Normal Distribution: Probability of a Negative Value

Question

The random variable $X$ can take negative and positive values. $X$ is distributed normally with mean $3$ and variance $4$. How can I find the probability that $X$ has a negative value?

Answer 1

My goodness! Such complications! This is a beginning problem about the normal distribution.

We have $X \sim Norm(\mu = 3, \sigma = 2).$ You want $P\{X < 0\},$ which can be found by 'standarization' and referring to printed normal tables, or by using statistical software.

Specifically, you should view this as $P\{X < 0\} = P\{Z < (0 - 3)/2\}.$ where $Z$ is standard normal.

In R statistical software, you would get more decimal places (not needed for your purposes) than from a printed table:

 pnorm(0, 3, 2)
 ##  0.0668072

Leave a Comment if you need more help. I will check in 10 hrs or so.

Answer 2

First you take the value of $\mu=3>0$ as starting point and let $x \geq 0$ . Then you add the posititve value of the difference between $-x$ and $\mu=3$. In total it is $\mu+|\mu-(-x)|=3+|3-(-x)|=3+|3+x|=6+x$. Because of the symmetry of the normal diststribution it is $f(-x)=f(6+x)$.

Now you can use the concept of the converse probability:

$P(X\leq -x)=1-P(X\leq 6+x)$, where $X\sim \mathcal N(3 , \ \sigma ^2)$