Wednesday, July 24, 2013

Normal Approximation to the Binomial Distribution]

Sppose that the probability a new vaccine will protect adults from a certain disease is 0.8. The vaccine is given to 200 adults. What is the probability that more than 170 of those adults will be protected by the vaccine?

This question is in the category of a binomial experiment with n trials equal to 200 and the probability of success p equal to 0.8, and the number of successes r equal to 170. There is a formula for the binomial distribution which can be used to calculate the probability. But this would be very time consuming and easy to make some errors in calculation. But there is a simpler way to solve this problem, and that is by using the normal distribution to approximate the binomial distribution. However, certain conditions must be present in order to use this method.

First, we must consider the binomial distribution with n = number of trials, p = probability of success on a single trial, q = 1 - p = probability of failure on a single trial, and r = number of successes. Then if np > 5 and nq > 5, then r has a binomial distribution which you can approximate with the normal distribution. The mean is approximated by np and the standard deviation is approximated by the square root of npq. This approximation becomes more accurate as the sample size n increases.

Now we can put this all together with an example. Suppose the owner of a hotel needs to install new air conditioner units to 25 of the rooms. From past experience with a noted brand, he knows that the air conditioner unit is guaranteed for 5 years, and the probability that it will last 10 years is 0.35. What is the probability that 12 or more of the units will last more than 10 years?

In this problem, n = 25, p = 0.35, q = 0.65. We want the probability that r is greater than 10. We can use the normal approximation to the binomial since np = 8.75, which is greater than 5. Also, nq = 16.25, which is greater than 5. The mean is 8.75 and the standard deviation is square root of npq, which is 2.38. Using the normal distribution, we calculate the z to be (9.5 - 8.75)/2.38 = 0.32. Note that we use 9.5 instead of 10 because of a continuity correction, which converts r to a continuous normal random variable x by subtracting 0.5, since r in this case is the left-point of the interval. Now we find the probability that z > 0.32, which is 0.3745. This value is found by using a table for the areas of a standard normal distribution.

This guide should help assist students having difficulties understanding the normal approximation to the binomial.

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