Download Normal Approximation to the Binomial Distribution

Normal Approximation to the Binomial Distribution
In order to quickly approximate binomial distribution questions involving large number of
trials we can use the normal distribution. If we have a binomial distribution with n trials
and probability of success p, then we knowp
it is approximated by a normal distribution with
mean µ = np and standard deviation σ = np(1 − p).
No matter what the distribution looks like, we will answer these questions by interpreting the probability as the area under the distribution curve between two values. For
example, suppose that the distribution of donut weights has mean µ = 5.5 and looks like
this:
Binomial Distribution
Normal Distribution
µ
µ
We know how to visualize binomial distribution probabilities as the area of the bars on the
binomial graph, and we know how to visualize normal distribution probabilities as the area
under the curve on the normal graph. The only question is how to translate from one to the
other. To do that we have to use what is called “continuity correction”.
Let’s consider the binomial probability P (3 ≤ r ≤ 5) shaded on the graph below.
3 4 5
Notice that the numbers 3 and 5 are in the middle of their respective bars, but we want
to include the area of the whole bars, so we need to approximate the area from 2.5 to 5.5.
That is the continuity correction, you subtract 1/2 from the left end of the interval and add
1/2 to the right end of the interval. The normal distribution probability we now need to find
is:
P (2.5 < x < 5.5)
We compute that as we have before – find the z score interval using the mean and standard
deviation from above, and use the built-in calculator.
The key is visualizing the probabilities as areas – draw the bars (or at least a couple
bars at the ends of the interval), and that should make it easy to decide whether to add or
subtract 1/2 for the continuity correction.
Here’s a complete worked example:
Example: Suppose that 52% of the population of U.S. voters favors a particular presidential
candidate. If a random sample of 130 voters is chosen, approximate the probability that at
most 60 favor this candidate. Use the normal approximation to the binomial with a correction
for continuity.
Solution: We are given a binomial distribution problem with n = 130 and p = 0.52. From
this we quickly compute
µ = 130 · 0.52 = 67.6
and
σ=
√
130 · 0.52 · 0.48 ≈ 5.696
The binomial probability we are interested in is “at most 60,” which means “60 or less,”
so the probability we want is
P (r ≤ 60)
Do we add or subtract 1/2 for the normal approximation? Let’s draw a few bars . . .
...
...
60
We want to include the bar for r = 60 and all the bars to the left. To include the whole bar
r = 60 we need to add 1/2 to get the normal probablility
P (x < 60.5)
Using the mean and standard deviation above, we translate this to the standard normal
probability
P (z < −1.246)
Using the built in calculator, we find that this probability is 0.1063822.
There is about a 10.6% chance that at most 60 of the voters favor the candidate.