Sanderson M. Smith

THE BINOMIAL SETTING
(It's nice when you have it!)

(1) Consider rolling a fair die 10 times and let x represent the number of 6's obtained.
If we are interested in the distribution of the random variable x, we have a perfect binomial setting.

___ ___ ___ ___ ___ ___ ___ ___ ___ ___

There are ten slots, and the probability of filling each slot is 1/6. We have N = 10 and p = 1/6.

(2) Consider a large population in which 1/6 of the people are Hispanic. We pick a random sample of 10 people from the population. Let x represent the number of Hispanics in the sample. We are interested in the distribution of the random variable x. We can again think of ten slots, as above. Since this would be a non-replacement selection process, the probability of filling the second slot would be dependent upon how the first slot was filled. We don't have independence, and hence we don't have a binomial setting. However, since the population is large, we don't lose much by assuming that the probability of filling each slot with a Hispanic individual is 1/6. In other words, we can get a reasonably good model of the distribution of x using a binomial setting with N = 10 and p = 1/6.

(3) Consider a small airline with 10-passenger planes. Assume that past research has shown that 1/6 of the people who purchase tickets for a flight don't show up for the flight. Let x represent the number of no-shows. If we make the assumption that there is a 1/6 probability that a given seat will be empty, we can model the distribution of x in a binomial setting with N = 10 and p = 1/6.

So, the following analysis of a binomial distribution with N = 10 and p = 1/6 could be used in an attempt to gain information about the random variable x in each of the scenarios described above.

The main purpose of this paper is to illustrate that the simple binomial formulas for mean, mean, and standard deviation will yield the same results that we would get by using the corresponding random variable formulas for the random variable x, where x is an element of the set {0,1,2,3,4,5,6,7,8,9,10}.

Here are the binomial calculations:

• Mean: mx = Np = (10)(1/6) = 1.666666667
• Standard deviation: sx = sqrt[Np(1-p)] = sqrt[(10)(1/6)(5/6)] = 1.178511302
• Variance: ( sx)2 = 1.388888889

Now the computations will be done using random variable formulas

• Mean: mx = sum[(xi)(pi)]
• Variance: ( sx)2 = sum[(xi- mx)2pi]

The computations in the following table were done on the TI-83. If you put the values of x in list L1, then the probabilities are easily calculated in a second list using (10 nCr L1)*(1/6)^L1*(5/6)^(10-L1) or binompdf(10,1/6,L1)

 x Probability (to 5 decimals) (xi)(pi) (to 5 decimals) (xi- mx)2pi (to 5 decimals) 0 0.16151 0.00000 0.44863 1 0.32301 0.32301 0.14356 2 0.29071 0.58142 0.03230 3 0.15505 0.46514 0.27564 4 0.05427 0.21706 0.29545 5 0.01302 0.06512 0.14371 6 0.00217 0.01302 0.04076 7 0.00025 0.00174 0.00706 8 0.00002 0.00015 0.00075 9 0.00000 0.00001 0.00004 10 0.00000 0.00000 0.00000 TOTALS 1.00000 1.66667 1.38889 This column total is 100%, as should be expected. This column total is the mean of the random variable x. Note that this is the same number obtained using the binomial mean formula. This column total is the variance of the random variable x. Note that this is the same number obtained using the binomial variance formula.

A binomial setting is extremely nice since the formulas for calculating mean, standard deviation, and variance are simple, and easy to use. The table above demonstrates that one obtains the same values from random variable formulas that are necessary to use when a binomial setting does not exist.