Two task series took ten minutes on average. Was their timing equally predictable? The mean describes the center; spread describes distances from it. Comparison needs both. Use roots, expectation and sample data.

Take invented teaching times in minutes: A=[9,10,10,10,11], B=[6,8,10,12,14]. Each has five values summing to 50.

Three ways to locate the center

The arithmetic mean is sum divided by count. Denote it m:

m = (x_(1) + ⋯ + x_(n))/(n).

Both give m=10 min. This describes observed data, without establishing the expectation of an unknown process. If we uniformly choose one of the five recorded values, the expectation of that separately defined model equals m.

The median of ordered numbers is the middle value for odd counts; for even counts it is the average of the two middle values. In [8,9,11,20] it is (9+11)/2=10, whereas the mean is 12. A and B both have median 10.

The mode is the most frequent value: 10 in A. In B each value appears once, so none stands out in frequency. There may be several modes; state the convention used for ties. Measures of center.

A · five values

9,10,10,10,11

051015m=10min

B · five values

6,8,10,12,14

051015m=10min
Each dot is one teaching observation in minutes; duplicates stack vertically, with no time encoded by height. Shared horizontal scale 0–15 min; dashed m=10.

How far from the center?

The range is maximum minus minimum: 2 min for A, 8 for B. It uses only extremes. For all deviations from m, compute the descriptive variance of these n numbers and its square root:

D = (1)/(n)((∑)_(i = 1))^(n)((x_(i) − m))^(2), σ = √(D).

Ordinary deviations sum to zero even for widely spread data; squares prevent signs cancelling. D has units min², while σ restores minutes.

D_(A) = (1 + 0 + 0 + 0 + 1)/(5) = 0.4, D_(B) = (16 + 4 + 0 + 4 + 16)/(5) = 8.

Thus σ_A=√0.4≈0.6325 min and σ_B=√8≈2.8284 min. This measures spread; it does not guarantee every value is within σ of the mean. OpenStax

Why sometimes divide by n−1?

To estimate population variance from an independent, identically distributed sample with finite variance, use the corrected estimator:

(s)^(2) = (1)/(n − 1)((∑)_(i = 1))^(n)((x_(i) − m))^(2) (n > 1).

For the same values, s²_A=2/4=0.5 min² and s²_B=40/4=10 min². This answers a different question from descriptive D with divisor n; identify your convention. The correction does not remove A046’s selection bias. Sample variance formula, NIST.

Finally replace only 14 in B with 50: the mean becomes 17.2, while the median stays 10. Check extreme values in context instead of deleting them automatically. Report count, center, spread, units and collection method: one number cannot hold them all.