Question

about 4% of the population has a particular genetic mutation. 800 people are randomly selected. find the standard deviation for the number of people with the genetic mutation in such groups of 800. round asnwer to two decimal places.

Explanation:

Step1: Identify the distribution type

This is a binomial distribution problem. Let $n = 800$ (number of trials/sample size) and $p=0.04$ (probability of success).

Step2: Recall the formula for standard - deviation of 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 $n = 800$ and $p = 0.04$ into the formula. First, calculate $1-p=1 - 0.04=0.96$. Then $np(1 - p)=800\times0.04\times0.96 = 30.72$.

Step4: Calculate the standard deviation

$\sigma=\sqrt{30.72}\approx5.54$.

Answer:

$5.54$