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.
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