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The normal distribution is a bell shape, symmetric about its mean $\mu$. Its spread is controlled by the standard deviation $\sigma$: a larger $\sigma$ makes the curve wider and flatter. We write $X\sim N(\mu ,{\sigma }^2)$ to say that the random variable $X$ follows a normal distribution with mean $\mu$ and variance ${\sigma }^2$. Note that the second number in the brackets is the variance (the square of the standard deviation). The total area between the curve and the x-axis is $1$, matching total probability.

There are infinitely many normal distributions, one for each pair $\mu$ and $\sigma$, but standardising collapses them all onto one standard normal distribution $N(0,1)$. The $z$-score re-measures a value $x$ as the number of standard deviations it lies from the mean. The sign of $z$ tells you whether $x$ is above or below the mean, and its size tells you how far. For example, $z=2$ means two standard deviations above the mean.

In the standard normal distribution, the probability of being at most $z$ is written $\Phi (z)$, read from a table (or a calculator). Useful values to know are $\Phi (1)=0.8413$, $\Phi (1.5)=0.9332$ and $\Phi (2)=0.9772$. To find $P(X<x)$, simply compute $z=\frac{x-\mu }{\sigma }$ and read off $\Phi (z)$. For a 'greater than' probability, use $P(X>x)=1-\Phi (z)$.

Standard normal tables usually list only $z>0$. For negative $z$, use the symmetry of the curve about $0$: $\Phi (-z)=1-\Phi (z)$. For example, $\Phi (-1)=1-0.8413=0.1587$. The probability of an interval, $P(a<X<b)$, is found by standardising both ends and computing $\Phi (z_b)-\Phi (z_a)$. For a symmetric interval about the mean, $P(\mu -k<X<\mu +k)$, this simplifies neatly to $2\Phi (z)-1$.

Some problems run the other way: given a probability, find the value $x$ — for instance, 'what value lies at the $97.72\%$ point from below?' The method reverses standardising. First look up the $z$ that gives the stated $\Phi (z)$ (e.g. $\Phi (z)=0.9772$ gives $z=2$), then rearrange $z=\frac{x-\mu }{\sigma }$ into $x=\mu +z\sigma$. Remember that if the probability is below $0.5$, the $z$-value is negative.