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The empirical rule is 68 - 95 - 99.7. 68% is the area for +/- 1 standard deviation (SD) from the mean, 95% is the area for +/- 2 SD from the mean; and 99.7% is the area for +/- 3 SD from the mean.

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Q: What is the empirical rule for 1 and 2 standard deviations?
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State the main reason for using the empirical rule rather than chebyshevs theorem?

The empirical rule can only be used for a normal distribution, so I will assume you are referring to a normal distribution. Chebyshev's theorem can be used for any distribution. The empirical rule is more accurate than Chebyshev's theorem for a normal distribution. For 2 standard deviations (sd) from the mean, the empirical rule says 95% of the data are within that, and Chebyshev's theorem says 1 - 1/2^2 = 1 - 1/4 = 3/4 or 75% of the data are within that. From the standard normal distribution chart, the answer for 2 sd from the mean is 95.44% So, as you can see the empirical rule is more accurate.


Is standard deviation the arithmetic mean of the squared deviations from the means?

No. Standard deviation is the square root of the mean of the squared deviations from the mean. Also, if the mean of the data is determined by the same process as the deviation from the mean, then you loose one degree of freedom, and the divisor in the calculation should be N-1, instead of just N.


What is the formula to determine the standard deviation of the resulting normal distribution when adding two normal distributions with different means and standard deviations?

s= bracket n over sigma i (xi-x-)^2 all over n-1 closed bracket ^ 1/2


What is the probability of randomly selecting values more then 2.57 standard deviations away from the mean in either a positive or negative direction in a normal distribution?

1% total 0.5% in either direction


What percent of data falls between 1 Standard deviation below and 2 stand deviations above the mean?

The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%