Normal Approximation to the Binomial

Some variables are continuous—there is no limit to the number of times you could divide their intervals into still smaller ones, although you may round them off for convenience. Examples include age, height, and cholesterol level. Other variables are discrete, or made of whole units with no values between them. Some discrete variables are the number of children in a family, the sizes of televisions available for purchase, or the number of medals awarded at the Olympic Games.

A binomial variable can take only two values, often termed successes and failures. Examples include coin tosses that come up either heads or tails, manufactured parts that either continue working past a certain point or do not, and basketball tosses that either fall through the hoop or do not.

You discovered that the outcomes of binomial trials have a frequency distribution, just as continuous variables do. The more binomial trials there are (for example, the more coins you toss simultaneously), the more closely the sampling distribution resembles a normal curve (see Figure 1). You can take advantage of this fact and use the table of standard normal probabilities (Table 2 in "Statistics Tables") to estimate the likelihood of obtaining a given proportion of successes. You can do this by converting the test proportion to a z‐score and looking up its probability in the standard normal table.

Figure 1.As the number of trials increases, the binomial distribution approaches the normal distribution.

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The mean of the normal approximation to the binomial is

μ = nπ

and the standard deviation is equation

where n is the number of trials and π is the probability of success. The approximation will be more accurate the larger the n and the closer the proportion of successes in the population to 0.5.

Example 1

Assuming an equal chance of a new baby being a boy or a girl (that is, π = 0.5), what is the likelihood that more than 60 out of the next 100 births at a local hospital will be boys?

equation

According to Table

, a z‐score of 2 corresponds to a probability of 0.9772. As you can see in Figure 2, there is a 0.9772 chance that there will be 60 percent or fewer boys, which means that the probability that there will be more than 60 percent boys is 1 – 0.9772 = 0.0228, or just over 2 percent. If the assumption that the chance of a new baby being a girl is the same as it being a boy is correct, the probability of obtaining 60 or fewer girls in the next 100 births is also 0.9772.
Figure 2.Finding a probability using a z‐score on the normal curve.

 
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