Question

A random variable of size 30 bin(20,0.6) ie binomial distribution.Find (x<12.2) (x>12.2(

Answer 1

Given:

$\begin{aligned} &X \sim \operatorname{Binomial}(20,0.6) \\ &N=20 \\ &p=0.06 \end{aligned}$

Checking for normality condition

$\begin{aligned} &N p=20 \times 0.6=12 \\ &N(1-p)=20 \times(1-0.6)=8 \end{aligned}$

Since both $\mathrm{Np}$ and $\mathrm{N}(1-\mathrm{p})$ is greater than 5, the given binomial distribution can be approximated to the normal distribution.

The normal approximation for the sample of size $\mathrm{n}=30$ is given as

$\begin{aligned} \bar{X} & \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \\ \mu_{X} &=N p \\ &=20 \times 0.6 \\ &=12 \\ \sigma_{X} &=\frac{\sqrt{N p(1-p)}}{\sqrt{n}} \\ &=\frac{\sqrt{20(0.6)(1-0.6)}}{\sqrt{30}} \\ &=0.4 \end{aligned}$

i)

$\begin{aligned} & \text{The required probability is calculated as}\\ & P(\bar{X}<12.2) =P\left(\frac{\bar{X}-\mu}{\sigma}-\frac{12.2-\mu}{\sigma}\right) \\ &=P\left(Z<\frac{12.2-12}{0.4}\right) \\ &=P(Z<0.5) \\ &=0.69146(\text { Using the standard normal table })\\ & \text{Thus, the required probability is 0.69146.} \end{aligned}$

ii)

The required probability is calculated as

$\begin{aligned} P(\bar{X}>12.2) &=1-P(\bar{X}<12.2) \\ &=1-P\left(\frac{\bar{X}-\mu}{\sigma}-\frac{12.2-\mu}{\sigma}\right) \\ &=1-P\left(Z<\frac{12.2-12}{0.4}\right) \\ &=1-P(Z<\frac{1}{2}) \\ &=1-0.69146(\text { Using the standard normal table }) \\ &=0.30854 \end{aligned}$

Thus, the required probability is $0.30854 .$