Normal approximation to the binomial

Introduction: Bridging the Gap

In your statistics journey, you have already met two heavyweights of the probability world: the Binomial Distribution (used for counting successes in fixed trials) and the Normal Distribution (that famous bell-shaped curve for continuous data). But what happens when a Binomial problem gets too big to handle? Imagine trying to calculate the probability of getting "at most 450 heads" in 1,000 coin flips using the Binomial formula—your calculator might just go on strike!

This chapter teaches you how to use the smooth, continuous Normal Distribution to approximate the chunky, discrete Binomial Distribution. It’s a mathematical "shortcut" that makes complex calculations much easier, provided certain conditions are met.

When Can We Use the Normal Approximation?

We can’t just use the Normal distribution for any Binomial problem. The Binomial distribution must look "Normal enough" (symmetrical and bell-shaped) for the approximation to be accurate. According to the Pearson Edexcel syllabus, you can use this approximation if either of the following sets of conditions is met:

Condition Set A:
1. The number of trials is large: \(n \ge 20\)
2. The probability of success is close to 0.5: \(p \approx 0.5\)

Condition Set B:
The expected number of successes or failures is high: \(np > 10\) or \(n(1-p) > 10\)

Memory Tip: Think of "The Power of 10." If you expect to see at least 10 successes (or 10 failures), the "staircase" of the Binomial bars starts to look like a smooth hill!

Setting Up the Parameters

If the conditions are met, we can move from a Binomial model \(X \sim B(n, p)\) to a Normal model \(Y \sim N(\mu, \sigma^2)\). But we need to know what mean (\(\mu\)) and variance (\(\sigma^2\)) to use for our new curve. We use the formulas from the Binomial distribution:

The Mean: \(\mu = np\)
The Variance: \(\sigma^2 = np(1-p)\)

So, our approximating distribution becomes:
\(Y \sim N(np, np(1-p))\)

The Golden Rule: Continuity Correction

This is the part where most students lose marks, so pay close attention! The Binomial distribution is discrete (you can have 5 successes or 6, but not 5.2). The Normal distribution is continuous (it covers every single tiny decimal in between).

To fix this, we use a continuity correction. We imagine that each whole number in the Binomial distribution actually occupies a "bin" that is 0.5 units wide on either side. For example, the number 10 actually covers everything from 9.5 to 10.5.

How to apply it:

Always draw a quick number line or sketch to visualize which "bins" you are including!

1. For "Equal to": \(P(X = 10)\) becomes \(P(9.5 < Y < 10.5)\)
2. For "Less than or equal to": \(P(X \le 10)\) includes the whole "10 bin," so it becomes \(P(Y < 10.5)\)
3. For "Strictly less than": \(P(X < 10)\) does not include 10, so it stops at the edge of the 9 bin: \(P(Y < 9.5)\)
4. For "Greater than or equal to": \(P(X \ge 10)\) includes the 10 bin, so it starts at the bottom edge: \(P(Y > 9.5)\)
5. For "Strictly greater than": \(P(X > 10)\) starts after the 10 bin is finished: \(P(Y > 10.5)\)

Common Mistake: Forgetting to adjust by 0.5. Without this, your probability will be slightly off, and in the exam, this usually means zero marks for that step!

Step-by-Step Guide to Solving Problems

Follow these steps every time to ensure accuracy:

Step 1: Check the conditions. State clearly that \(n\) is large and \(p\) is near 0.5 (or \(np > 10\)).
Step 2: Calculate the parameters. Find \(\mu = np\) and \(\sigma^2 = np(1-p)\).
Step 3: Apply the continuity correction. Change your discrete value (like \(X \ge 25\)) to a continuous range (like \(Y > 24.5\)).
Step 4: Use your calculator. Use the Normal Distribution mode (Ncd) on your calculator with your \(\mu\) and \(\sigma\) (remembering that \(\sigma = \sqrt{np(1-p)}\)).

Example Walkthrough

Question: A fair coin is tossed 100 times. Find the probability of getting more than 55 heads.

1. Model: \(X \sim B(100, 0.5)\).
2. Conditions: \(n=100\) (large) and \(p=0.5\). Approximation is valid.
3. Parameters:
\(\mu = 100 \times 0.5 = 50\)
\(\sigma^2 = 100 \times 0.5 \times 0.5 = 25\)
So, \(Y \sim N(50, 25)\).
4. Continuity Correction: We want \(P(X > 55)\). This means we don't include 55; we start at 56. The "bin" for 56 starts at 55.5. So we calculate \(P(Y > 55.5)\).
5. Final Calculation: Using a calculator for \(P(Y > 55.5)\) with \(\mu = 50\) and \(\sigma = 5\), we get approximately \(0.1357\).

Key Takeaways for Revision

• Why? We approximate when the Binomial trials (\(n\)) are too large for standard tables or direct calculation.
• Conditions: \(n \ge 20, p \approx 0.5\) OR \(np > 10\) or \(n(1-p) > 10\).
• The Mean: \(\mu = np\).
• The Variance: \(\sigma^2 = np(1-p)\).
• Continuity Correction: Always add or subtract 0.5 to move from discrete to continuous boundaries.

Quick Review: If you are asked to find \(P(X \le 15)\) using a Normal approximation, what value would you use for the continuous boundary? (Answer: 15.5, because we want to include the entire "15 bin").

Did you know? The Normal approximation is also the basis for hypothesis testing of a Binomial proportion when the sample size is large. You will use these same \(\mu\) and \(\sigma^2\) values when conducting significance tests in Paper 2!