Normal probability plot

Sort the data, turn each rank $i$ out of $n$ into a proportion with a plotting position such as $\frac{i - 0.375}{n + 0.25}$, and convert that proportion to $z_i$ (the z-score, or standard normal score, at that proportion). Plotting the pairs $(x_{(i)}, z_i)$ with the data on the horizontal axis gives a normal probability plot. If the data came from a normal model then $z = \frac{x - \mu}{\sigma}$, a straight line in $x$, so the check is straightness.

Ten service times in minutes: 2.1, 2.2, 2.3, 2.4, 2.6, 2.8, 3.1, 3.6, 4.5, 7.0, with mean 3.26 and standard deviation 1.5072. The largest value's plotting position is $\frac{10 - 0.375}{10.25} = 0.939$, so it is plotted at $z = 1.55$. Standardizing it instead gives $\frac{7.0 - 3.26}{1.5072} = 2.48$. The observation is 2.48 standard deviations out where the model expects the largest of ten to be about 1.55, so the point sits well to the right of the line and the top of the plot flattens away.

That flattening is the direction people reverse. With the data horizontal and the z-scores vertical, right skew makes the plot concave down: steep on the left, flattening toward the right, because the largest values run farther out than a normal model predicts. Left skew is the mirror image, concave up. The sentence to avoid is "the plot flattens at the top, so the tail is on the left." The flattening marks the end where the values are stretched out, and the stretched-out end is the long tail, so a plot that flattens at the high end is skewed right.

Two boundaries. Some software puts the data on the vertical axis, which mirrors the whole picture and reverses the concavity, so read the axis labels before reading a direction. And a small sample wanders off the line by chance, so straightness is a reason not to worry rather than a proof: no plot can establish normality.