# Normally distributed random variables and probability?

• Let $X$ be a normally distributed random variable with mean μ equal to the standard deviation . Then $P (X > 0)$ is $A$ $0$, $B$ $0.3413$, $C$ $0.5$, $D$ $0.8413$

the answer is $D=0.8413$ .. So far what I got was:

$z= \frac{0-μ}{\sqrt{σ}}$, which gives me $z=-\sqrt{σ}$ ? Where should I proceed from there?

Also if $D=0.8413$, $P (X > 0)= 1- P ( X < 0)= 0.8413$ so $P ( X < 0) =0.1587$ which is $Z= - 1.00$ from the tables? but again... how do I get there?

Remember that $\mu = \sigma$ for this question: $$P(X>0)=P\left(\frac{X-\mu}{\mu}>\frac{-\mu}{\mu}\right)=P(Z>-1)$$

• ohhhh I completely mistook the standard deviation for the variance my bad. thank you! Mar 31 '17 at 15:52

The $Z$-score of a value of a normal variable can be interpreted as "number of standard deviations above the mean."

For example, if the mean is $10$ and the standard deviation is $2,$ then $12$ has a $Z$-score of $1$ (it is exactly one standard deivation above the mean) whereas $7$ has a $Z$-score of $-1.5$ (it is $1.5$ standard deviations below the mean).

Now you have a mean $\mu$ and standard deviation $\mu,$ so $0$ is (how many) standard deviations above the mean?

Notice that we always get a negative $Z$ for anything below the mean; and $\mu$ must be positive, since it is the standard deviation of the distribution as well as its mean.