Question

About 5% of the population has a particular genetic mutation. 1000 people are randomly selected.

Answer 1

Answer:

$\simeq 6.89202$

Step-by-step explanation:

Let, X = Number of people with the genetic mutation in such groups of 1000

Here, we get a random variable X with binomial distribution with parameters

n = 1000 and p = 5 % = 0.05

So,

X $\sim$ bin ( 1000 , 0.05)

so V(X) or Variance(X) = $n \times p \times (1 -p)$

                                     = 47.5

So, SD(X) = Standard Deviation(X) = $\sqrt (47.5)$

                                                        $\simeq 6.89202$