# Individuals who have a particular gene have a 0.49 probability of contracting a certain disease....

## Question:

Individuals who have a particular gene have a 0.49 probability of contracting a certain disease. Suppose that 1,101 individuals with the gene participate in a lifetime study. What is the standard deviation of the number of people who eventually contract the disease? Round to 2 decimal places.

## Binomial Distribution:

Considering a binomial distribution, we know that the mean of the distribution is greater than its variance. Also, for a large number of trials, the binomial distribution tends to the normal distribution.

In this question, we will use the standard deviation of the binomial distribution to get the standard deviation of the number of people who eventually contract the disease.

Given that,

The probability of contracting a certain disease, {eq}p = 0.49 {/eq}

Number of individuals with the gene that participate in the lifetime study, {eq}n = 1101 {/eq}

We know that:

The standard deviation ({eq}\sigma {/eq} or {eq}Sd {/eq}) of the binomial distribution is defined as:

{eq}Sd = \sqrt{np(1-p)} {/eq}

Now,

{eq}Sd = \sqrt{1101\times 0.49\times (1-0.49)}\\ Sd = 16.58 {/eq}

The standard deviation of the number of people who eventually contract the disease is 16.58.