Question 1151501: Suppose that 25% of the people in a certain population have low hemoglobin levels. The experiment is to choose 5 people at random from this population. Let the discrete random variable X be the number of people out of 5 with low hemoglobin levels.
A. Find the probability distribution of X.
B. Find the probability that at least 2 people have low hemoglobin levels.
C. Find the probability that at most 3 people have low hemoglobin levels.
D. Find the expected number of people with low hemoglobin levels out of the 5 people.
E. Find the variance of the number of people with low hemoglobin levels out of the 5 people.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a binomial distribution with n=5 and p=0.25
probability of 0 people being low is 0.75^5=0.2373
of 1 is 5*0.25*0.75^4=0.3955
Of 2 is 5C2*0.25^2*0.75^3=10*0.0625*.75^3=0.2637
of 3 is 10*0.25^3*0.75^2=0.0879
of 4 is 5*.25^4*.75=0.0146
of 5 is 0.25^5=0.001
at least 2 is 1-P(0,1)=1-0.6328=0.3672
at most 3 is P(0,1,2,3)=0.9844
E(X)=np=1.25
V(X)=np(l-p)=1.25*0.75=0.9375
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