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I couldn't figure out this question: What is the $P( |X-10| > 2)$ of a normal distribution when the mean is 10, and the 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 $\Pr(Z\le\frac{2}{6})$.2012-10-25

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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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    @user133466 take $X=-10$, then |-10-10|=20>2. Assume $\cal{N}$ is discrete, if 2 then you dont account for $P(X=10)$ although $-10$ satisfies P(|X-10|>2)2012-10-25