\begin{gathered} P\left(9.9014\le x\le 18.0986\right)\approx P\left(10\le x\le 18\right) \\ P\left(10\le x\le 18\right)=\sum _{r=10}^{18}\left(\begin{array}{c}20\\ r\end{array}\right){\left(0.7\right)}^r{\left(1-0.7\right)}^{20-r} \\ =\left(0.03082+0.06537+0.1144+1.16426+0.19164+0.17886+0.0716+0.13042+0.02785\right) \\ =0.97522 \end{gathered}$$

Hence, the probability that x is in the given interval is 0.97522 .