The bell curve and the 68-95-99.7 rule

What you will be able to do

• read a density curve, where an area is a share of the values;
• describe a normal curve by its mean and standard deviation;
• use the 68-95-99.7 rule, and say where its numbers come from.

A histogram that looks like a bell

A made-up school measures the heights of $400$ students of the same age. Draw a histogram of the shares, so that the bars have a total area of $1$. A smooth bell shape appears. Most heights are near the middle, and fewer lie further out on either side.

The red curve over the bars is a normal curve with mean $160\ \text{cm}$ and standard deviation $8\ \text{cm}$. It fits the bars well. A curve like this, with a total area of $1$ under it, is a density curve. The share of values between two heights is the area under the curve between them.

A height is a continuous random variable. So the probability of one exact height, such as exactly $160\ \text{cm}$, is $0$. Only ranges of heights have probabilities.

Two numbers fix the curve

Every normal curve has the same shape. Only two numbers change it.

• The mean $\mu$ is the centre. The curve is symmetric about it, so the mean, the median and the mode are all equal.
• The standard deviation $\sigma$ sets the width. On each side, the curve changes from bending down to bending up at one standard deviation from the mean.

We write $X \sim N(\mu, \sigma)$. Some books write $N(\mu, \sigma^2)$ with the variance in place of the standard deviation, so always check which one a book means.

The 68-95-99.7 rule

For every normal curve, the share within $k$ standard deviations of the mean is the same. It depends only on $k$. Python's math.erf function gives these areas exactly.

• Within $1\sigma$: $0.6827$, about $68\%$.
• Within $2\sigma$: $0.9545$, about $95\%$.
• Within $3\sigma$: $0.9973$, about $99.7\%$.

For the made-up heights, about $68\%$ lie from $152\ \text{cm}$ to $168\ \text{cm}$, about $95\%$ from $144\ \text{cm}$ to $176\ \text{cm}$, and almost all from $136\ \text{cm}$ to $184\ \text{cm}$.

The rule also splits into pieces, because the curve is symmetric. Half of $68\%$ is $34\%$, between the mean and one standard deviation above it. The share above $\mu + \sigma$ is $\dfrac{100\% - 68\%}{2} = 16\%$. The share above $\mu + 2\sigma$ is about $2.5\%$.

You should find that the share inside depends only on $k$. Change the mean or the standard deviation, and the shaded part moves or stretches, but its area stays the same for the same $k$.

Try it in Python

The program finds the three areas of the rule from math.erf, and the three ranges for the made-up heights.

Your turn

The usual mistakes

• Reading the height of the curve as a probability. The probability is the area under the curve, not its height.
• Using $95\%$ for one side. The $95\%$ is split, $47.5\%$ on each side of the mean.
• Forgetting that the rule is rounded. The exact shares are $0.6827$, $0.9545$ and $0.9973$.
• Mixing up $N(\mu, \sigma)$ and $N(\mu, \sigma^2)$. Check whether the second number is a standard deviation or a variance.

Remember these

• A density curve has a total area of $1$, and areas are shares.
• A normal curve is fixed by $\mu$ and $\sigma$. It is symmetric about $\mu$.
• About $68\%$, $95\%$ and $99.7\%$ lie within $1$, $2$ and $3$ standard deviations of the mean.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What share of a normal distribution lies above $\mu + 2\sigma$?

2. The made-up heights have $\mu = 160\ \text{cm}$ and $\sigma = 8\ \text{cm}$. Between which heights do about $99.7\%$ of students lie?

3. What is the total area under any density curve?

4. Use the rule to find the share of the made-up heights between $152\ \text{cm}$ and $176\ \text{cm}$.

5. A normal curve is changed from $\sigma = 8\ \text{cm}$ to $\sigma = 4\ \text{cm}$. What happens to its shape?

References

• OpenStax, Introductory Statistics, Chapter 6, sections on the standard normal distribution and the empirical rule.