Finding the Mean, Variance, and Standard Deviation of a Binomial Distribution About 8% of the population has a particular genetic mutation. Suppose 1000 people are randomly selected and the number with the particular genetic mutation are counted. 1) Find the mean number of people who will have the genetic mutation in such groups of 1000 people. 2) Find the variance of the number of people who will have the genetic mutation in such groups of 1000 people. Round answer to 2 decimal places. 3) Find the standard deviation of number of people who will have the genetic mutation in such groups of 1000 people. Round answer to 2 decimal places.
Finding the Mean, Variance, and Standard Deviation of a Binomial Distribution
About 8% of the population has a particular genetic mutation. Suppose 1000 people are randomly selected and the number with the particular genetic mutation are counted.
1) Find the mean number of people who will have the genetic mutation in such groups of 1000 people.
2) Find the variance of the number of people who will have the genetic mutation in such groups of 1000 people. Round answer to 2 decimal places.
3) Find the standard deviation of number of people who will have the genetic mutation in such groups of 1000 people. Round answer to 2 decimal places.
Solution :
Given : n = 1000
p = 0.08
q = 1- p = 0.92
For Binomial distribution ,
Mean = np
Variance = npq
Standard deviation =
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