What is the $P( |X-10| > 2)$ of a normal distribution when mean is 10, and standard deviation is 6?

Question

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?

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 $\Pr(Z\le\frac{2}{6})$.

Seyhmus Güngören · Accepted Answer · 2012-10-25 19:37:58Z

up vote 2 down vote accepted

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

answered Oct 25 '12 at 19:37

Seyhmus Güngören
2,4301314

	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

1 Answer