
Statistical Analysis
Statistical analysis sounds like some harrowing, difficult
course your older brother took his first semester in college. You
may be asking yourself, “If college students have difficulty with
statistical analysis, how can the SAT expect me, a high school student,
to answer these items correctly?” Relax: SAT statistical analysis
and collegelevel statistical analysis are two completely different
worlds, and we’re not just talking about the better parties.
Statistical analysis items on the SAT always include a data
set, which is a collection of measurements or quantities.
An example of a data set is the set of math test scores for 15 students
in Ms. Kronhorst’s sixthgrade class:
88, 78, 84, 90, 94, 90, 68, 80, 94, 98, 84, 90,
74, 92, 86
You will be asked to find one or more of the following
values:
 Arithmetic mean
 Median
 Mode
 Range
Arithmetic Mean, a.k.a. Average
When people throw out the word average in
everyday conversation, they are talking about arithmetic mean.
Both terms mean the same thing. Mean is the most
commonly tested statistical analysis concept on the SAT. The actual
formula for finding the average is not complicated. It’s the sum
of all the elements contained in a data set, divided by the number
of elements in the set:
Take another look at the test scores of the 15 students
in Ms. Kronhorst’s class. We’ve sorted scores in her class from
lowest to highest:
68, 74, 78, 80, 84, 84, 86, 88, 90, 90, 90, 92,
94, 94, 98
To find the arithmetic mean of this data set, sum the
scores, then divide by 15, because there are 15 students in her
class:
Unfortunately, the SAT is not going to be this direct
when it tests mean. The SAT likes to trick you. Just remember the
formula above and the three pieces it is divided into:
 Mean
 Sum of elements
 Number of elements
To solve for any one piece, the other two pieces must
be supplied in the item, one way or another. Let’s take a look at
an example:

This item appears to be some type of timeconsuming, trialanderror monster
in which you begin to choose random numbers to see whether the average
will equal 28. But let’s take a look at the pieces provided to us in
the item. Both the mean and the number of elements are
given, and by placing these pieces into the formula above, we come
up with the equation:
In solving this equation, we discover that the sum of
the elements must equal 284, or 112. So if the sum is 112,
we can then solve for the fourth number by writing out:
11 + 32 + 50 + fourth number = 112
Solving for the fourth number is easy. All you have to
do is subtract the sum of 11, 32, and 50 from 112:
fourth number = 112 – (11 + 32 + 50) = 112 – 93
= 19.
That’s choice C.
Always keep in mind that all arithmetic mean items revolve
around the three pieces of the formula. Two pieces must
always be given in some form in order to solve for the
third. Once you get this down, you can solve pretty much every item
that asks about arithmetic mean on the SAT.
Look at this example:

This item seems especially difficult because we aren’t
given any of the values of the numbers being averaged. So what?
We don’t need those numbers to solve this item. Because we are looking
for the average of two numbers, we only need to know their sum.
If you remember the formula, you’ll only need to
look at what information is given to you in the item. If the average
of six numbers is 9.5, then the sum of the six numbers must equal
9.56, or 57. The other information given
to you is that the average of four numbers is 7. Therefore, the
sum of those four numbers must equal 74, or 28. Now, if the sum of all
six numbers is 57 and the sum of four of those numbers is 28, then
the sum of the remaining two numbers must be 28 + x =
57, or 29. If the sum of two numbers is 29, then their average will
be , or 14.5. That’s choice E.
Tricky, but doable.
Another tricky mean item the SAT likes to use is the changing
mean. Those items look like this:

Remembering the formula, use the information provided
in the item to solve for the unknowns. If the average age of 8 members
is 18, then the sum of their ages must be 188, or 144. When one more member is added,
the number of members increases to 9 and our mean increases to 22.
Therefore, the new sum of ages must be 229, or 198. Because the addition of
the new member raised the sum of ages from 144 to 198, the new member’s
age must be the difference between the two sums, or 54, choice D.
Kind of old to be joining a Dungeons & Dragons club,
but some people are just late bloomers.
Commit the arithmetic mean (average) formula to memory.
Remember, two of the three pieces must be supplied to you in the
question. Find out what they are and solve for the third.
Median
Have you ever been on a road trip with your family and
noticed a highway sign that reads “Keep Off the Median”? Did you
ever wonder what that meant? On a highway, the median is that strip
of grass in the middle separating the two directions of traffic.
The median serves the same purpose in statistical analysis.
It is the middle number in any given data set. Let’s look at the
hypothetical set Q, which equals {12, 5, 7, 4,
5}. Rearrange the numbers by order of value, and you get:
Set Q = {4, 5, 5, 7, 12}
Because there is an odd number of elements, the median
is the number directly in the middle, 5.
If a set has an even number of elements, simply take the
average of the middle two. For example:
Set R = {4, 5, 5, 7, 9,
12}
In this case, the middle number is between 5 and
7, so we simply add the two numbers together and divide by 2: . Your
median is 6, even though the number 6 doesn’t appear in the set.
Mode
The mode is the number within a set that
appears most frequently. In the set {6, 8, 6, 3, 10}, the mode is
6 because it appears twice, and all the other numbers appear only
once. It is possible to have more than one mode if two or more numbers
appear at the same, highest frequency. The set {23, 14, 21, 14,
18, 23, 20} has modes of 14 and 23, because both appear twice. In
any given set, if all the elements appear an equal number of times,
then there is no mode.
Range
The range measures the difference between
the smallest and largest element of a given data set. Remember
the test scores in Ms. Kronhorst’s class:
{68, 74, 78, 80, 84, 84, 86, 88, 90, 90,
90, 92, 94, 94, 98}
The range is 98 – 68 = 30.
