In: Statistics and Probability
QUESTION PART A: About 1% of the population has a particular
genetic mutation. 900 people are randomly selected.
Find the mean for the number of people with the genetic mutation in
such groups of 900.
QUESTION PART B: About 4% of the population has a particular
genetic mutation. 400 people are randomly selected.
Find the standard deviation for the number of people with the
genetic mutation in such groups of 400. Round your answer to three
decimal places
Part A)
About 1% of the population has a particular genetic mutation. This is same as the probability that a randomly selected person from this population has the genetic mutation is 0.01
Let X be the number of with the genetic mutation in such groups of 900. We can say that X has a Binomial distribution with parameters, number of trials (number of persons randomly selected) n=900 and success probability (the probability that a randomly selected person from this population has the genetic mutation) p=0.01
Using the formula for Binomial expectation, we can say that the expectation of X is
ans: the mean for the number of people with the genetic mutation in such groups of 900 is 9
Part B)
About 4% of the population has a particular genetic mutation. This is same as the probability that a randomly selected person from this population has the genetic mutation is 0.04
Let X be the number of with the genetic mutation in such groups of 400. We can say that X has a Binomial distribution with parameters, number of trials (number of persons randomly selected) n=400 and success probability (the probability that a randomly selected person from this population has the genetic mutation) p=0.04
Using the formula for Binomial standard deviation, we can say that the standard deviation of X is
ans: the standard deviation for the number of people with the genetic mutation in such groups of 400 is 3.919