Normal Approximation to the Binomial
Prerequisites
Binomial
Distribution, History
of the Normal Distribution, Areas
of Normal Distributions
In the section on the history of the normal distribution,
we saw that the normal distribution can be used to approximate
the binomial distribution. This section shows how to compute
these approximations.
Lets begin with an example. Assume you have a fair
coin and wish to know the probability that you would get 8 heads
out of 10 flips. The binomial distribution has a mean of μ
= Nπ = (10)(0.5) = 5 and a variance
of σ2 = Nπ(1π)=
(10)(0.5)(0.5) = 2.5. The standard deviation is therefore 1.5811.
A total of 8 heads is (8  5)/1.5811 =1.8973 standard deviations
above the mean of the distribution. The question then is, "What
is the probability of getting a value exactly 1.8973 standard
deviations above the mean?" You may be surprised to learn
that the answer is 0: The probability of any one specific point
is 0. The problem is that the binomial distribution is a discrete
probability distribution whereas the normal distribution is a
continuous distribution.
The solution is to round off and consider any value
from 7.5 to 8.5 to represent an outcome of 8 heads. Using this
approach, we figure out the area under a normal curve from 7.5
to 8.5. The area in green in Figure 1 is an approximation of the
probability of obtaining 8 heads.
The solution is therefore to compute this area.
First we compute the area below 8.5 and then subtract the area
below 7.5.
The results of using the normal area calculator
to find the area below 8.5 are shown in Figure 2. The results
for 7.5 are shown in Figure 3.
The differences between the areas is 0.044 which
is the approximation of the binomial probability. For these parameters,
the approximation is very accurate. The demonstration in the next
section allows you to explore its accuracy with different parameters.
If you did not have the normal area calculator,
you could find the solution using a table of the standard normal
distribution (a Z table) as follows:
 Find a Z score for 7.5 using the formula Z = (7.5  5)/1.5811
= 1.58.
 Find the area below a Z of 1.58 = 0.943.
 Find a Z score for 8.5 using the formula Z = (8.5  5)/1.5811
= 2.21.
 Find the area below a Z of 2.21 = 0.987.
 Subtract the value in step 2 from the value in step 4 to
get 0.044.
The same logic applies when calculating the probability
of a range of outcomes. For example, to calculate the probability
of 8 to 10 flips, calculate the area from 7.5 to 10.5.
The accuracy of the approximation depends on
the values of N and π. A rule of thumb
is that the approximation is good if both Nπ and N(1π)
are both greater than 10.
