OPTIMA Statistics

Subsection 3.4.1 Visualizing the Binomial Distribution

Before we even start talking about how we can approximate binomial probabilities using the normal distribution, let's think a little about why we can. Below are three probability histograms for a binomial random variable \(X\) resulting from \(n = 10\) trials. The first shows the distribution of \(X\) when \(p = 0.1\text{,}\) the middle when \(p = 0.5\text{,}\) and the right when \(p = 0.9\text{.}\)

Which of these distributions would we call mound–shaped? The one in the middle appears to be the most mound-shaped of the three. The other two are skewed either to the right or to the left. Note that the one in the middle has a probability of \(0.5\text{.}\) The binomial distribution looks the most like the normal distribution when \(p = 0.5\text{.}\) However, as \(n\) increases, the value of \(p\) becomes less important. Consider the distributions below with the same values of \(p\text{,}\) but with \(n = 80\text{.}\)

Notice that with the larger value of \(n\text{,}\) all three of these probability histograms look pretty mound shaped. Also notice that as \(n\) increases, the number of bars increases as well, and the distribution of probabilities starts to look less stair-stepped, and more like a smooth curve. Try playing with this yourself by performing the following steps.

1. Open the interactive binomial distribution page.

2. Change the value of \(p\) (in the bottom right-hand corner) to several different percents to see what happens (for example, try 25, 50, and 75).

3. Change the value of \(n\) (in the bottom left-hand corner) to several different numbers to see what happens (for example, try \(n=10\text{,}\) \(30\text{,}\) \(50\text{,}\) and so on).

4. Try different combinations of \(n\) and \(p\) and notice how mound-shaped or skewed the distribution looks.

5. Finally, click the “Show Normal Curve” button to see how the normal curve “fits” on top of the binomial probability histogram.

Hopefully you have noticed that the larger \(n\) is and the closer \(p\) is to \(0.5\text{,}\) the less “gap” there is between the normal curve and the bars. That is, the less of the bar sticks up above, or does not reach up to the normal curve. The smaller this “gap” is, the better our approximation will be.

Subsection 3.4.2 When can We Approximate?

If then the binomial distribution can have so many different shapes depending on the parameters \(n\) and \(p\text{,}\) when is is enough like the mound–shaped normal distribution to allow us to use the normal distribution to approximate probabilities?

Of course the larger \(n\times p\) and \(n\times q\) get, the better the approximation will be. However, the criteria above tells us when an approximation will be “good enough” for us to use. Notice that this criteria has nothing to do with the specific probability that we want to compute. It doesn't matter if we are looking for the probability that \(X\) is greater than a number, less than a number, or between two numbers. What matters is how large \(n\times p\) and \(n\times q\) are. This is because, as we saw on the last page, these two parameters control the shape of the binomial probability histogram, and this histogram is what either matches a normal distribution well or does not match well.

Subsection 3.4.3 Continuity Correction

There is one final issue to address before we are ready to start approximating binomial probabilities with the normal distribution. This issue has to do with the fact that we are using a continuous probability density curve to approximate a discrete random variable. To see why this may be a problem, consider the following picture. Suppose that the binomial random variable \(X\) (shown by the bar) is being approximated using a normal random variable \(Y\) (shown by the curve).

In the normal distribution, the probability \(P(Y=10)\) is exactly zero because it is a single line. In the binomial distribution, however, the entire green bar represents \(P(X=10)\text{.}\) If we wish to use the normal distribution to find \(P(X=10)\) in the binomial distribution, we need to translate this bar into a range of \(Y\) values. That range will extend from the bottom end of the bar, which is at \(10 - 0.5\text{,}\) to the top end of the bar at \(10 + 0.5\text{.}\) Therefore,

When we change a whole number into a range like this we are correcting for the fact that we use a continuous random variable in place of a discrete random variable.

In the following examples, we will apply the continuity correction to translate a probability statement for a discrete random variable \(X\) into an approximately equivalent statement for a continuous random variable \(Y\text{.}\)

One caution on using the continuity correction. It should only be used when we are approximating a binomial distribution with a normal distribution. If we start with a normal distribution, then the variable is already continuous, and no correction is needed.

Subsection 3.4.4 Normal Approximations

tools in place to use a normal distribution to approximate probabilities for a binomial distribution. Recall that a binomial random variable has mean and standard deviation as shown below.

So when we use a normal distribution to approximate a binomial probability, the mean of that normal distribution will be \(\mu = n\times p\) and the standard deviation will be \(\sigma = \sqrt{n\times p\times q}\text{.}\) Putting this together with the criteria for approximation and the continuity correction, we can solve examples such as the following.

A binomial process involves \(400\) trials in which the probability of a success is \(p = 0.35\text{.}\) What is the probability that there are at least \(168\) successes in this process?

Solution

We first must be sure that we can use a normal approximation. To assess this, we check \(n\times p\) and \(n\times q\text{.}\)

\begin{equation*} n\times p = 400(0.35) = 140 \quad \text{ and } \quad n\times q = 400(0.65) = 260\text{.} \end{equation*}

Since both of these are greater than \(5\text{,}\) we can continue with a normal approximation.

Next, we need to compute the mean and standard deviation to use for the normal distribution. We have actually already found the mean above, but we repeat this computation together with that of the standard deviation below.

\begin{align*} \mu \amp = n\times p = 400(0.35) = 140\\ \sigma \amp = \sqrt{n\times p\times q} = \sqrt{400(0.35)(0.65)} \approx 9.54 \end{align*}

Our final task is to use the normal distribution given by this mean and standard deviation to find \(P(X \geq 168)\text{.}\) We draw a picture to help us visualize the appropriate continuity correction, zooming in on the right tail of the distribution since 168 is above the mean of 140.

The picture and the z-score formula help us make the following computation.

\begin{align*} P(X \geq 168) \amp = P(Y > 167.5)\\ \amp = P\left(\frac{Y-\mu}{\sigma} > \frac{167.5 - 140}{9.54}\right)\\ \amp = P(Z > 2.88)\\ \amp = 1 - P(Z\lt 2.88)\\ \amp = 1 - 0.9980\\ \amp = 0.0020\text{.} \end{align*}

Observe that in this question, we are actually using three different distributions. There is the binomial distribution for which we are actually wanting to find probabilities, represented by the variable \(X\text{.}\) Then there is the normal distribution we use to approximate it, represented by the variable \(Y\text{.}\) Finally, there is the standard normal distribution to which we convert in order to use the standard normal distribution table, represented by the variable \(Z\text{.}\) Our next example picks up where Example 3.4.1 left off.

A recent study has determined that 32.2% of Americans are obese. A research group wishing to study this phenomena samples \(12,000\) individuals in a large metropolitan area. What is the probability that no more than \(3750\) of these individuals are obese? Use a normal approximation.

Solution

Checking our criteria for approximating yields

\begin{equation*} n\times p = 12000(.322) = 3864 \quad \text{ and } \quad n\times q = 12000(0.678) = 8136\text{.} \end{equation*}

Since both of these are greater than \(5\text{,}\) we may approximate this probability using a normal distribution.

That normal distribution will have a mean and standard deviation of

\begin{align*} \mu \amp = 12000(0.322) = 3864\\ \sigma \amp = \sqrt{12000(0.322)(0.678)} = 51.1839\text{.} \end{align*}

We note that “no more than 3750” means we want \(X\) to be less than or equal to 3750. So we draw the picture shown below to help us correctly apply the continuity correction, this time zooming in on the left tail since 3750 is less than the mean of 3864.

This gives us the following probability computation

\begin{align*} P(X \leq 3750) \amp = P(Y \lt 3750.5)\\ \amp = P\left(\frac{Y-\mu}{\sigma} \lt \frac{3750.5-3864}{51.1839}\right)\\ \amp = P(Z \lt -2.22)\\ \amp = 0.0132\text{.} \end{align*}

Therefore, the probability of no more than \(3750\) obese individuals is approximately \(0.0132\text{.}\)

Subsection 3.4.5 How Good are These Approximations?

We have made a point of identifying the probabilities we get from a normal distribution as approximations for the probabilities from a binomial distribution. Any time we are approximating, the question naturally arises, how good is that approximation? In the next several examples we will look at both the normal approximation and, with the help of a computer, the binomial probability to see how good these approximations really are.

Notice that in this case, \(n\times p\) and \(n\times q\) were very close to 5. Therefore, the approximation was acceptable, but not great. In the next example, let's look at what happens when \(n\times p\) and \(n\times q\) are much larger than 5.