91Ó°ÊÓ

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.

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

Hence,$P\left(x≤12\right)=0.22773$

c.

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

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

d.

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

Hence,$P\left(9≤x≤18\right)=0.98723$

e.

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

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

f.

$$\begin{gathered} \mathrm{μ}=np \\ =20×0.7 \\ =14 \\ {\mathrm{σ}}^2=npq \\ =20×0.7×0.3 \\ =4.2 \\ \mathrm{σ}=\sqrt{npq} \\ =\sqrt{4.2} \\ =2.0493 \end{gathered}$$

Hence,$\mathrm{μ}=14,{\mathrm{σ}}^2=4.2and\mathrm{σ}=2.0493$

g.

$$\begin{gathered} \mathrm{μ}+2\mathrm{σ}=14+\left(2×2.0493\right) \\ =14+4.0986 \\ =18.0986 \\ \mathrm{μ}-2\mathrm{σ}=14-\left(2×2.0493\right) \\ =14-4.0986 \\ =9.9014 \end{gathered}$$

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

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

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