Normal Probability Plot vs Histogram

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.

A histogram cuts the number line into bins of equal width and draws a bar over each bin whose height is the count of values inside it, or the relative frequency. The bars touch because the axis is a continuous scale rather than a list of labels. Every histogram needs a boundary rule, and the usual one puts a value in the bin whose lower bound it meets or exceeds and whose upper bound it falls below.

Thirty exam scores binned by tens give 9 scores in 70 to 80 and 12 in 80 to 90, and three of those students scored exactly 80. Switch to the other convention, where a value belongs to the bin whose upper bound it equals, and those three move down: the counts become 12 and 9, the tallest bar changes, and so does the shape you would report. Nothing in the drawn picture tells you which rule was used.

The misreading is "twelve students scored 85." The bar over 80 to 90 says twelve students scored somewhere in that interval and nothing more. The individual values have been thrown away, which is also why you cannot read a median off a histogram, only the bin the median lands in.

Equal width is doing real work in the definition. When bins differ in width, height stops encoding frequency fairly: a bin of width 20 holding 12 values and a bin of width 10 holding 9 draw bars of height 12 and 9, so the wide one looks bigger, while the densities are $12 / 20 = 0.6$ and $9 / 10 = 0.9$ values per unit and the narrow bin is the more crowded one. With unequal bins the vertical axis has to carry density and you read area, not height.

A histogram also destroys the order the values arrived in, which is what a time plot keeps, and its shape depends on the bin width you choose. Histograms are topic 1.5.