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?
What is the P( |X-10| > 2) of a normal distribution when the mean is 10, and the standard deviation is 6?
2
$\begingroup$
probability
statistics
-
0We 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
1 Answers
2
$P(|X-10| > 2)=P(X>12)+P(X<8)=\int_{12}^\infty\cal{N}(x)dx+\int_{-\infty}^8\cal{N}(x)dx$ where $\cal{N}$ is the Gaussian density.
-
0@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