Question

about 10% of the population has a particular genetic mutation. 600 people are randomly selected. find the standard deviation for the number of people with the genetic mutation in such groups of 600. round answer to two decimal places.

Explanation:

Step1: Identify the binomial distribution parameters

This is a binomial distribution problem. Let $n = 600$ (number of trials/sample size) and $p=0.1$ (probability of success - having the mutation).

Step2: Recall the formula for the standard - deviation of a binomial distribution

The formula for the standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$.

Step3: Substitute the values of $n$ and $p$ into the formula

Substitute $n = 600$ and $p = 0.1$ into the formula:
\[

$$\begin{align*} \sigma&=\sqrt{600\times0.1\times(1 - 0.1)}\\ &=\sqrt{600\times0.1\times0.9}\\ &=\sqrt{54} \end{align*}$$

\]

Step4: Calculate the value and round

$\sqrt{54}\approx7.35$

Answer:

$7.35$