Missing Values and Outliers

The mean works backward neatly because it comes from a sum. The median, range and mode do not, so go back to what each one depends on.

• Median: it depends on the order of the values. Put the values in order with $x$ in its place, then write what the middle is.
• Range: it depends only on the two end values. If the given range is greater than the range of the known values, $x$ must be a new greatest value or a new least value. That leaves two candidates, so you need one more clue to choose.
• Mode: it depends on how often each value appears. Take $5, 8, 2, 8, 5, x$ with exactly one mode, $5$. In the known values, $5$ and $8$ each appear twice, which is a tie. If $x = 5$, then $5$ appears three times and is the only mode. If $x = 8$, the mode is $8$. Any other $x$ leaves at least two values tied for most often. So $x = 5$.

The table gives one short example for each measure.

Now turn to unusual values. An outlier is a data value that is much greater or much less than the other values in its data set. The word much is vague, so this lesson uses one common check built from the quartiles, which you already know how to find:

• Order the data. Split it at the median into a lower half and an upper half. When the count is odd, the median is in neither half.
• $Q_1$, the first quartile, is the median of the lower half. $Q_3$, the third quartile, is the median of the upper half.
• The interquartile range is $\text{IQR} = Q_3 - Q_1$. It is the spread of the middle half of the data.

The check uses two boundary values called fences. The lower fence is $1.5 \times \text{IQR}$ below $Q_1$, and the upper fence is $1.5 \times \text{IQR}$ above $Q_3$, as the next two equations show. Under this fence check, a value is an outlier when it is less than the lower fence or greater than the upper fence. A value on a fence or between the fences is not an outlier.

Read down the table. The mean climbs as the longest time grows: $16$, then $19$, then $22$. In the last row the total is $96 + 58 = 154$. The median moves from $15.5$ to $16$ and then stays there. The mode stays at $15$.

Here is why each measure of center acts this way:

• The mean uses the size of every value, so an extreme value shifts the total, and all the values share that shift. Including a value greater than every other value always raises the mean. Including a value less than every other value always lowers it.
• The median uses only the order of the values. As long as the outlier is not a middle value, making it more extreme does not change the median. Taking it away shifts the middle by half a position, which is why the median went from $16$ to $15.5$.
• The mode uses only how often each value appears. Here $37$ appears once and $15$ appears twice, so the mode is $15$ with or without the outlier.

With the outlier, the mean of $19$ minutes is as long as or longer than six of the seven trips, so here it no longer describes a typical trip well. The median of $16$ minutes is still the middle trip.

Takeaway. A measure follows a rule, and you can use that rule in both directions.

• Missing value: write the rule with $x$ in the data set, find the value that fits, and test it. For the mean, the total counts $x$ too.
• Outlier: find the fences from $Q_1$, $Q_3$ and the IQR, using the lower and upper halves of the ordered data (the median is in neither half when the count is odd). Then find each measure of center with and without the outlier.

The mean uses the size of every value, so an outlier greater than every other value raises it, and one less than every other value lowers it. The median uses only order, so an outlier that is not a middle value can become more extreme without changing the median.