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Lesson 3 of 5

The normal distribution.

About 25 minutes · 5 practice questions

A histogram shows how data spread out. On a Galton board: the mode is the tallest bar, the standard deviation grows like the square root of the number of rows, and the $68$–$95$ rule: how much of the data lies within one and two standard deviations.

Step 1 of 5 · A histogram

Step 1 of 5

A histogram

When there are many readings, it is hard to look at each one. A histogram splits the values into bins of equal width, and over each bin stands a bar whose height is the number of values that fell in it. At a glance you see where most of the data are, how far they spread, and whether they have a tail to one side.

A Galton board builds a histogram by itself: each ball bounces 1010⁠ times, and here only 25%25\%⁠ of the time to the right. A bin’s number is how many times the ball went right. The amber bars are the expected count in each bin, and the blue curve is the bell curve that approximates them.

The histogram leans left, with a longer tail to the right. Find its mode.

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Step 2 of 5

The width of the bell

In the probability Unit you saw that the board’s mean is n⋅pn \cdot p⁠: the number of rows times the chance of bouncing right. The standard deviation has a formula too:

σ=n⋅p⋅(1−p)\sigma = \sqrt{n \cdot p \cdot (1 - p)}⁠

On a fair board, p=0.5p = 0.5⁠, that is n4\sqrt{\frac{n}{4}}⁠. The Greek letter σ\sigma⁠ (sigma) is the standard deviation of a whole distribution, as μ\mu⁠ (mu) is its mean.

There are 88⁠ rows now. Change the number of rows.

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Step 3 of 5

One standard deviation

With many rows, the histogram is almost the same as the smooth bell curve. A curve like this is called a normal distribution, and it turns up wherever many small, independent effects add up: people’s heights, measuring errors, the weight of fruit. Every normal distribution has the same shape, and only μ\mu⁠ and σ\sigma⁠ change: where the centre is and how wide it is.

So one question answers for all of them. Here μ=8\mu = 8⁠ and σ=2\sigma = 2⁠.

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Step 4 of 5

Two standard deviations

About 68%68\%⁠ within one standard deviation: about two thirds. And if the range widens to two standard deviations each side?

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Step 5 of 5

Real data

These are the three numbers of the 6868⁠–9595⁠–99.799.7⁠ rule: in a normal distribution, about 68%68\%⁠ of the data lie within μ±σ\mu \pm \sigma⁠, 95%95\%⁠ within μ±2σ\mu \pm 2\sigma⁠, and 99.7%99.7\%⁠ within μ±3σ\mu \pm 3\sigma⁠. A value farther out than three standard deviations is very rare.

The rule works for every normal distribution, so knowing the mean and the standard deviation tells you a lot. Try it on people’s heights.

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Interactive scene

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Each step sets up the Scene for you, and a control marked “fixed” stays put for that step.

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Practice · Calculate

The board has 1010⁠ rows, and the chance of bouncing right is 25%25\%⁠. A bin’s number is how many times the ball went right, from 00⁠ on the left. What is the mode: which bin has the tallest bar?

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Practice · Reach the target

On a fair board the standard deviation is n⋅p⋅(1−p)\sqrt{n \cdot p \cdot (1 - p)}⁠ bins, where nn⁠ is the number of rows and p=0.5p = 0.5⁠. Change Rows of pegs so the standard deviation is exactly 22⁠ bins, then press Check.

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Practice · Predict, then look

The board has 1616⁠ fair rows: the mean is bin 88⁠ and the standard deviation 22⁠ bins. What share of the balls lands between bin 66⁠ and bin 1010⁠, within one standard deviation of the mean?

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Practice · Predict, then look

And what share lies within two standard deviations, between bin 44⁠ and bin 1212⁠?

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Practice · Calculate

Adult women’s heights are roughly normal, with a mean of 162 cm and a standard deviation of 6 cm. About what per cent of women are between 150 cm and 174 cm tall? Write the number only.