A Level Statistics: Normal and Binomial Distributions

The binomial and normal distributions are the two most important probability models you meet at A Level and in intermediate statistics. The binomial counts successes in a fixed number of trials, while the normal describes measurements such as heights, marks and weights that cluster around an average. Knowing when to use each one, and how they connect, is the key to the whole topic.

The Binomial Distribution

When Does It Apply?

A random variable \( X \) follows a binomial distribution, written \( X \sim B(n, p) \), when all four of these conditions hold:

1. There is a fixed number of trials, \( n \).
2. Each trial has only two outcomes, called success and failure.
3. The probability of success, \( p \), is the same on every trial.
4. The trials are independent.

Examples include the number of heads in 20 coin tosses, or the number of faulty bulbs in a box of 50 when each has the same chance of being faulty. If any condition fails, for example when drawing cards without replacement, the binomial model is not exact.

The Formula

The probability of exactly \( r \) successes is

$$ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}, \qquad r = 0, 1, \dots, n $$

where \( \binom{n}{r} = \frac{n!}{r!(n - r)!} \) counts the ways of arranging the successes. The mean and variance are

$$ E(X) = np, \qquad \text{Var}(X) = np(1 - p) $$

Worked Example 1: Exam Guessing

A student guesses on 10 multiple-choice questions. Each question has a probability 0.3 of being answered correctly by luck. Let \( X \) be the number correct, so \( X \sim B(10, 0.3) \).

(a) Find \( P(X = 3) \).

Step 1. Compute the coefficient: \( \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \).

Step 2. Compute the powers: \( 0.3^3 = 0.027 \) and \( 0.7^7 = 0.0823543 \).

Step 3. Multiply:

$$ P(X = 3) = 120 \times 0.027 \times 0.0823543 = 0.2668 $$

(b) Find \( P(X \le 1) \).

Step 1. Add the probabilities for 0 and 1:

$$ P(X = 0) = 0.7^{10} = 0.028248 $$

$$ P(X = 1) = 10 \times 0.3 \times 0.7^9 = 10 \times 0.3 \times 0.040354 = 0.121061 $$

Step 2. Total: \( P(X \le 1) = 0.028248 + 0.121061 = 0.1493 \).

(c) Find \( P(X \ge 3) \). Use the complement: \( P(X \ge 3) = 1 - P(X \le 2) \). We need \( P(X = 2) = 45 \times 0.09 \times 0.7^8 = 45 \times 0.09 \times 0.057648 = 0.233474 \). So \( P(X \le 2) = 0.1493 + 0.2335 = 0.3828 \), and \( P(X \ge 3) = 1 - 0.3828 = 0.6172 \).

Mean and variance: \( E(X) = 10 \times 0.3 = 3 \), and \( \text{Var}(X) = 10 \times 0.3 \times 0.7 = 2.1 \). It makes sense that 3 correct answers is the single most likely outcome.

The Normal Distribution

Its Shape and Properties

A continuous random variable \( X \sim N(\mu, \sigma^2) \) has a symmetric, bell-shaped curve centred on the mean \( \mu \), with spread controlled by the standard deviation \( \sigma \). Useful facts:

• About 68% of values lie within one standard deviation of the mean.
• About 95% lie within two standard deviations.
• About 99.7% lie within three standard deviations.
• For a continuous variable, \( P(X = a) = 0 \), so \( P(X < a) \) and \( P(X \le a) \) are the same.

Standardising

Every normal distribution can be converted to the standard normal \( Z \sim N(0, 1) \) using

$$ Z = \frac{X - \mu}{\sigma} $$

You then read probabilities from tables of \( \Phi(z) = P(Z < z) \) or from a calculator. Because the curve is symmetric, \( \Phi(-z) = 1 - \Phi(z) \).

Worked Example 2: Student Heights

The heights of students at a college are normally distributed with mean 165 cm and standard deviation 6 cm.

(a) Find the probability that a student is taller than 174 cm.

Step 1. Standardise: \( z = \frac{174 - 165}{6} = 1.5 \).

Step 2. Use tables: \( \Phi(1.5) = 0.9332 \).

Step 3. \( P(X > 174) = 1 - 0.9332 = 0.0668 \).

So about 6.7% of students are taller than 174 cm.

(b) Find the height exceeded by the tallest 10% of students.

Step 1. We need \( a \) such that \( P(X > a) = 0.10 \), so \( P(Z < z) = 0.90 \).

Step 2. From inverse tables, \( z = 1.2816 \).

Step 3. Reverse the standardisation:

$$ a = \mu + z\sigma = 165 + 1.2816 \times 6 = 165 + 7.69 = 172.7 \text{ cm} $$

Sanity check: 172.7 cm is between the mean and 174 cm, which fits, because 174 cm cut off only 6.7% while we needed 10%.

Connecting the Two: The Normal Approximation

When \( n \) is large and \( p \) is not too close to 0 or 1, the binomial distribution looks very like a normal curve. A common rule of thumb is that \( np > 5 \) and \( n(1 - p) > 5 \). Then

$$ B(n, p) \approx N\big(np,\; np(1 - p)\big) $$

Because the binomial is discrete and the normal is continuous, we apply a continuity correction: each whole number \( r \) is treated as the interval from \( r - 0.5 \) to \( r + 0.5 \).

Worked Example 3: A Large Survey

In a town, 40% of adults use a particular mobile wallet. A random sample of 100 adults is taken. Estimate the probability that at most 45 use the wallet.

Step 1. \( X \sim B(100, 0.4) \). Check the conditions: \( np = 40 \) and \( n(1 - p) = 60 \), both greater than 5.

Step 2. Mean \( = 40 \), variance \( = 100 \times 0.4 \times 0.6 = 24 \), so \( \sigma = \sqrt{24} = 4.899 \).

Step 3. Apply the continuity correction: \( P(X \le 45) \approx P(Y < 45.5) \), where \( Y \sim N(40, 24) \).

Step 4. Standardise: \( z = \frac{45.5 - 40}{4.899} = 1.123 \).

Step 5. \( \Phi(1.123) \approx 0.869 \).

The exact binomial answer, found with software, is 0.8689, so the approximation is excellent. Without the continuity correction you would get \( \Phi(1.021) \approx 0.846 \), which is noticeably worse.

Choosing the Right Model

• Counting successes in a fixed number of independent trials: binomial.
• Measuring a continuous quantity that clusters symmetrically around a mean: normal.
• A binomial with large \( n \) and moderate \( p \): normal approximation with continuity correction.

Did You Know?

• The normal curve was studied by Abraham de Moivre around 1733, precisely as an approximation to the binomial for many coin tosses.
• Carl Friedrich Gauss later used the same curve to model errors in astronomical measurements, which is why it is often called the Gaussian distribution.
• A Galton board, where balls bounce left or right off pins, produces a binomial pattern that looks strikingly like a bell curve.

Key Takeaways

• Use \( B(n, p) \) only when trials are fixed, independent, two-outcome and have constant probability.
• For a binomial, the mean is \( np \) and the variance is \( np(1 - p) \).
• Standardise normal values with \( Z = \frac{X - \mu}{\sigma} \) and use symmetry to handle negative \( z \).
• For inverse problems, find \( z \) first, then use \( x = \mu + z\sigma \).
• The normal approximation to the binomial needs a continuity correction of 0.5.

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