The binomial distribution models the number of successful outcomes across a fixed sequence of independent trials, where each trial has only two possible outcomes (success or failure) and a constant probability of success. It is denoted by:

X \sim B(n, p)

Where:

n is the number of trials ( $n \in \mathbb{N}$ ).
p is the constant probability of success in each trial ( $0 < p < 1$ ).
The probability of failure is $q = 1 - p$ .

When the number of trials n becomes very large, calculating binomial probabilities directly using the formula:

P(X = x) = \binom{n}{x} p^x (1 - p)^{n-x}

becomes computationally intensive, and cumulative binomial tables cannot cover every combination of n and p.

Under specific conditions, the discrete binomial distribution smoothly approaches the continuous, symmetrical bell curve of the normal distribution:

Y \sim N(\mu, \sigma^2)

This substitution is called the normal approximation to the binomial distribution.

Illustration of the normal approximation to the binomial distribution with bar graph and smooth curve graph — Image Source: Gianpiero Placidi

Why the Normal Approximation Matters

Real-World Applications: In political polling, sample surveys of thousands of voters are modelled using normal approximations to generate margin-of-error estimates. In manufacturing quality control, batches containing tens of thousands of items are inspected for defect rates using normal approximations. In medical trials, large patient cohorts are analysed to determine the efficacy rate of treatments against a baseline.
Academic Relevance: The approximation is a direct manifestation of the De Moivre–Laplace Theorem (a specialised form of the Central Limit Theorem), linking discrete combinatorics directly to continuous integral calculus and Gaussian probability density functions.

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Core Theory

1. Conditions for Validity

A normal approximation cannot be applied indiscriminately. For the approximation to be sufficiently accurate, two key criteria must be satisfied:

The sample size n must be large.
The probability of success p must be close to 0.5 (ensuring the binomial distribution is reasonably symmetrical rather than heavily skewed).

Practical Numerical Criteria:

UK exam specifications accept the following rule of thumb:

np > 5 \quad \text{and} \quad n(1 - p) > 5 \quad (\text{or } nq > 5)

If n is large but p is very small ( $p < 0.1$ ) such that $np \le 5$ , the distribution remains heavily positively skewed. In such cases, a Poisson approximation ( $X \sim \text{Po}(\lambda)$ , with $\lambda = np$ ) is appropriate instead.
If p is close to 1, the distribution is negatively skewed; swapping the definition of "success" and "failure" allows testing the complementary condition.

2. Parameters of the Approximating Normal Distribution

A binomial distribution $X \sim B(n, p)$ has theoretical mean and variance given by:

Mean ( $\mu$ ): $\mu = E(X) = np$
Variance ( $\sigma^2$ ): $\sigma^2 = \text{Var}(X) = np(1 - p) = npq$
Standard Deviation ( $\sigma$ ): $\sigma = \sqrt{np(1 - p)} = \sqrt{npq}$

Therefore, the continuous variable Y used to approximate X is modelled as:

Y \sim N\left(np, \; np(1 - p)\right)

3. The Continuity Correction

The binomial distribution is discrete (the random variable X can only take integer values: $0, 1, 2, \dots, n$ ). In contrast, the normal distribution is continuous (the random variable Y can take any real value on a continuous spectrum).

On a continuous scale, the probability of taking an exact single value is zero: $P(Y = k) = 0$ .

To bridge this gap, each discrete integer k is represented as a continuous interval of unit width centred on k:

[k - 0.5, \; k + 0.5]

When converting an inequality from discrete X to continuous Y, always inspect whether the boundary value is included or excluded:

Exact Value: $P(X = k) \approx P(k - 0.5 < Y < k + 0.5)$
At Least k (Inclusive Lower Bound $\ge$ ): $P(X \ge k) \approx P(Y > k - 0.5)$
Strictly More Than k (Exclusive Lower Bound >): $P(X > k) \iff P(X \ge k + 1) \approx P(Y > k + 0.5)$
At Most k (Inclusive Upper Bound $\le$ ): $P(X \le k) \approx P(Y < k + 0.5)$
Strictly Less Than k (Exclusive Upper Bound <): $P(X < k) \iff P(X \le k - 1) \approx P(Y < k - 0.5)$

4. Standardisation and Calculating Probabilities

Once the continuity correction has been applied to find the continuous boundary y, standardise to the standard normal distribution $Z \sim N(0, 1)$ using:

Z = \dfrac{Y - \mu}{\sigma} = \dfrac{Y - np}{\sqrt{np(1 - p)}}

Probabilities are then evaluated using standard normal cumulative distribution tables ( $\Phi(z)$ values) or statistical calculator functions:

$P(Z < z) = \Phi(z)$
$P(Z > z) = 1 - \Phi(z)$
By symmetry: $\Phi(-z) = 1 - \Phi(z)$

Discrete Binomial Expression	Verbal Description	Continuous Normal Equivalent (with Continuity Correction)
$\text{[math]}$	Exactly k successes	$\text{[math]}$
$\text{[math]}$	At least k / No fewer than k	$\text{[math]}$
$\text{[math]}$	More than k / Strictly greater than k	$\text{[math]}$
$\text{[math]}$	At most k / No more than k	$\text{[math]}$
$\text{[math]}$	Fewer than k / Strictly less than k	$\text{[math]}$
$\text{[math]}$	Between a and b inclusive	$\text{[math]}$
$\text{[math]}$	Strictly between a and b	$\text{[math]}$

Worked Examples

Score:

In a large secondary school, 35% of students participate in an extracurricular sports club. A random sample of 800 students is chosen.

(a) Justify why a normal distribution can be used to approximate the number of students in the sample who participate in sports.

(b) Calculate the mean and standard deviation of the approximating normal distribution.

Solution

Step 1: State the binomial distribution model

Let X be the number of students who participate in sports:

$\text{[math]}$

Here $\text{[math]}$ , $\text{[math]}$ , and $\text{[math]}$ .

Step 2: Justify the normal approximation

Check the validity conditions:

$\text{[math]}$

Since both $\text{[math]}$ and $\text{[math]}$ (and n is large with p reasonably close to 0.5), a normal approximation is valid.

Step 3: Calculate the normal parameters

Mean:

$\text{[math]}$

Variance:

$\text{[math]}$

Standard deviation:

$\text{[math]}$

Thus, $\text{[math]}$ .

Step 4: Apply the continuity correction

We want the probability of at least 250 students:

$\text{[math]}$

Because the value 250 is included, extend the continuous interval 0.5 units to the left:

$\text{[math]}$

Step 5: Standardise to find the z-score

$\text{[math]}$

Step 6: Evaluate the probability using standard normal tables

$\text{[math]}$

From standard normal distribution tables:

$\text{[math]}$

The probability that at least 250 students participate is 0.9881 (or 98.81%).

A manufacturing process produces ball bearings where 5% are defective. A quality assurance team inspects a random batch of 400 ball bearings. Calculate the probability that at most 24 ball bearings are defective.

Solution

Step 1: Define the distribution

Let X be the number of defective ball bearings:

$\text{[math]}$

Step 2: Confirm validity

$\text{[math]}$

The approximation is valid.

Step 3: Calculate parameters

$\text{[math]}$

Approximating model: $\text{[math]}$

Step 4: Apply the continuity correction

We require at most 24 defectives:

$\text{[math]}$

Because 24 is included on the upper boundary, expand 0.5 units to the right:

$\text{[math]}$

Step 5: Standardise to find the z-score

$\text{[math]}$

Step 6: Determine the probability

$\text{[math]}$

From normal distribution tables:

$\text{[math]}$

The probability that at most 24 ball bearings are defective is 0.8485 (or 84.85%).

A fair six-sided die is rolled 180 times. Use a normal approximation to estimate the probability of rolling exactly 30 sixes.

Solution

Step 1: Set up the binomial parameters

Let X be the number of sixes rolled:

$\text{[math]}$

Step 2: Verify validity and compute parameters

$\text{[math]}$

Approximating model: $\text{[math]}$

Step 3: Apply the continuity correction for an exact single value

An exact discrete outcome of 30 is represented by the continuous interval from 29.5 to 30.5:

$\text{[math]}$

Step 4: Standardise both boundaries

Lower boundary:

$\text{[math]}$
Upper boundary:

$\text{[math]}$

Step 5: Calculate the area between the z-scores

$\text{[math]}$

From tables, $\text{[math]}$ :

$\text{[math]}$

The probability of rolling exactly 30 sixes is approximately 0.0796 (or 7.96%).

An airline knows from historic records that 80% of passengers booked on a regional flight show up. A carrier books 200 passengers for a flight series. Find the probability that between 155 and 170 passengers (inclusive) show up.

Solution

Step 1: Set up the distribution

Let X be the number of passengers who show up:

$\text{[math]}$

Step 2: Confirm conditions and calculate parameters

$\text{[math]}$

Approximating model: $\text{[math]}$

Step 3: Apply continuity correction to the interval

We want:

$\text{[math]}$

Expand both sides to include the integer bars:

$\text{[math]}$

Step 4: Standardise both limits

Lower boundary:

$\text{[math]}$
Upper boundary:

$\text{[math]}$
Step 5: Compute the probability

$\text{[math]}$

From tables: $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

The probability is 0.8026 (or 80.26%).

A seed merchant guarantees that 60% of seeds germinate. A gardener plants 150 seeds. Use a normal approximation to find the probability that strictly between 85 and 100 seeds germinate.

Solution

Step 1: Set up the model

Let X be the number of germinating seeds:

$\text{[math]}$

Step 2: Calculate normal parameters

$\text{[math]}$

Approximating model: $\text{[math]}$

Step 3: Convert strict inequalities to inclusive integer bounds

"Strictly between 85 and 100" means:

$\text{[math]}$

Step 4: Apply the continuity correction

Expand the inclusive integer range by 0.5 at each end:

$\text{[math]}$

Step 5: Standardise to find z-scores

Lower boundary:

$\text{[math]}$

Upper boundary:

$\text{[math]}$

Step 6: Compute the final probability

$\text{[math]}$

From tables: $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

The probability is 0.7163 (or 71.63%).

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Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.

Normal Approximation Definition Explained with Example Problems

Why the Normal Approximation Matters

Core Theory

1. Conditions for Validity

Practical Numerical Criteria:

2. Parameters of the Approximating Normal Distribution

3. The Continuity Correction

4. Standardisation and Calculating Probabilities

Worked Examples

Theory

Mode

Measures of Position in Statistics: Theory and Practice Questions

Confidence Interval

Events

Solved Problem of Probabilty 1

Solved Problem of Probabilty 13

Solved Problem of Probabilty 15

Confidence Interval for the Mean

Contingency Tables

Multiplication Rule

Standard Normal Table

Solved Problem of Probabilty 10

Combinatorics and Probability

Confidence Interval for the Proportion

Normal Distribution

Percentiles

Conditional Probability Word Problems

Using the Z Table

Pie Charts

Solved Problem of Probability 18

Tree Diagram Examples with Solutions

Probability Theory

Bayes’ Theorem

Conditional Probability

Standard Normal Distribution

Solved Problems of Probability 4

Solved Problems of Probability 17

Solved Problems of Probability 16

Solved Problems of Probability 3

Solved Problem of Probabilty 9

S1 and S2 distributions cheat sheet

What Is The Law of Total Probability?

Solved Problems of Simple and Compound Probability

Probability Problems: Worked Examples and Practice Problems

How to Calculate the Median Class: Worked Examples

Conditional Probability Example Problems and Solutions

Properties of Probability: Theory and Worked Examples

Normal Approximation: A Clear Guide with Examples

Formulas

Probability Formulas

Probability Formulas

Exercises

Confidence Interval Problems

Arithmetic Mean Problems

Median Worksheet

Normal Distribution Word Problems

Mode Worksheet

Probability Worksheet

Probability Word Problems And Solutions

Arithmetic Mean Worksheet

Standard Deviation Problems

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Events

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