6.5: Approximating Binomial with the Normal Distribution
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We’ve now seen that sample means form bell-shaped distributions under the Central Limit Theorem, even if the population isn’t normal. The CLT also can help us with categorical data, especially when working with the binomial distribution.
Sometimes, calculating binomial probabilities directly (especially with large \( n \)) can be slow or impossible without technology. This is where the normal approximation to the binomial provides a valuable shortcut. Although access to computational power makes binomial calculations much easier today, this approximation was an incredibly valueable historical tool.
Logical Connection
Suppose we run a binomial experiment with:
- \( n \): number of independent trials
- \( p \): probability of success
- \( X \): number of successes
The binomial random variable \( X \) is a sum of independent Bernoulli variables:
\( X = X_1 + X_2 + \cdots + X_n \), where \( X_i \sim \text{Bernoulli}(p) \)
The mean and standard deviation of a binomial distribution are:
- \( \mu = np \)
- \( \sigma = \sqrt{np(1 - p)} \)
Since \( X \) is a sum of many independent random variables, the Central Limit Theorem tells us that that \( x \) is approximately normal, with corresponding means and standard deviations.
This approximation gets better as \( n \) increases, particularly it is advised to use this approximation when:
- \( np \geq 10 \)
- \( n(1 - p) \geq 10 \)
Normal Approximation to the Binomial
If \( X \sim \text{Binomial}(n, p) \), and \( np \geq 10 \), \( n(1 - p) \geq 10 \), then:
\( X \sim N\left(np, \sqrt{np(1 - p)}\right) \)
We can approximate \( P(X \leq k) \) by computing the corresponding z-score:
\[ z = \frac{k + 0.5 - np}{\sqrt{np(1 - p)}} \]
We add the \(0.5\) as a continuity corrector, which guarantees a more accurate result when using a continuous distribution to approximate a discrete.
Example 1: Approximating a Binomial Probability
A fair die is rolled 60 times. What is the probability of getting at most 8 sixes?
- Let \( X \) be the number of sixes. Then \( X \sim \text{Bin}(n = 60, p = 1/6) \)
- \( \mu = 60 \cdot \frac{1}{6} = 10 \)
- \( \sigma = \sqrt{60 \cdot \frac{1}{6} \cdot \frac{5}{6}} \approx 2.89 \)
- Use continuity correction: \( P(X \leq 8) \Rightarrow P(X \leq 8.5) \)
- Standardize: \( z = \frac{8.5 - 10}{2.89} \approx -0.52 \)
- From the normal table, \( P(Z \leq -0.52) \approx 0.3015 \)
Answer: \( P(X \leq 8) \approx 0.302\) using the normal approximation.
Using a binomial distribution, we find the actual probability to be \( P(X \leq 8) = 0.312\).
Coming up in Chapter 7
In the next chapter, we’ll use what we’ve learned to construct confidence intervals, helping us estimate unknown population parameters with a known degree of certainty.