Assume that xis a binomial random variable with n = 100

and p = 5. Use the normal probability distribution to approximate

the following probabilities:

a.$\mathit{P}\left(x\le 48\right)$

b.$\mathit{P}\mathbf{(}\mathbf{50}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{65}\mathbf{)}$

c.$\mathit{P}\mathbf{(}\mathbf{x}\mathbf{\ge }\mathbf{70}\mathbf{)}$

d.$\mathit{P}\mathbf{(}\mathbf{55}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{58}\mathbf{)}$

e.$\mathit{P}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{62}\mathbf{)}$

f.$\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{49}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{x}\mathbf{\ge }\mathbf{72}\mathbf{)}$

X is a binomialrandom variable.

$$\begin{gathered} n=100 \\ p=0.5 \end{gathered}$$

For Normal approximation

$$\begin{gathered} \mu =np \\ =100\times 0.5 \\ =50 \\ \sigma =\sqrt{npq} \\ =\sqrt{100\times 0.5\times 0.5} \\ =5 \end{gathered}$$

a.

The interval is

$$\begin{gathered} \mathrm{\mu }\pm 3\sigma =50\pm \left(3\times 5\right) \\ =50\pm 15 \\ =\left(35,65\right) \\ P\left(x\le 48\right)=P\left(x\le 48.5\right) \\ \approx P\left(z<\frac{\left(48.5\right)-\mu }{\sigma }\right) \\ =P\left(z<\frac{48.5-50}{5}\right) \\ =P\left(z<\frac{-1.5}{5}\right) \\ =P\left(z<-0.3\right) \\ =1-P\left(z<0.3\right) \\ =1-0.6179 \\ =0.3821 \end{gathered}$$

Hence,$P\left(x\le 48\right)=0.3821$ .

b.

$$\begin{gathered} P\left(50\le x\le 65\right)=P\left(49.5<x<65.5\right) \\ \approx P\left(\frac{49.5-50}{5}<z<\frac{65.5-50}{5}\right) \\ =P\left(\frac{49.5-50}{5}<z<\frac{65.5-50}{5}\right) \\ =P\left(\frac{49.5-50}{5}<z<\frac{65.5-50}{5}\right) \\ =P\left(-0.1<z<3.1\right) \\ =P\left(z<3.1\right)-P\left(-0.1<z\right) \\ =0.999-1+0.5398 \\ =0.5388 \end{gathered}$$

Hence,$P\left(50\le x\le 65\right)=0.5388$ .

c.

$$\begin{gathered} P\left(x=70\right)=P\left(69.5<x<70.5\right) \\ \approx P\left(\frac{69.5-50}{5}<z<\frac{70.5-50}{5}\right) \\ =P\left(3.9<z<4.1\right) \\ =1-1 \\ =0 \end{gathered}$$

Hence ,$P\left(x=70\right)=0$ .

d.

$$\begin{gathered} P\left(55\le x\le 58\right)=P\left(54.5\le x\le 58.5\right) \\ \approx P\left(\frac{54.5-50}{5}<z<\frac{58.5-50}{5}\right) \\ =P\left(0.9<z<1.7\right) \\ =0.9554-0.8159 \\ =0.1395 \end{gathered}$$

Hence,$P(55\le x58)=0.1395$.

e.

$$\begin{gathered} P\left(x=62\right)=P\left(61.5<x<62.5\right) \\ \approx P\left(\frac{61.5-50}{5}<z<\frac{62.5-50}{5}\right) \\ =P\left(2.3<z<2.5\right) \\ =0.9938-0.9893 \\ =0.0045 \end{gathered}$$

Hence,$P\left(x=62\right)=0.0045$.

f.

$$\begin{gathered} P\left(x\le 49\right)=P\left(x>49.5\right) \\ \approx P\left(z>\frac{49.5-50}{5}\right) \\ =P\left(z<-0.1\right) \\ =1-0.5398 \\ =0.4602 \\ P\left(x\le 72\right)=P\left(x<72.5\right) \\ \approx P\left(z<\frac{72.5-50}{5}\right) \\ =1-P\left(z<4.5\right) \\ =1-1 \\ =0 \end{gathered}$$

$P\left(x\le 49\mathrm{or}\mathrm{x}\ge 72\right)=0.4602$

Hence,$P\left(x\le 49\mathrm{or}\mathrm{x}\ge 72\right)=0.4602$ .