One out of every 5000 individuals in a population carries

Step 1 of 6

Here one out of every 5,000 individual in a population carries a certain defective gene

So, p = 1/5000

         = 0.0002

The sample size is 1000

So, n = 1000

Let X be the number of individuals who carry the gene

Here \(\mathrm{X}\sim\mathrm{B}(1000,\ 0.0002)\)

Since ‘n’ is large and ‘p’ is small

The Binomial distribution will be converted to the Poisson distribution

The pmf of poisson distribution is \(\mathrm{P}(\mathrm{X})=\frac{e^{-\lambda}\lambda^x}{x!},\mathrm{\ X}=0,\ 1,\ 2,\ldots\ldots\ldots\ldots\) 

So,

\(\begin{aligned} \lambda & =\mathrm{np} \\ & =1000(0.0002) \\ & =0.2 \end{aligned}\)

Hence \(\mathrm{X} \sim \text { Poisson }(0.2)\)