The Normal Distribution Explained

The normal distribution is the famous bell-shaped curve that appears again and again in nature and in data: heights, exam marks, measurement errors and much more. Because so many quantities follow it, learning to work with the normal distribution is one of the most useful skills in statistics. This guide explains its properties, the z-score method and how to solve typical exam questions.

What Is the Normal Distribution?

A continuous random variable \( X \) is normally distributed with mean \( \mu \) and standard deviation \( \sigma \), written

$$ X \sim N(\mu, \sigma^2) $$

if its probability density function is

$$ f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2\sigma^2}} $$

You will rarely need to use this formula directly, but it is good to recognise it.

Key Properties

• The curve is bell-shaped and symmetric about the mean \( \mu \).
• The mean, median and mode are all equal to \( \mu \).
• The total area under the curve is 1, and probabilities are areas under the curve.
• The curve never touches the x-axis, but it gets extremely close beyond about 3 standard deviations.
• A larger \( \sigma \) gives a wider, flatter curve; a smaller \( \sigma \) gives a taller, narrower one.

Because \( X \) is continuous, \( P(X = a) = 0 \) for any single value, so \( P(X < a) \) and \( P(X \le a) \) are the same.

The Empirical (68–95–99.7) Rule

For any normal distribution:

• About 68% of values lie within \( \mu \pm \sigma \).
• About 95% lie within \( \mu \pm 2\sigma \).
• About 99.7% lie within \( \mu \pm 3\sigma \).

For instance, if IQ scores follow \( N(100, 15^2) \), roughly 95% of people score between 70 and 130.

The Standard Normal Distribution and Z-Scores

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

$$ z = \frac{x - \mu}{\sigma} $$

The z-score tells you how many standard deviations a value lies above (positive) or below (negative) the mean. Probabilities for \( Z \) are given in tables or on calculators as

$$ \Phi(z) = P(Z < z) $$

Useful facts from symmetry:

• \( P(Z > z) = 1 - \Phi(z) \)
• \( \Phi(-z) = 1 - \Phi(z) \)
• \( P(a < Z < b) = \Phi(b) - \Phi(a) \)

Some commonly used values: \( \Phi(1) = 0.8413 \), \( \Phi(1.5) = 0.9332 \), \( \Phi(2) = 0.9772 \).

Worked Example 1: Finding Probabilities

Question: The heights of adult men in a city are normally distributed with mean 170 cm and standard deviation 8 cm. Find the probability that a randomly chosen man is (a) shorter than 182 cm, (b) between 162 cm and 186 cm.

Part (a)

Step 1: Standardise.

$$ z = \frac{182 - 170}{8} = \frac{12}{8} = 1.5 $$

Step 2: Read the table.

$$ P(X < 182) = P(Z < 1.5) = \Phi(1.5) = 0.9332 $$

About 93.3% of men are shorter than 182 cm.

Part (b)

Step 1: Standardise both limits.

$$ z_1 = \frac{162 - 170}{8} = -1, \qquad z_2 = \frac{186 - 170}{8} = 2 $$

Step 2: Use symmetry for the negative value.

$$ \Phi(-1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587 $$

Step 3: Subtract.

$$ P(162 < X < 186) = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185 $$

So about 81.9% of men have heights in this range.

Worked Example 2: Finding a Cut-Off Value (Inverse Normal)

Question: Exam marks are normally distributed with mean 60 and standard deviation 12. The top 10% of students receive an A grade. What is the minimum mark for an A?

Step 1: Translate into a probability statement. We need the mark \( a \) such that

$$ P(X > a) = 0.10 \quad\Longleftrightarrow\quad P(X < a) = 0.90 $$

Step 2: Find the z-value. From the inverse normal table, \( \Phi(z) = 0.90 \) when

$$ z \approx 1.2816 $$

Step 3: Convert back to marks. Rearranging \( z = \frac{x - \mu}{\sigma} \) gives \( x = \mu + z\sigma \):

$$ a = 60 + 1.2816 \times 12 = 60 + 15.38 = 75.38 $$

A student needs about 75.4 marks (76 if marks are whole numbers) to get an A.

Check: \( z = \frac{75.38 - 60}{12} \approx 1.28 \), and \( \Phi(1.28) \approx 0.90 \). Correct.

Where the Normal Distribution Is Used

• Natural measurements: heights, weights, blood pressure.
• Education: standardised test scores and grading curves.
• Manufacturing: quality control limits for product sizes.
• Physics and chemistry: random measurement errors in experiments.
• Statistics: the Central Limit Theorem says sample means are approximately normal for large samples, which underpins confidence intervals and hypothesis tests.

Common Mistakes

• Dividing by the variance instead of the standard deviation. If you are given \( N(170, 64) \), then \( \sigma = 8 \), not 64.
• Forgetting to subtract from 1 for "greater than" questions.
• Ignoring negative z-values. Most tables only list positive \( z \), so use \( \Phi(-z) = 1 - \Phi(z) \).
• Using the wrong tail in inverse problems. "Top 10%" means \( P(X < a) = 0.90 \), not 0.10.
• Not sketching. A quick bell curve with the shaded region prevents most errors.

Tips

• Always sketch the curve and shade the area you want.
• Write down \( \mu \) and \( \sigma \) clearly before standardising.
• Check that probabilities make sense: values above the mean should give \( \Phi(z) > 0.5 \).

Key Takeaways

• The normal distribution is symmetric and bell-shaped, centred on \( \mu \) with spread \( \sigma \).
• About 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean.
• Convert any normal value to a z-score with \( z = \frac{x - \mu}{\sigma} \) and use \( \Phi(z) \) tables.
• For cut-off questions, find \( z \) first, then use \( x = \mu + z\sigma \).

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