Range vs Standard Deviation

The range reports spread as one number: the maximum minus the minimum. It carries the units of the data, so a range of 36 points and a range of 36 seconds are different statements, and it can never be negative. Only two observations enter the calculation, so everything between them is invisible to it.

Take five test scores: 62, 71, 78, 84, 98. The maximum is 98, the minimum is 62, and the range is $98 - 62 = 36$ points. Now replace the middle three scores with 63, 64, and 97. The range is still 36, even though the data now sit in two clumps at the ends instead of spreading out evenly. The range cannot tell those two classes apart.

The answer that costs marks is an interval. "The range is 62 to 98" says where the data live, which is a fine sentence in ordinary English and is not the statistic. The range is 36. If a question asks for the range and you write two numbers, you have reported the minimum and the maximum, which a boxplot already shows.

Two consequences of using only the extremes are worth holding onto. Adding an observation can never shrink the range, only leave it alone or widen it, so a larger sample from the same population tends to report a larger range even when nothing about the spread has changed. That makes the range a weak way to compare data sets of different sizes. And one mistyped value sets it outright: change that 98 to 980 and the range jumps to 918 while the median stays at 78.

The range is one of the summary statistics for a single quantitative variable in Unit 1 topic 1.7, alongside the interquartile range and the standard deviation. Both of those survive an extreme value far better, which is why the range is usually quoted as context rather than relied on as the measure of spread.

Standard deviation reports spread as a typical distance between a value and the mean, carried in the units of the data. For a sample it is $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ are the observations, $\bar{x}$ (x-bar) is the sample mean, and $n$ is how many values there are. For a whole population the symbol becomes $\sigma$ (sigma), the mean becomes $\mu$ (mu), and the divisor is $n$ instead of $n - 1$. The word typical is loose on purpose: $s$ is the square root of an average squared distance, not the average of the distances.

Take the five values 4, 8, 11, 13, 14. The mean is $50/5 = 10$, so the deviations are -6, -2, 1, 3, and 4. Squared they are 36, 4, 1, 9, and 16, which sum to 66. Divide by $n - 1 = 4$ for the variance $s^2 = 16.5$, then take the square root: $s = 4.06$. Treating the same five numbers as a whole population instead gives $\sigma = \sqrt{66/5} = 3.63$, so settle that question before starting.

The sentence to unlearn is "the standard deviation is the average distance from the mean." For those five values the average distance really is $(6 + 2 + 1 + 3 + 4)/5 = 3.2$, which is the mean absolute deviation, and it is not 4.06. Squaring before averaging gives far-out values more weight, so unless every value sits the same distance out, $s$ lands above the plain average distance.

A standard deviation is never negative, and it equals 0 in exactly one case: every value in the set is identical, so every deviation is 0. It is also not resistant. Change that 14 to a 44 and the median stays at 11 while $s$ goes from 4.06 to 16.02, because one distance of 28 becomes 784 inside the sum and swamps the other four.

Report it with the variable and the units attached, never as a bare number. In the Fall 2026 course this sits in Unit 1, whose topic 1.7 is titled Summary Statistics for One Quantitative Variable.

Finance uses this exact statistic as its measure of risk. The standard deviation of an investment's period returns is computed by the same steps used here and is the number quoted as an asset's risk: standard deviation of returns.