Probability of random variable in Normal Distribution I've been talking to my lecturer about choosing random values from a Normal Distribution and he says the following: "Roughly 68% of expected values  $ \in (\mu-\sigma,\mu + \sigma)$ does not imply that every value within this range is equally probable." He says that I need to substitute x,mu and sigma into the PDF to get the Probability of x, where x is some random value. Does this make sense?
Thanks
 A: Yes, in general the probability of a random variable $X$ being in the interval from $x_0$ to $x_1$ is given by:
$$ P(x_0 < X < x_1) = \int_{x_0}^{x_1} f(x)\;dx $$
This represents the area under the graph of the pdf $f(x)$. Also note that it doesn't make sense with a continuous distribution to talk about the probability $P(X=x)$. This is zero because the limits of the integral are the same.
In the case of the normal distribution, the pdf is shown below for the standard normal distribution $\mu=0,\sigma=1$, denoted by $Z\sim N(0,1)$.

Here the area between the red lines $\approx0.68$ representing $P(-1 < Z < 1)$. The area between the green lines represents the probability $P(-0.8 < Z < -0.6) \approx 0.06$. Whereas the area between the blue lines represents the probability $P(-0.1 < Z < 0.1) \approx 0.08$. You can see graphically and numerically that these last two areas / probabilities are not the same, despite having the same width of $x$ values, justifying your lecturer's statement.
The probability can also be expressed in terms of the cumulative distribution function, and very often is for the normal distribution.
$$\begin{align}
P(x_0 < X < x_1)
&= \int_{-\infty}^{x_1} f(x)\;dx - \int_{-\infty}^{x_0} f(x)\;dx\\
&= F(x_1) - F(x_0)
\end{align}$$
where $F(x)$ is the cumulative distribution function of $X$. In the case of the normal distribution, the cumulative distribution function is $\Phi(x)$ and it is usual to use tables to lookup probabilities. If you plot $\Phi(x)$, you can see that it isn't a straight line, so the probability doesn't increase uniformly, once again justifying your lecturer's statement.

