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An old textbook says the range of data can be estimated as 6 times the standard deviation. If the data is normally distributed what percentage of the data is within the range?

By 'range of data', does the question mean biggest - smallest? In that case, how do I write an expression for it if it's normally distributed?

Jim
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  • The range is supposed to contain 'all' the data. But if your data is normally distributed, only 'few' points will lie outside of your mean plus or minus three standard deviations. Thus it is an estimate and not the actual range of data. – Bruce Zheng Oct 08 '14 at 19:20

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Let $X$ be normally distributed with mean $\mu$ and standard deviation $\sigma$. Then $$\Pr(\mu-3\sigma\le X\le \mu+3\sigma)=\Pr\left(\left|\frac{X-\mu}{\sigma}\right|\le 3\right)=\Pr(|Z|\le 3),$$ where $Z$ is standard normal. From tables, we find that $\Pr(Z\le 3)\approx 0.99865$. Thus $$\Pr(\mu-3\sigma\le X\le \mu+3\sigma)\approx 2(0.99865-0.5)=0.9973.$$ Informally, the probability that $X$ is more than $3$ standard deviation units away from the mean is $\approx 2.7\times 10^{-3}$, quite small. Thus only a small proportion of observations will lie outside the interval $\mu\pm 3\sigma$, which has width $6\sigma$.

André Nicolas
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  • why did you write 2* (0.99865-0.5) in line 4. May I ask how you got the 0.5? @André Nicolas – CCC Jan 03 '16 at 04:22
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    We have $\Pr(Z\le3)\approx 0.99865$. So $\Pr(0\le Z\le 3)\approx 0.99865-0.5$, since $\Pr(Z\lt 0)=0.5$. So by symmetry $\Pr(-3\le Z\le 3)\approx2(0.99865-0.5)$. The reason I did it this way is that the usual tables for the standard normal only give $\Pr(Z\le k)$ for $k$ from $0$ to about $3.5$. Of course if you are using software you can usually get the answer directly without going through this extra calculation. – André Nicolas Jan 03 '16 at 04:35