Wim Ruitenburg's Spring 2010 MATH 1300-101

Repeated coin tosses produce better and better approximations to the normal distribution. Some basic facts about fair coin tossing are:

Suppose we toss a fair coin once, where we score a point each time when heads turns up. Then we have chance 1/2 to score 0 points and chance 1/2 to score 1 point. The distribution looks like:

0 1
Suppose we toss a fair coin twice, where we score a point each time when heads turns up. Then we have chance 1/4 to score 0 points, a chance 2/4 to score 1 point, and a chance 1/4 to score 2 points. The distribution looks like:

0 1 2
Suppose we toss a fair coin three times, where we score a point each time when heads turns up. Then we have chance 1/8 to score 0 points, a chance 3/8 to score 1 point, a chance of 3/8 to score 2 points, and a chance 1/8 to score 3 points. The distribution looks like:

0 1 2 3
Suppose we toss a fair coin four times, where we score a point each time when heads turns up. Then we have chance 1/16 to score 0 points, a chance 4/16 to score 1 point, a chance of 6/16 to score 2 points, a chance 4/16 to score 3 points, and a chance 1/16 to score 4 points. The distribution looks like:

0 1 2 3 4
Keep this up by considering more and more coin tosses, and the area in red will look more and more like the normal distribution.

We get the number sequences like 1, 4, 6, 4, 1 for the red graph columns heights above through the so-called Pacal triangle:

When we toss a fair coin n times, then the mean equals n/2 and the standard deviation equals sqrt(n)/2.

Some basic numerical facts about the normal distribution are:

The mean falls exactly on the line of symmetry of the curve.
The median falls exactly on the line of symmetry of the curve.
The standard deviation is exactly the horizontal distance from the line of symmetry to the inflection point of the curve.
Given standard deviation s, the quartiles lie at approximately (0.675)s above and below the line of symmetry of the curve.
There is approximately 0.68 or 68% chance that an outcome lies within one standard deviation s from the mean. By symmetry this implies that if one scores 1 standard deviation s above average, then one falls approximately in the 84% percentile.
There is approximately 0.95 or 95% chance that an outcome lies within two standard deviations 2s from the mean. By symmetry this implies that if one scores 2 standard deviations 2s above average, then one falls approximately in the 97.5% percentile.
There is approximately 0.997 or 99.7% chance that an outcome lies within three standard deviations 3s from the mean. By symmetry this implies that if one scores 3 standard deviations 3s above average, then one falls approximately in the 99.85% percentile.

Last updated: March 2010
Comments and suggestions: Email wimr@mscs.mu.edu