About 10% of the population has a particular genetic mutation. 500 people are randomly selected.
Find the standard deviation for the number of people with the genetic mutation in such groups of 500
To find the standard deviation for the number of people with the genetic mutation in groups of 500, we will assume that the presence of this genetic mutation in each person is an independent event and use the binomial distribution to model this situation. The binomial distribution is appropriate here because we have a fixed number of trials (n = 500), two possible outcomes for each trial (mutation or no mutation), and the probability of success (having the mutation) is the same for each trial.
Given that about 10% of the population has the genetic mutation, we can express this as a probability of success (p) of 0.10. The probability of failure (q), which is not having the mutation, is 1 - p = 0.90.
The formula for the standard deviation (σ) of a binomial distribution is:
σ = sqrt(n * p * q)
Where:
- n is the number of trials (500 people in our case)
- p is the probability of success (0.10)
- q is the probability of failure (0.90)
Now, we can calculate the standard deviation:
σ = sqrt(500 * 0.10 * 0.90)
σ = sqrt(50 * 0.90)
σ = sqrt(45)
σ ≈ 6.71
Therefore, the standard deviation for the number of people with the genetic mutation in groups of 500 is approximately 6.71.