- One can easily verify that the mean for a single binomial trial, where S(uccess) is scored as 1 and F(ailure) is scored as 0, is p; where p is the probability of S. Hence the mean for the binomial distribution with n trials is np.
- One can easily verify that the variance for a single binomial trial, where S is scored as 1 and F is scored as 0, is p(1-p). Hence the variance for the binomial distribution with n trials is np(1-p). This provides that the standard deviation is (np(1-p))^.5.

Example: If 10% of men are bald, what is the probability that fewer than 100
in a random sample of 818 men are bald?

Form the z-score, for which purpose it is necessary
to have the mean (*mu*) and standars deviation (*sigma*)

*mu* = np = 818 × .1 = 81.8.

*sigma* = (np(1-p))^.5 = (818 × .1 × .9)^.5 = 8.5802

z = (n-*mu*)/*sigma* = (100-81.8)/8.58 = 2.12

Since we are interested in fewer than (draw a picture), from the normal table
we find that 98.3% of the time there will be fewer than 100 bald men.

The validity of the normal approximation is illustrated if you click here.

Simulation with a binomial experiment is one way to generate a normal distribution.

N.B.: Either do all the calculations with count data as we have done here, or convert everything (including the standard deviation) to proportions.

**Applets:** The normal approximation to the binomial is illustrated by David Lane (this employs the continuity correction factor). A cruder version is also available. The classic falling ball model for the binomial convergence to the normal distribution can be seen at Davidson University or a .com (The classical model has each yellow ball going to the adjacent slot to the right or left with probability .5 when it hits a green ball, but these simulations look like more horizontal travel is possible).

**Competencies:** If n=25 and p=.2, calculate the mean, variance, and standard deviation of the binomial distribution.

If n=200 and p = .67, estimate the probability that the number of successes is greater than 140.