Step 9: The Normal Distribution

From bar charts to smooth curves

Measure the heights of thousands of adults and draw a histogram. You will see one hump in the middle, sloping shoulders on each side and thin tails. Do the same for exam scores, birth weights or the small errors a machine makes when it cuts pipe, and the same bell shape keeps appearing. Because one curve fits so many things, learning to use it once lets you answer questions about all of them.

Heights are continuous: a person could be 170.2 cm or 170.23 cm. There are endless possible values, so asking for the chance that someone is exactly 170.000000 cm tall gives an answer of essentially zero. Sensible questions are about ranges, such as “between 165 and 175 cm”.

Imagine a histogram with more and more data and narrower and narrower bars. Its jagged top smooths into a curve. That curve is a density curve. It never dips below zero, and the total area under it is exactly 1. So the chance of landing between two values is the area under the curve between them.

One useful result follows. A single point has no width, so it has no area. That means \( P(X < 170) \) and \( P(X \le 170) \) are the same for a continuous variable. This was not true for the discrete variables of Step 7, where \( P(X \le 3) \) includes the whole bar at 3.

The normal curve and the empirical rule

The normal distribution is a family of density curves that are symmetric, single-peaked and bell-shaped, with the mean, median and mode all in the middle. Two numbers fix each one: the mean \( \mu \), which says where the centre is, and the standard deviation \( \sigma \), which says how wide it is. We write \( X \sim N(\mu, \sigma) \).

Changing \( \mu \) slides the curve left or right without changing its shape. A small \( \sigma \) gives a tall, narrow curve; a large \( \sigma \) gives a low, wide one. The total area always stays 1, so a wider curve must also be flatter.

Say adult heights are \( N(170, 8) \) in centimetres. Two standard deviations is 16 cm, so about 95% of adults are between 154 and 186 cm. The other 5% is split evenly between the two tails, so about 2.5% are taller than 186 cm. The rule is quick, but it only works at whole numbers of standard deviations. For anything else you need a table.

z-scores and the standard normal table

There is a different normal curve for every pair of \( \mu \) and \( \sigma \), so nobody could print a table for each one. The fix is to standardise: rewrite each value as the number of standard deviations it sits from the mean.

After this change, every normal variable becomes the standard normal, \( N(0, 1) \), with mean 0 and standard deviation 1. One table then serves every problem. The table below gives the area to the left of each z.

The area left of 0 is exactly 0.5, because the curve is symmetric. Symmetry also means the area left of −1.00 (0.1587) equals the area right of +1.00, which is \( 1 – 0.8413 = 0.1587 \).

Finding probabilities, and working backwards

Almost every question is one of three types:

Use the same four steps every time: write down \( \mu \) and \( \sigma \), sketch the curve and shade what you want, convert the boundaries to z-scores, then combine the table values.

Sometimes you start with a percentage and need the value. For example: what score marks the top 10%? Find the z with the right area to its left in the table, then reverse the z formula:

An entrance exam is \( N(500, 100) \), and a scholarship goes to the top 2.28%. Then \( 1 – 0.0228 = 0.9772 \) of the area lies below the cut-off, which matches \( z = 2.00 \). So the cut-off is \( 500 + 2.00 \times 100 = 700 \).

Before you touch the table, decide which way you are going. If you are given a value and need a proportion, standardise first. If you are given a proportion and need a value, find z first.

Combining normals, checking normality and approximating the binomial

Sums and differences. Step 7 showed that means add and independent variances add. The normal adds one more fact: a sum or difference of independent normal variables is itself normal. Its mean is \( \mu_X \pm \mu_Y \) and its standard deviation is \( \sqrt{\sigma_X^2 + \sigma_Y^2} \). The variances add even for a difference. For example, if two independent tasks take \( N(30, 6) \) and \( N(20, 8) \) minutes, the total is normal with mean 50 and standard deviation \( \sqrt{36 + 64} = 10 \). The chance the total exceeds 65 minutes is found from \( z = 15/10 = 1.50 \), giving \( 1 – 0.9332 = 0.0668 \).

Is the data really normal? Many real variables are not. Incomes and waiting times are right-skewed, and a normal model can even give a probability to impossible negative values. Use two checks. First, a histogram: look for one peak, rough symmetry and no long tail. Second, a normal quantile plot (also called a QQ plot): if the data is normal, the points lie close to a straight line. A steady bend away from the line means skew. Small wobbles are fine.

Normal approximation to the binomial. As \( n \) grows, binomial bars look more and more like a bell. When both \( np \ge 10 \) and \( n(1-p) \ge 10 \), you can treat a binomial as normal with \( \mu = np \) and \( \sigma = \sqrt{np(1-p)} \). If 50 people answer a yes/no question and the true “yes” rate is 0.4, then \( np = 20 \) and \( n(1-p) = 30 \), both at least 10. So \( \mu = 20 \) and \( \sigma = \sqrt{12} \approx 3.46 \). If either condition fails, use the exact binomial formula instead.

Practise: 6 questions

Try each one on paper first, then check the worked answer. Use only the table above.

Check yourself

The short version

• For continuous variables, probability is area under a density curve with total area 1, so single exact values have probability 0.
• The normal curve \( N(\mu, \sigma) \) is symmetric: \( \mu \) places it and \( \sigma \) sets its width. The empirical rule gives 68%, 95% and 99.7% within 1, 2 and 3 standard deviations.
• Standardise with \( z = (x – \mu)/\sigma \) and read the area to the left. Right tail = 1 − table value; between = larger minus smaller. To find a cut-off, use \( x = \mu + z\sigma \).
• Sums and differences of independent normals are normal, with variances adding. Check normality with a histogram and a quantile plot, and approximate a binomial by a normal only when \( np \ge 10 \) and \( n(1-p) \ge 10 \).
• The full chapter in Statistics Made Simple has many more worked examples and a full set of tiered practice questions with a complete answer key.