Lesson #13 The Binomial Distribution
If X follows a Binomial distribution, with parameters n and p, we use the notation X ~ B(n, p) p x (1-p) (n-x) f(x) = x = 0, 1, …, n E(X) = np Var(X) = np(1-p)
If X = # obese, then X ~ B(5,.4). P(no obese people)= P(X = 0)= f(0) = (1)(1)(.07776)=.0778 x = 0, 1, 2, 3, 4, 5
P(one obese person)= P(X = 1)= f(1) = (5)(.4)(.1296)=.2592 P(two obese people)= P(X = 2)= f(2) = (10)(.16)(.216)=.3456
f(3) = (10)(.064)(.36)=.2304 f(4) = (5)(.0256)(.6)=.0768 f(5) = (1)(.01024)(1)=.0102
x012345x f(x) F(x) P(no more than 2 obese) = P(X < 2)= F(2) =.6826 P(at least 4 obese) = 1 - P(X < 3)= 1 - F(3) = P(X > 4) = =.0870
x012345x f(x) F(x) P( 2 to 3, inclusive, obese) = P(2 < X < 3) = F(3) - F(1) = P(X < 3) = = P(X < 1) E(X) = (5)(.4) = 2
P(2 < X < 3)
P(2 < X < 3)
P(2 < X < 3) = P(X < 3)
P(2 < X < 3) = P(X < 3) P(X < 1)
If X = # who passed, X ~ B(10,.9) Let Y = # who did not pass, Y ~ B(10,.1) X + Y = 10, so Y = 10 - X E(X) = (10)(.9) = 9
P(at least 7 passed) = P(X > 7) X Y
P(at least 7 passed) = P(X > 7) X Y
P(at least 7 passed) = P(X > 7) X Y = P(Y < 3)= F(3)=.9872
P(at most 4 passed) = P(X < 4) X Y
P(at most 4 passed) = P(X < 4) X Y = P(Y > 6)= 1 - P(Y < 5) = 1 - F(5)= =.0001