Mean, Median, Mode, and Standard Deviation Explained

• Mean: the average. Add all values and divide by how many there are.

$$\bar{x} = \frac{\sum x_i}{n}$$

• Median: the middle value after you sort the data. With an even number of values, average the two middle ones.
• Mode: the value that appears most often. A data set can have one mode, more than one mode, or no mode.

Quick median example with an odd count: for 3, 7, 1, 9, 5, sort to get 1, 3, 5, 7, 9. The median is 5.

Eight students scored these points on a 10-point quiz:

$$2, \ 4, \ 4, \ 4, \ 5, \ 5, \ 7, \ 9$$

Mean

1. Add the values: $2 + 4 + 4 + 4 + 5 + 5 + 7 + 9 = 40$.
2. Divide by the count: $40 \div 8 = 5$.

Median

1. The data is already sorted. There are 8 values, so the middle two are the 4th and 5th values: 4 and 5.
2. Average them: $(4 + 5) \div 2 = 4.5$.

Mode

4 appears three times, more than any other value, so the mode is 4.

Standard deviation measures the typical distance of values from the mean. A small value means the data is bunched together. A large value means it is spread out.

Using the same quiz scores (mean = 5), treat the eight students as the whole population:

1. Subtract the mean from each value: $-3, -1, -1, -1, 0, 0, 2, 4$.
2. Square each difference: $9, 1, 1, 1, 0, 0, 4, 16$.
3. Add the squares: $9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32$.
4. Divide by $N$ to get the variance: $32 \div 8 = 4$.
5. Take the square root: $\sigma = \sqrt{4} = 2$.

The population standard deviation is 2 points. Most scores fall within about 2 points of the mean of 5.

The only difference is what you divide by in step 4.

$$\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} \qquad s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$$

• Use population ($\sigma$, divide by $N$) when your data includes every member of the group you care about.
• Use sample ($s$, divide by $n - 1$) when your data is a sample used to estimate a bigger group. This is the more common case.

Dividing by $n - 1$ makes the sample standard deviation slightly larger, which corrects for the fact that a sample usually underestimates the spread.

Quiz scores as a sample: $32 \div 7 \approx 4.571$, so $s = \sqrt{4.571} \approx 2.14$.

Another example: Find both for 10, 12, 14.

1. Mean: $36 \div 3 = 12$.
2. Differences: $-2, 0, 2$. Squares: $4, 0, 4$. Sum: $8$.
3. Population: $\sigma = \sqrt{8 \div 3} = \sqrt{2.667} \approx 1.63$.
4. Sample: $s = \sqrt{8 \div 2} = \sqrt{4} = 2$.

The mean gets pulled toward extreme values (outliers). The median does not, so it is the better measure of center for skewed data.

Example: Six people's yearly incomes, in thousands of dollars: 30, 32, 35, 38, 40, 250.

1. Mean: $(30 + 32 + 35 + 38 + 40 + 250) \div 6 = 425 \div 6 \approx 70.83$.
2. Median: the middle two values are 35 and 38, so $(35 + 38) \div 2 = 36.5$.

Five of the six people earn 40 or less, yet the mean is almost 71. The median of 36.5 describes a typical person much better. That is why reports on home prices and incomes usually use the median.

Use the mean for roughly symmetric data without outliers. Use the median for skewed data or data with outliers. Use the mode for categories, like the most popular shoe size.