OVERVIEW: The binomial distributionis frequently useful in situations where there are two outcomes ofinterest, such as success or failure. It is often used to modelreal-life situations, and it finds its way into many extremely usefuland important statistical applications and computations.
The binomial setting:
(1) Each observation is in one of twocategories: success or failure.
(2) A fixed number, N, of observations.
(3) Observations are independent
. (Knowing the result of oneobservation tells you nothing about the other observations.)(4) The probability of success is the same foreach observation.
If a count, X, has a binomial distribution withnumber of observations, N, and probability of success, p, then
Mean(X) = mx
= Np Standard Deviation(X) = sx
= sqrt[Np(1-p)] The probability that one will get exactly ksuccesses is N
Ck pk (1-p)N-k .
Example:
I roll a single die 60 times. If Xrepresents the number of times I roll a "3", then
m
x = 60(1/6) = 10.s
x = sqrt[60(1/6)(5/6)] =2.89.The probability that I will roll exactly ten 3'sis 60
C10 (1/6)10 (5/6)50 = .1370 = 13.7%. On the TI-83,
binompdf( 100,1/6,10) =.1370
Example (Small sample size from large population. Use of binomial distribution is appropriate.):
Assume that 30% of a population isHispanic. A random sample of size 4 is chosen from this population. If X is the number of Hispanics in the sample, then
m
x = (4)(.3) = 1.2s
x = sqrt[(4)(.3)(.7)] =0.9165Prob(X=0) =
4 C0 (.3)0 (.7)4 = 0.2401Prob(X=1) =
4 C1 (.3)1 (.7)3 = 0.4116Prob(X=2) =
4 C2 (.3)2 (.7)2 = 0.2646Prob(X=3) =
4 C3 (.3)3 (.7)1 = 0.0756Prob(X=4) =
4 C4 (.3)4 (.7)0 = 0.0081The probability that a sample would contain two orfewer Hispanics is Prob(X=0) + Prob(X=1) + Prob(X=2) = 0.2401 +0.4116 + 0.2646 = 0.9163 = 91.63%.
The TI-83 can be very useful in calculatingbinomial probabilities. For instance, the probability that thesample contains exactly 2 Hispanics is
binompdf( 4,.3,2) = .2646.The probability that the sample contains 2 orfewer Hispanics is
binomcdf( 4,.3,2) = .9163.
NOTE: It is important to understand when one hasa binomial setting, and when one doesn't have such a setting. Forinstance, consider a regular shuffled deck of 52 cards.
Setting #1: I pick a card at randomand note whether or not it is a heart. I put the card back in thedeck, thoroughly shuffle the deck, and then randomly pick a card. Again, I note whether or not it is a heart. I repeat this processten times. If X is the total number of hearts I obtained in the tentrials, then this is a binomial setting. Possible values for X are0,1,2,3,4,5,6,7,8,9,10. [N = 10, p = 0.25, and each observation isindependent of the previous ones.]
Setting #2: Basically the same situation, exceptthat I do not put the randomly picked card back in the deck beforeeach reshuffling. After ten trials, possible values for X are0,1,2,3,4,5,6,7,8,9,10. However, this is not a binomial setting. Each observation after the first is not independent of the previousone.