You roll a die 1000 times, you add up the numbers you have rolled, and you get 3698. Do you think it is a fair die?

The expected value for a die roll is [math]\frac{ 1+2+3+4+5+6}{6} = 3.5[/math]. The variance is [math]\frac{ (1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2}{6} = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} - (3.5)^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} = 2.91666666\ldots[/math]

The sum of independent random variables will have an expected value equal to a sum of the individual expected values  and a variance equal to the sum of the individual variances, so  the mean of 1000 die rolls will have

• expected value [math]1000\cdot  3.5 = 3500[/math]
• variance [math]1000\cdot \frac{35}{12}= \frac{8750}{3} = 2916.6666\ldots[/math]

In addition, the distribution of this sum will be approximately a normal distribution, with mean 3500 and standard deviation equal to the square root of the variance, i.e. [math]\sqrt{\frac{8750}{3}} = 54.00617\ldots[/math], almost exactly 54.

With a normal distribution, you expect results within one standard deviation (above or below) the mean about 68% of the time, within two standard deviations about 95% of the time, and within three standard deviations about 99.7% of the time, see   68–95–99.7 rule

 3698 is 198 above the mean, which is [math]3.666\ldots [/math]standard deviations above the mean. You'd expect something that far away from the mean (or more) only about .025% (that's[math] \frac{1}{40} [/math]of one percent) of the time, and if you restrict yourself to only OVERshooting the mean (i.e. 3698 or higher), it would only happen half as often, once every eight thousand or so tries -- that is, if you repeatedly ran the experiment of rolling a fair die 1000 times and recorded the sum, you'd expect to get a sum as large or larger than 3698 about once every 8000 experiments, though you'd also expect to have a sum of 3302 or smaller once every 8000 experiments, too.
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