Center and Spread

One of my favorite statistical questions is also one of the least useful without context: What is the average?

Consider five annual incomes, measured in thousands of dollars:

$$40,\quad 45,\quad 50,\quad 55,\quad 60.$$

The arithmetic mean adds the observations and divides by their number:

$$\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i =\frac{40+45+50+55+60}{5}=50.$$

The mean is the balancing point of the values. It uses every observation, which is both its strength and its vulnerability.

The median tells another story

The median is the middle value after sorting the observations. Here it is also 50.

But imagine the fifth person wins a tremendous prize. Replace 60 with 600. Our data become:

$$40,\quad45,\quad50,\quad55,\quad600.$$

The mean is now 158. The median remains 50. Neither calculation is wrong, but they answer slightly different questions.

The mean describes a mathematical balance point and is strongly affected by extreme observations. The median describes the middle of the ordered observations and is generally more resistant to outliers.

If there are an even number of observations, the median is conventionally the average of the two middle values.

Spread: how different are the observations?

A center without a description of variation is often misleading.

The simplest measure of spread is the range:

$$\text{range}=\max(x)-\min(x).$$

A more sophisticated measure is the sample standard deviation, $s$:

$$s=\sqrt{\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}}.$$

The square root returns the answer to the original units. Dividing by $n-1$, rather than $n$, produces the familiar sample variance estimator used to estimate population variance under standard assumptions.

Standard deviation is not the average distance from the mean. It is the square root of the average squared deviation with that sample correction.

Standard deviation, like the mean, can be strongly affected by extreme values. For heavily skewed data, the median and interquartile range may give a more representative description of a typical observation.

Experiment with an outlier

Below, change the final observation. The mean, median, and sample standard deviation will update immediately.

Try setting the final value to 60, then to 600. Ask yourself which summary best represents a typical member of this tiny group, and why.

Check your understanding

1. For $2,4,6,8$, calculate the mean and median.
2. If the largest observation increases by 100, what happens to the mean of a five-observation dataset?
3. Can two datasets share the same mean but have dramatically different variability?

Suggested responses

1. Both are 5. 2. The mean increases by 20. 3. Yes: compare 5, 5, 5 with 0, 5, 10. Both have mean 5, but the latter is more variable.

Why the mean and median solve different problems

For any proposed center $c$, expand $x_i-c=(x_i-\bar x)+(\bar x-c)$. Summing squared deviations gives

$$\sum_i(x_i-c)^2=\sum_i(x_i-\bar x)^2+n(\bar x-c)^2,$$

because centered deviations sum to zero. The mean uniquely minimizes squared error. A median instead minimizes $\sum_i|x_i-c|$; moving a proposed center across an ordered observation changes how many distances increase and decrease. For an even sample, any point between the two middle values minimizes absolute error; the midpoint is the conventional reported median. Neither center is universally correct without specifying what kind of loss or description matters.

For the original five incomes, squared deviations from 50 are $100,25,0,25,100$, summing to 250. The sample variance is $250/4=62.5$ and the standard deviation is $\sqrt{62.5}\approx7.9057$ thousand dollars. The variance has squared-income units; the standard deviation returns to income units. For a descriptive population consisting of exactly these five people, dividing by five instead gives population variance 50. The choice of denominator reflects the intended quantity, not a universal preference for one button.

The $n-1$ sample variance is unbiased for population variance under independent identically distributed finite-variance sampling. Its square root is not generally an unbiased estimator of population standard deviation. That distinction matters when interpreting the word “unbiased.”

Additional practice

1. Calculate the mean, median, sample variance, and range for $[1,2,3,10]$.
2. Add 100 to every observation. What changes? Multiply every value by $-2$ instead; how does the standard deviation change?
3. Explain which summary you would report for typical income and which for total income per person.

Worked additional solutions

1. Mean 4, median 2.5, squared deviations $9+4+1+36=50$, sample variance $50/3$, and range 9.
2. Adding 100 shifts centers but leaves spread unchanged. Multiplying by $-2$ multiplies standard deviation by 2 and variance by 4; it also reverses order.
3. Median may better describe a middle person’s skewed income. The mean multiplied by population size reproduces the total. Report distributional context rather than asking one statistic to answer both questions.