# Calculating $P(X-E(X)\geq \mathrm{SD})$ for a normal random variable $X$

If you'd have to calculate the following probability of an $X\sim \mathrm{Normal}$: $$P(X-E(X)\geq \mathrm{SD}),$$

how would you calculate it? And if the calculation will be for $4\cdot\mathrm{SD}$, how does it matter?

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Hint: If $X\sim\mathcal{N}(\mu,\sigma^2)$, then $$\frac{X-\mu}{\sigma}\sim\mathcal{N}(0,1).$$

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ok thanks, got that. so i'm getting the 1-Ф(1) for the SD. what are the news in the same calcultaion for 4*SD?it there anything differnt? –  adamco Jan 11 '13 at 16:31
The exact same calculations applies: $$P(X-\mu\geq 4\sigma)=P\left(\frac{X-\mu}{\sigma}\geq 4\right)=1-\Phi(4)$$ –  Stefan Hansen Jan 11 '13 at 16:33
thank you very much! –  adamco Jan 11 '13 at 16:36
@adamco One difference is that many tables of the values of $\Phi(x)$ give values only for $x \in (0, 3.0)$ or $x \in (0, 3.5)$ and so you may need to get more extensive tables or look for a calculator (there are many available, even on-line) that will tell you the value of $\Phi(4)$. –  Dilip Sarwate Jan 11 '13 at 18:20
Note, for large values in $\Phi$ (i.e. > 3) asymptotic approximations start to be pretty accurate for most uses. –  John Jan 11 '13 at 20:46