Normal approximation to the binomial distribution

If the number of trials, n, is large, the binomial distribution is approximately equal to the normal distribution. (This is nice, since we really do not want to explicitly calculate binomial probabilities when n > 100.)

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.

return to index