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The binomial distribution applies to binary random processes. Democrat or Republican. Win or lose. Success or failure. Heads or tails.
Concerns repeated, independent trials, each with the same probability of success. If winning one game means that you are more likely to win the next game, it is not binomial, because they are not independent. If for the Orioles the probability of beating the Yankees is different from the probability of beating the Royals, then you would not describe a game against the Yankees followed by two games against the Royals as a binomial setting.
Suppose that Kobe is an 80 percent free throw shooter and he takes 15 free throws, and the results of the free throws are independent of one another. What is the average number of free throws that he makes? What is the standard deviation? What is the probability that Kobe makes exactly 12? What is the probability that he makes no more than 12?
The average number he makes is 80 percent of 15, or (0.8)(15) = 12. If he took 12 free throws, the average would be (0.8)(12) = 9.6 (!)
The standard deviation of the number that he makes in 15 attempts is sqrt [(0.8)(1 - 0.8)(15)] = sqrt(2.4) = 1.55
The probability of making exactly 12 can be found on the calculator using distributions binompdf(15,0.8,12) = .25
The probability of making no more than 12 can be found using binomcdf(15,0.8,12) = .60
For a binomial process with a probability of success p and n trials, we have:
mean = np
standard deviation = sqrt[np(1-p)]
probability of exactly k successes = binompdf(n,p,k)
probability of no more than k successes = binomcdf(n,p,k)
Consider one trial. Let X be the number of successes. Then we have
| X | P(X) | X-mean | P(X)(X-mean)2 |
|---|---|---|---|
| 0 | 1-p | 0-p | (1-p)p2 |
| 1 | p | 1-p | (p)(1-p)2 |
mean = 0(1-p) + 1(p) = p
variance = (1-p)p2 + (p)(1-p)2 = (1-p)p(p+1-p) = p(1-p)
standard deviation is the square root of the variance
What about two trials? The number of successes in two trials is W. W = X + Y, where X and Y are each a binomial trial.
mean of W = mean of X + mean of Y = 2p
standard deviation of W = sqrt(variance of X + variance of Y) = sqrt[2p(1-p)]
If you keep adding more trials, you get mean = np, standard deviation = sqrt[np(1-p)]
Where does binompdf() come from? The probability of getting k successes in n trials is pk(1-p)(n-k) times the binomial coefficient. The binomial coefficient is (n!)/[k!(n-k)!].
if n = 15, k = 12, and p = .8, then the binomial coefficient is [(15)(14)(13)]/[(3)(2)(1)] = 455. Multiply this by (.8)12(.2)3 to get .25
To do binomcdf by hand would be very tedious
Binomial processes have two possible outcomes, success or failure. There are n independent trials, each with the same probability of success, p.
mean = np
standard deviation = sqrt[np(1-p)]
probability of exactly k successes = binompdf(n,p,k)
probablity of no more than k successes = binomcdf(n,p,k)
To do binompdf by hand, use the binomial coefficient times pk(1-p)(n-k)
If you are curious to understand more about the binomial coefficient, see Pascal's Triangle.