SAT statistics: mean, median, spread, and outliers

01Calculate the center from the actual data

For the data $2,4,4,6,9$, the mean is $(2+4+4+6+9)/5=5$. The median is 4 because it is the middle value in the sorted list. The mode is 4 because it occurs most often. The range is $9-2=7$.

These measurements answer different questions. The mean balances the whole list. The median marks its center by position. The mode identifies the most frequent value. A set may have several modes or no uniquely most frequent value.

With an even number of observations, average the two middle values. For $2,4,6,9$, the median is $(4+6)/2=5$. Sort first; the middle entries in an unsorted display aren't necessarily the middle of the data.

02Use mean times count to recover a total

Six quiz scores have a mean of 82, so their total is $6(82)=492$. After a seventh quiz, the mean is 84, so the new total is $7(84)=588$. The seventh score is $588-492=96$.

This is easier than making up the six individual scores. The mean gives enough information about their combined total. Keep both counts in the setup; using six for the new total would ignore the added quiz.

If a question asks how much an added value changes the mean, remember that both the total and the denominator change. Adding a value above the old mean raises it. Adding the old mean leaves it unchanged.

03Weight group means by their group sizes

One class has 10 students with a mean score of 70. Another has 30 students with a mean score of 90. The combined mean is $\frac{10(70)+30(90)}{40}=85$.

Averaging 70 and 90 to get 80 would give the two classes equal weight, even though the second has three times as many students. The combined mean should be closer to 90, which gives you a quick reasonableness check.

In a frequency table, multiply each value by its frequency before adding. The count is the sum of the frequencies, not the number of table rows. Three rows may represent forty observations.

04Read a frequency-table median by position

For this table, there are 10 observations. The fifth and sixth observations are both 3, so the median is 3. The mean is $(1\cdot 2+3\cdot 5+8\cdot 3)/10=4.1$.

05An outlier can move the mean without moving the median

Start with $2,4,4,6,9$. Replacing 9 with 99 raises the mean from 5 to 23, but the median remains 4. The middle observation did not move. The range rises from 7 to 97.

That example explains why you might prefer the median when describing a typical value in a data set with an extreme observation. It doesn't mean the median can never change when an outlier changes. Its behavior depends on which positions are affected.

For two groups with equal means, compare their spread. The sets $4,5,6$ and $1,5,9$ both have mean 5, but the second set lies farther from 5 and has the larger standard deviation. You can make that comparison without calculating a formula.

06Know what shifting and scaling do

Adding 7 to every observation raises both mean and median by 7. It leaves range and standard deviation unchanged because all distances between values stay the same.

Multiplying every observation by 3 multiplies the mean and median by 3, and multiplies range and standard deviation by 3. For a negative multiplier, the order reverses; spread scales by the multiplier's absolute value.

Use statistics practice for both numerical and visual questions. For a dot plot, count the dots before calculating an average. For a histogram, check whether the bars describe exact values or intervals: an interval doesn't tell you each observation's exact value. Name what the graph actually supports before doing arithmetic.