Suppose that in a large population the proportion of people that have a certain disease
is 0.01. Use the Poisson approximation to find the probability that at least four people will
have the disease in a group of 400 randomly selected people.
3 answers
lol are you taking stat 2507 were all in the same boat
SAME CLASS FML
X~Bin(400,0.01)
np= 4
P(x>=4)= 1-P(Y<=3)
1-[P(X=3)+P(X=2)+P(X=1)+P(X=0)]
1- ((e^-4)*(4^3))/3! +((e^-4)*(4^2))/2! + ((e^-4)*(4^1))/1! +((e^-4)*(4^0))/0!=
1- 0.4330= 0.567
np= 4
P(x>=4)= 1-P(Y<=3)
1-[P(X=3)+P(X=2)+P(X=1)+P(X=0)]
1- ((e^-4)*(4^3))/3! +((e^-4)*(4^2))/2! + ((e^-4)*(4^1))/1! +((e^-4)*(4^0))/0!=
1- 0.4330= 0.567
1 answer