91Ó°ÊÓ

x is a binomial random variable withp= 0.4 and n = 25

Yes. If,$npâ‰\yen 5andn(1-p)â‰\yen 5$ , the x would be approximate the normal distribution.

So,

$$\begin{gathered} np=25×0.4 \\ =10 \end{gathered}$$

And,

$$\begin{gathered} n(1-p)=25×\left(1-0.4\right) \\ =25×0.6 \\ =15 \end{gathered}$$

So, here$npâ‰\yen 5andn(1-p)â‰\yen 5$

Therefore, x would be approximated by normal distribution.

b.

Assuming x is the normal distribution, then the mean and variance of approximating normal distribution is,

$$\begin{gathered} Mean=np \\ =25×0.4 \\ =10 \\ Variance=np\left(1-p\right) \\ =25×0.4×\left(1-0.4\right) \\ =25×0.4×0.6 \\ =6 \end{gathered}$$

Therefore, the mean of x is 10 and the variance of x is 6.

$$\begin{gathered} P\left(xâ‰\yen 9\right)=1-p\left(x<9\right) \\ =1-0.425\left[Bybinomialtable\right] \\ =0.575 \end{gathered}$$

Therefore, $P\left(xâ‰\yen 9\right)=0.575$

If $x~N\left(0.4,25\right)$

Then,

$$\begin{gathered} \mathrm{μ}=0.4×25 \\ =10 \end{gathered}$$

And,

$\begin{array}{l}\mathrm{σ}=\sqrt{np(\left(1-p\right)}\\ =\sqrt{25×0.4×(1-0.4)}\\ =\sqrt{25×0.4×0.6}\\ =2.4495\end{array}$

Then,

role="math" localid="1660280899974" $\begin{array}{l}z=\frac{x-\mathrm{μ}}{\mathrm{σ}}\\ =\frac{9-10}{2.4495}\\ =-0.4082\end{array}$

$$\begin{gathered} P(zâ‰\yen -0.4082)=1-P(z<-0.4082) \\ =1-(1-P\left(z<0.4082\right)) \\ =1-(1-0.6591) \\ =0.6591 \end{gathered}$$

$$\begin{gathered} P(xâ‰\yen 90)=P\left(zâ‰\yen -0.4082\right) \\ =0.6591 \end{gathered}$$

Therefore, by the normal approximation,$P(xâ‰\yen 9)=0.6591$