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. But there’s something even better: the CLT also helps 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.
Logical Connection (Informal Proof)
Suppose we run a binomial experiment with:
- \( n \): number of trials (fixed)
- \( p \): probability of success
- \( X \): number of successes in \( n \) independent trials
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:
\( X \sim \text{Approximately Normal}(\mu = np, \sigma = \sqrt{np(1 - p)}) \)
This approximation gets better as \( n \) increases — particularly if:
- \( np \geq 10 \)
- \( n(1 - p) \geq 10 \)
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 with a continuity correction:
\[ z = \frac{k + 0.5 - np}{\sqrt{np(1 - p)}} \]
Note: We use \( k + 0.5 \) when going from a discrete count (binomial) to a continuous distribution (normal). This is called the continuity correction.
Example 1: Approximating a Binomial Probability
A fair die is rolled 60 times. What is the probability of getting at most 8 sixes?
- Let \( X = \) 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 \( P(X \leq 8) \approx 0.30 \) using the normal approximation.
Example 2: Checking Validity of the Approximation
Suppose a survey finds that 85% of people support a local measure. You survey 120 people and ask: What is the probability that at least 100 people support the measure?
- Let \( X \sim \text{Bin}(n = 120, p = 0.85) \)
- Check: \( np = 102 \), \( n(1 - p) = 18 \) → both ≥ 10 ✅
- \( \mu = 102 \), \( \sigma = \sqrt{120 \cdot 0.85 \cdot 0.15} \approx 3.91 \)
- We want: \( P(X \geq 100) \Rightarrow P(X \geq 99.5) \)
- \( z = \frac{99.5 - 102}{3.91} \approx -0.64 \)
- \( P(Z \geq -0.64) = 1 - P(Z < -0.64) \approx 1 - 0.2611 = 0.7389 \)
Answer There's a 73.9% chance at least 100 people support the measure.
Summary: What We Learned in Chapter 6
In this chapter, we introduced the concept of sampling distributions — a way to explore how statistics behave when drawn from repeated random samples.
- We saw how sample statistics (like means and proportions) have a distribution of their own.
- We defined the standard error as a measure of variability between sample statistics.
- We explored the Central Limit Theorem, which allows us to use the normal model for many sampling distributions, even when the original population isn’t normal.
- We extended the CLT to apply to both sample means and sample proportions.
- We finished by showing how binomial probabilities can be approximated using the normal distribution, effectively linking discrete and continuous thinking.
This chapter creates the blueprint for making smart statistical decisions. 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.


