# Lesson 28: Approximations for Discrete Distributions

Lesson 28: Approximations for Discrete Distributions

## Overview

In the previous lesson, we explored the Central Limit Theorem, which states that if $$X_1, X_2, \dots , X_n$$ is a random sample of "sufficient" size n from a population whose mean is $$\mu$$ and standard deviation is $$\sigma$$, then:

$$Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{\sum\limits_{i=1}^n X_i-n\mu}{\sqrt{n}\sigma} \stackrel {d}{\longrightarrow} N(0,1)$$

In that lesson, all of the examples concerned continuous random variables. In this lesson, our focus will be on applying the Central Limit Theorem to discrete random variables. In particular, we will investigate how to use the normal distribution to approximate binomial probabilities and Poisson probabilities.

## Objectives

Upon completion of this lesson, you should be able to:

• To learn how to use the normal distribution to approximate binomial probabilities.
• To learn how to use the normal distribution to approximate Poisson probabilities.
• To be able to apply the methods learned in this lesson to new problems.

# 28.1 - Normal Approximation to Binomial

28.1 - Normal Approximation to Binomial

As the title of this page suggests, we will now focus on using the normal distribution to approximate binomial probabilities. The Central Limit Theorem is the tool that allows us to do so. As usual, we'll use an example to motivate the material.

## Example 28-1

Let $$X_i$$ denote whether or not a randomly selected individual approves of the job the President is doing. More specifically:

• Let $$X_i=1$$, if the person approves of the job the President is doing, with probability $$p$$
• Let $$X_i=0$$, if the person does not approve of the job the President is doing with probability $$1-p$$

Then, recall that $$X_i$$ is a Bernoulli random variable with mean:

$$\mu=E(X)=(0)(1-p)+(1)(p)=p$$

and variance:

$$\sigma^2=Var(X)=E[(X-p)^2]=(0-p)^2(1-p)+(1-p)^2(p)=p(1-p)[p+1-p]=p(1-p)$$

Now, take a random sample of $$n$$ people, and let:

$$Y=X_1+X_2+\ldots+X_n$$

Then $$Y$$ is a binomial($$n, p$$) random variable, $$y=0, 1, 2, \ldots, n$$, with mean:

$$\mu=np$$

and variance:

$$\sigma^2=np(1-p)$$

Now, let $$n=10$$ and $$p=\frac{1}{2}$$, so that $$Y$$ is binomial($$10, \frac{1}{2}$$). What is the probability that exactly five people approve of the job the President is doing?

#### Solution

There is really nothing new here. We can calculate the exact probability using the binomial table in the back of the book with $$n=10$$ and $$p=\frac{1}{2}$$. Doing so, we get:

\begin{align} P(Y=5)&= P(Y \leq 5)-P(Y \leq 4)\\ &= 0.6230-0.3770\\ &= 0.2460\\ \end{align}

That is, there is a 24.6% chance that exactly five of the ten people selected approve of the job the President is doing.

Note, however, that $$Y$$ in the above example is defined as a sum of independent, identically distributed random variables. Therefore, as long as $$n$$ is sufficiently large, we can use the Central Limit Theorem to calculate probabilities for $$Y$$. Specifically, the Central Limit Theorem tells us that:

$$Z=\dfrac{Y-np}{\sqrt{np(1-p)}}\stackrel {d}{\longrightarrow} N(0,1)$$.

Let's use the normal distribution then to approximate some probabilities for $$Y$$. Again, what is the probability that exactly five people approve of the job the President is doing?

#### Solution

First, recognize in our case that the mean is:

$$\mu=np=10\left(\dfrac{1}{2}\right)=5$$

and the variance is:

$$\sigma^2=np(1-p)=10\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=2.5$$

Now, if we look at a graph of the binomial distribution with the rectangle corresponding to $$Y=5$$ shaded in red:

we should see that we would benefit from making some kind of correction for the fact that we are using a continuous distribution to approximate a discrete distribution. Specifically, it seems that the rectangle $$Y=5$$ really includes any $$Y$$ greater than 4.5 but less than 5.5. That is:

$$P(Y=5)=P(4.5< Y < 5.5)$$

Such an adjustment is called a "continuity correction." Once we've made the continuity correction, the calculation reduces to a normal probability calculation:

Now, recall that we previous used the binomial distribution to determine that the probability that $$Y=5$$ is exactly 0.246. Here, we used the normal distribution to determine that the probability that $$Y=5$$ is approximately 0.251. That's not too shabby of an approximation, in light of the fact that we are dealing with a relative small sample size of $$n=10$$!

Let's try a few more approximations. What is the probability that more than 7, but at most 9, of the ten people sampled approve of the job the President is doing?

#### Solution

If we look at a graph of the binomial distribution with the area corresponding to $$7<Y\le 9$$ shaded in red:

we should see that we'll want to make the following continuity correction:

$$P(7<Y \leq 9)=P(7.5< Y < 9.5)$$

Now again, once we've made the continuity correction, the calculation reduces to a normal probability calculation:

By the way, you might find it interesting to note that the approximate normal probability is quite close to the exact binomial probability. We showed that the approximate probability is 0.0549, whereas the following calculation shows that the exact probability (using the binomial table with $$n=10$$ and $$p=\frac{1}{2}$$ is 0.0537:

$$P(7<Y \leq 9)=P(Y\leq 9)-P(Y\leq 7)=0.9990-0.9453=0.0537$$

Let's try one more approximation. What is the probability that at least 2, but less than 4, of the ten people sampled approve of the job the President is doing?

#### Solution

If we look at a graph of the binomial distribution with the area corresponding to $$2\le Y<4$$ shaded in red:

we should see that we'll want to make the following continuity correction:

$$P(2 \leq Y <4)=P(1.5< Y < 3.5)$$

Again, once we've made the continuity correction, the calculation reduces to a normal probability calculation:

\begin{align} P(2 \leq Y <4)=P(1.5< Y < 3.5) &= P(\dfrac{1.5-5}{\sqrt{2.5}}<Z<\dfrac{3.5-5}{\sqrt{2.5}})\\ &= P(-2.21<Z<-0.95)\\ &= P(Z>0.95)-P(Z>2.21)\\ &= 0.1711-0.0136=0.1575\\ \end{align}

By the way, the exact binomial probability is 0.1612, as the following calculation illustrates:

$$P(2 \leq Y <4)=P(Y\leq 3)-P(Y\leq 1)=0.1719-0.0107=0.1612$$

Just a couple of comments before we close our discussion of the normal approximation to the binomial.

(1) First, we have not yet discussed what "sufficiently large" means in terms of when it is appropriate to use the normal approximation to the binomial. The general rule of thumb is that the sample size $$n$$ is "sufficiently large" if:

$$np\ge 5$$ and $$n(1-p)\ge 5$$

For example, in the above example, in which $$p=0.5$$, the two conditions are met if:

$$np=n(0.5)\ge 5$$ and $$n(1-p)=n(0.5)\ge 5$$

Now, both conditions are true if:

$$n\ge 5\left(\frac{10}{5}\right)=10$$

Because our sample size was at least 10 (well, barely!), we now see why our approximations were quite close to the exact probabilities. In general, the farther $$p$$ is away from 0.5, the larger the sample size $$n$$ is needed. For example, suppose $$p=0.1$$. Then, the two conditions are met if:

$$np=n(0.1)\ge 5$$ and $$n(1-p)=n(0.9)\ge 5$$

Now, the first condition is met if:

$$n\ge 5(10)=50$$

And, the second condition is met if:

$$n\ge 5\left(\frac{10}{9}\right)=5.5$$

That is, the only way both conditions are met is if $$n\ge 50$$. So, in summary, when $$p=0.5$$, a sample size of $$n=10$$ is sufficient. But, if $$p=0.1$$, then we need a much larger sample size, namely $$n=50$$.

(2) In truth, if you have the available tools, such as a binomial table or a statistical package, you'll probably want to calculate exact probabilities instead of approximate probabilities. Does that mean all of our discussion here is for naught? No, not at all! In reality, we'll most often use the Central Limit Theorem as applied to the sum of independent Bernoulli random variables to help us draw conclusions about a true population proportion $$p$$. If we take the $$Z$$ random variable that we've been dealing with above, and divide the numerator by $$n$$ and the denominator by $$n$$ (and thereby not changing the overall quantity), we get the following result:

$$Z=\dfrac{\sum X_i-np}{\sqrt{np(1-p)}}=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\stackrel {d}{\longrightarrow} N(0,1)$$

The quantity:

$$\hat{p}=\dfrac{\sum\limits_{i=1}^n X_i}{n}$$

that appears in the numerator is the "sample proportion," that is, the proportion in the sample meeting the condition of interest (approving of the President's job, for example). In Stat 415, we'll use the sample proportion in conjunction with the above result to draw conclusions about the unknown population proportion p. You'll definitely be seeing much more of this in Stat 415!

# 28.2 - Normal Approximation to Poisson

28.2 - Normal Approximation to Poisson

Just as the Central Limit Theorem can be applied to the sum of independent Bernoulli random variables, it can be applied to the sum of independent Poisson random variables. Suppose $$Y$$ denotes the number of events occurring in an interval with mean $$\lambda$$ and variance $$\lambda$$. Now, if $$X_1, X_2,\ldots, X_{\lambda}$$ are independent Poisson random variables with mean 1, then:

$$Y=\sum\limits_{i=1}^\lambda X_i$$

is a Poisson random variable with mean $$\lambda$$. (If you're not convinced of that claim, you might want to go back and review the homework for the lesson on The Moment Generating Function Technique, in which we showed that the sum of independent Poisson random variables is a Poisson random variable.) So, now that we've written $$Y$$ as a sum of independent, identically distributed random variables, we can apply the Central Limit Theorem. Specifically, when $$\lambda$$ is sufficiently large:

$$Z=\dfrac{Y-\lambda}{\sqrt{\lambda}}\stackrel {d}{\longrightarrow} N(0,1)$$

We'll use this result to approximate Poisson probabilities using the normal distribution.

## Example 28-2

The annual number of earthquakes registering at least 2.5 on the Richter Scale and having an epicenter within 40 miles of downtown Memphis follows a Poisson distribution with mean 6.5. What is the probability that at least 9 such earthquakes will strike next year? (Adapted from An Introduction to Mathematical Statistics, by Richard J. Larsen and Morris L. Marx.)

#### Solution.

We can, of course use the Poisson distribution to calculate the exact probability. Using the Poisson table with $$\lambda=6.5$$, we get:

$$P(Y\geq 9)=1-P(Y\leq 8)=1-0.792=0.208$$

Now, let's use the normal approximation to the Poisson to calculate an approximate probability. First, we have to make a continuity correction. Doing so, we get:

$$P(Y\geq 9)=P(Y>8.5)$$

Once we've made the continuity correction, the calculation again reduces to a normal probability calculation:

\begin{align} P(Y\geq 9)=P(Y>8.5)&= P(Z>\dfrac{8.5-6.5}{\sqrt{6.5}})\\ &= P(Z>0.78)=0.218\\ \end{align}

So, in summary, we used the Poisson distribution to determine the probability that $$Y$$ is at least 9 is exactly 0.208, and we used the normal distribution to determine the probability that $$Y$$ is at least 9 is approximately 0.218. Not too shabby of an approximation!

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