Question

About 9% of the population has a particular genetic mutation. 900 people are randomly selected. Find the standard deviation for the number of people with the genetic mutation in such groups of 900.

Answer 1

Answer:

The standard deviation is of 8.586.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a genetic mutation, or they do not. The probability of a person having the mutation is independent of any other person, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

About 9% of the population has a particular genetic mutation.

This means that $p = 0.09$

900 people are randomly selected.

This means that $n = 900$

Find the standard deviation for the number of people with the genetic mutation in such groups of 900.

$\sqrt{V(X)} = \sqrt{900*0.09*0.91} = 8.586$

The standard deviation is of 8.586.