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The normal curve as a model of data

Statistics and data · level 7, lesson 2 of 7

A question to start

Roll one die and the totals 1 to 6 are equally likely: a flat histogram. Roll two dice and add them: 7 is the most likely total, and the histogram is a triangle. Roll five dice and add them, and something strange happens: the histogram is already a smooth hill, low at both ends, high in the middle, the same on both sides.

1 die: totals 1 to 6 2 dice: totals 2 to 12 5 dice: totals 5 to 30

The same hill appears in the heights of adults, the weights of apples from one orchard and the errors in a thousand careful measurements of the same distance. Why does one shape keep appearing in such different places, and what can you do with it once you have it?

The idea

From histogram to smooth curve

Take a histogram of a very large data set and make the bins narrower and narrower. The jagged tops of the bars blur into a smooth curve. If you scale the bars so that the total area of all the bars is 1, each bar's area is the fraction of the data in that bin.

A smooth curve used this way is called a density curve. Its two rules are:

  1. It never dips below the horizontal axis.
  2. The total area under it is exactly 1 (that is, 100% of the data).

Then the area under the curve between two values \(a\) and \(b\) is the fraction of the data between \(a\) and \(b\). Reading a density curve is reading areas, not heights.

The normal curve

The most important density curve is the normal curve, also called the bell curve. A normal curve is:

  • single-peaked and symmetric: the peak is at the mean, which is also the median and the mode;
  • completely fixed by two numbers: the mean \(\mu\) (Greek "mu"), which sets where the centre is, and the standard deviation \(\sigma\) (Greek "sigma"), which sets how wide it is;
  • bending in a particular way: near the peak the curve bends downwards like a dome; further out it bends upwards like a bowl. The points where the bending changes, called inflection points, are exactly at \(\mu - \sigma\) and \(\mu + \sigma\). So you can see one standard deviation on the picture;
  • never quite touching zero: the tails go on forever, but they get thin very fast.
130 140 150 160 170 180 190 200 210 μ = 160, σ = 6 μ = 175, σ = 10

Changing \(\mu\) slides the curve along the axis without changing its shape. Changing \(\sigma\) stretches it: a bigger \(\sigma\) makes it wider, and, because the area must stay 1, also lower.

The curve's equation is

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

You will not need to work with it here. Two things are worth noticing. The square \((x - \mu)^2\) makes the curve symmetric: a value 5 above the mean gives the same height as a value 5 below. And the minus sign in the power of \(e\) makes the curve fall off very quickly as you move away from \(\mu\); that is why values more than 3 standard deviations from the mean are rare.

Why the bell keeps appearing

Look again at the dice. Each total is a sum of many small random pieces. Extreme totals need every piece to be extreme in the same direction (five 6s, or five 1s), which can happen in only one way out of \(6^5 = 7776\). Middle totals can be made in hundreds of ways, because highs and lows cancel. So sums pile up in the middle, and the pile takes the bell shape.

A person's height is the result of many genes and many years of meals and sleep, each nudging it a little up or down. A measurement error is the sum of many tiny disturbances: a small shake of the hand, a little heat, a slight misreading. When many small, independent effects add up, the result is close to normal, whatever the shape of each small effect. This is a real theorem, the central limit theorem, proved in Probability, level 8. It is why the normal curve is a good model for so many kinds of data.

A model is a simple, exact description that fits real data well enough to be useful. Real heights are not perfectly normal (nobody is 3 metres tall, though the curve allows it), but the normal model predicts percentages of heights very well.

The 68–95–99.7 rule

For every normal curve, whatever \(\mu\) and \(\sigma\) are:

  • about 68% of values lie within 1 standard deviation of the mean, between \(\mu - \sigma\) and \(\mu + \sigma\);
  • about 95% lie within 2 standard deviations;
  • about 99.7% lie within 3 standard deviations.
68% 13.5% 13.5% 2.35% 2.35% μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ

(More exactly: 68.27%, 95.45% and 99.73%.) Why is the rule the same for every normal curve? Because every normal curve is the same shape, only slid and stretched. Measure distances in units of \(\sigma\) and all normal curves become one curve. That idea gets a name, the z-score, in the next lesson.

Because the curve is symmetric, the rule splits into pieces:

  • from \(\mu\) to \(\mu + \sigma\): half of 68%, that is 34%;
  • from \(\mu + \sigma\) to \(\mu + 2\sigma\): \((95 - 68) \div 2 =\) 13.5%;
  • from \(\mu + 2\sigma\) to \(\mu + 3\sigma\): \((99.7 - 95) \div 2 =\) 2.35%;
  • above \(\mu + 2\sigma\): \((100 - 95) \div 2 =\) 2.5%; above \(\mu + 3\sigma\): 0.15%.

Is the normal model right for these data?

Before using the rule, check that the data look normal. Three checks:

  1. Draw a histogram. It should have one hump and be roughly symmetric, with no long tail.
  2. Count. Find what fraction of the data is within 1, 2 and 3 standard deviations of the mean. It should be near 68%, 95% and 99.7%.
  3. Use a normal probability plot (technology draws it). It plots each value against where it would be if the data were exactly normal. Normal data give points close to a straight line; skewed data give a curve.

Not all data are normal

Incomes are the classic counter-example. They are strongly skewed right: most people earn a moderate amount, and a few earn many times more. Mean minus two standard deviations is often a negative income, which is impossible. Waiting times, the sizes of earthquakes, the populations of towns and the number of followers people have online are all skewed right. For them the 68–95–99.7 rule gives wrong answers, and the median and IQR describe the data better.

Worked examples

The heights of the adult women in a town are close to normal with mean \(\mu = 163\) cm and standard deviation \(\sigma = 7\) cm. (These are made-up but realistic numbers.)

1. A middle band. What percent are between 149 cm and 177 cm?

\(149 = 163 - 14 = \mu - 2\sigma\) and \(177 = 163 + 14 = \mu + 2\sigma\). Within 2 standard deviations: about 95%.

2. One tail. What percent are taller than 170 cm?

\(170 = \mu + \sigma\). Within 1 standard deviation is 68%, so 32% is outside, half on each side: 16% taller.

3. A lopsided band. What percent are between 156 cm and 177 cm?

\(156 = \mu - \sigma\) and \(177 = \mu + 2\sigma\). Add the pieces: from \(\mu - \sigma\) to \(\mu\) is 34%, from \(\mu\) to \(\mu + \sigma\) is 34%, from \(\mu + \sigma\) to \(\mu + 2\sigma\) is 13.5%. Total: \(34 + 34 + 13.5 =\) 81.5%.

4. A machine that fills bags. A machine fills flour bags with a mean of 505 g and a standard deviation of 2.5 g, normally distributed. The label says 500 g. What fraction of bags are underweight?

\(500 = 505 - 5 = \mu - 2\sigma\). Below \(\mu - 2\sigma\) is about 2.5%, so about 1 bag in 40 is underweight. To make underweight bags rarer, the factory can raise the mean (costing flour) or reduce \(\sigma\) (a more precise machine). Setting the mean to \(500 + 3 \times 2.5 = 507.5\) g makes underweight bags about 0.15%: fewer than 2 in 1000.

5. Testing the model on real-looking data. The clinic waiting times from the last lesson were

1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 8, 12, 19

Their mean is about 5.54 and their standard deviation (the sample version, level 6) is about 4.99.

  • Within 1 standard deviation means between \(5.54 - 4.99 = 0.55\) and \(5.54 + 4.99 = 10.53\). That includes 11 of the 13 values, which is 85%, not 68%.
  • Mean minus 2 standard deviations is \(5.54 - 9.98 \approx -4.4\) minutes: a negative waiting time.

The data are skewed right, and the normal model does not fit. Use the median and IQR instead.

Try it

IQ-style test scores are scaled to be normal with mean 100 and standard deviation 15.

  1. What percent of people score between 85 and 115? Between 70 and 130?
  2. What percent score above 115? Above 145?
  3. What percent score between 70 and 115?
  4. A normal distribution has \(\mu = 40\). About 2.5% of values are above 52. What is \(\sigma\)?
  5. Sketch two normal curves on the same axis: one with \(\mu = 50, \sigma = 5\) and one with \(\mu = 50, \sigma = 10\). Which is taller? Why must it be?
  6. Explain why the 68–95–99.7 rule is the same for every normal curve.
  7. A data set of 200 times has mean 12 minutes and standard deviation 9 minutes, and no time is negative. Explain why these data cannot be close to normal.
  8. Among 1000 values from a normal distribution, about how many would you expect more than 3 standard deviations from the mean? If you saw 40 such values, what would you conclude?
Answers
  1. 85 to 115 is \(\mu \pm \sigma\): 68%. 70 to 130 is \(\mu \pm 2\sigma\): 95%.
  2. Above 115 (\(\mu + \sigma\)): \((100 - 68) \div 2 =\) 16%. Above 145 (\(\mu + 3\sigma\)): \((100 - 99.7) \div 2 =\) 0.15%.
  3. From 70 to 100 is half of 95%, that is 47.5%. From 100 to 115 is 34%. Total 81.5%.
  4. Above \(\mu + 2\sigma\) is 2.5%, so \(52 = 40 + 2\sigma\), giving \(2\sigma = 12\) and \(\sigma = 6\).
  5. The \(\sigma = 5\) curve is narrower and taller. Both curves must enclose area 1; the wider curve spreads that area over more of the axis, so it has to be lower. (In fact its peak is exactly half as high.)
  6. Every normal curve is the same shape, only moved along (by \(\mu\)) and stretched (by \(\sigma\)). Moving and stretching do not change what fraction lies within "1 stretch-unit" of the centre, so the fractions are the same for all of them.
  7. If they were normal, about 2.5% of them (5 times) would be below \(\mu - 2\sigma = 12 - 18 = -6\) minutes, and even more would be below 0. But no time is below 0. The lowest possible value, 0, is only \(12 \div 9 \approx 1.3\) standard deviations below the mean, while the times can reach far above it: the data must be bunched near 0 with a long tail to the right, that is skewed right.
  8. About 0.3% of 1000, that is 3 values. Seeing 40 means the tails are much heavier than normal: the normal model does not fit (or something unusual, such as a measuring fault, is in the data).

Common mistake

Using the 68–95–99.7 rule on any data at all. The rule belongs to the normal curve. For skewed data it gives wrong percentages and can predict impossible values. Look at the shape first.

Reading heights instead of areas. The height of a density curve at one value is not a percentage. Percentages are areas under the curve, between two values.

Where it leads