GretaGarbo (01-20-2013)
We can use the normal distribution as a close approximation to the binomial distribution whenever n*p >= 5 and nq >= 5.
The mean of the binomial distribution is n*p, where n is the sample size and p is the probability of "success" or the probability of the event we're interested in.
The standard deviation of the binomial distribution is the square root of n*p*q, where q = 1-p.
Transforming the information given in the problem into a z score requires the formula:
z = (x - np) / sqrt(npq)
We then use this formula in the same way we would solve a problem using the normal distribution tables (z tables)
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Sample Problem
Let's say that in a town of 20,000 people, 44% support an upcoming referendum vote. If you conducting a pre-vote poll of the eligible voters in the town and surveyed 100 people, what is the probability that the survey will show that the referendum will pass?
In order for a referendum to "pass," it requires a majority vote, or 51%. So, this is the same as asking "what is the probability of getting a sample survey of 100 that shows support for the referendum."
z = (x - np) / sqrt(npq)
x = 51
np = 100*.44 = 44
npq = 100*.44*.56
z = (51 - 44) / sqrt(24.64)
z = 7 / 4.96
z = 1.41
Now, we determine the proportion of the area under the normal curve above z = 1.41 and we find that it is 0.079
So, there is a 7.9% chance of getting 51 out of 100 to indicate "in favor" of the referendum when in fact 44% of the population favors it.
GretaGarbo (01-20-2013)
Thanks for the example that really helps.
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