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I couldn't figure out this question: What is the $P( |X-10| > 2)$ of a normal distribution when mean is 10, and standard deviation is 6?

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We want the probability of being more than $\frac{2}{6}$ "standard deviation units" away from the mean. This is $2\Pr(Z\gt \frac{2}{6})$, where $Z$ is standard normal. tables, or equivalent software. The table will probably give $\Pr(Z\le\frac{2}{6})$. –  André Nicolas Oct 25 '12 at 19:52
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1 Answer 1

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$$P(|X-10| > 2)=P(X>12)+P(X<8)=\int_{12}^\infty\cal{N}(x)dx+\int_{-\infty}^8\cal{N}(x)dx$$ where $\cal{N}$ is the Gaussian density.

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for the second term are you sure it's $P(X<8)$ not $P(2<X<8)$? –  user133466 Oct 25 '12 at 19:39
    
@user133466 take $X=-10$, then $|-10-10|=20>2$. Assume $\cal{N}$ is discrete, if $2<X<8$ then you dont account for $P(X=10)$ although $-10$ satisfies $P(|X-10|>2)$ –  Seyhmus Güngören Oct 25 '12 at 20:20
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