Recover totals and weighted means

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Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Problem Solving and Data Analysis
Techniques: Weighted-meansFrequency-tablesRecover-totalsMissing-valuesCombined-groups

What you’ll learn

Read a frequency table as a list of repeated values.
Find a total from a mean and a count.
Find a missing value when you know the mean.
Combine groups that have different sizes or means.
Solve for an unknown frequency or group size.

Prerequisites

Find and interpret center0%
Use mean as total divided by count and interpret it as a fair-share value.

Why this matters on the SAT

Rebuild the total before you average

Sometimes the SAT hides the data. A frequency table packs repeated values into a few rows, and a group mean packs a whole group into one number. Either way, don’t average the numbers you can see as if each one showed up once. First work out how much each row or group adds to the total. In short: rebuild the total, then divide.

Solution to the example

The table stands for three trips of $12$ minutes, five trips of $15$ minutes and two trips of $21$ minutes. Together they take

12(3)+15(5)+21(2)=36+75+42=153.

There are $3+5+2=10$ trips, so the mean is

\frac{153}{10}=15.3\text{ minutes}.

Choice B is correct. The table has only three rows, but they stand for $10$ trips.

SAT example

The table summarizes the route times, in minutes, for $10$ bus trips.

Route time (minutes)	Frequency
$12$	$3$
$15$	$5$
$21$	$2$

What is the mean route time, in minutes?

A
$15$
B
$15.3$
C
$16$
D
$16.5$

Keep total, count, and mean connected

Everything here starts from one relationship:

\text{mean}=\frac{\text{total of the values}}{\text{number of values}}.

Multiply both sides by the number of values, and you get the form you’ll use most:

\boxed{\text{total of the values}=\text{mean}\times\text{number of values}}.

Think of it as three linked amounts:

The mean is the size of one equal share.
The count is how many shares there are.
The total is what all the shares add up to.

Say $8$ measurements have a mean of $13$ . Then their total is

13(8)=104.

You never needed to see the eight measurements themselves.

Check your understanding:

A team of $14$ players has a mean age of $16$ years. What is the sum of all $14$ ages?

Common mistake:

A mean is one equal share, not the whole group. When you’re given a mean and a count, multiply them before you combine that group with any other values. A quick size check helps: a total for many observations should usually be larger than a single observation.

Read a frequency table as repeated values

A frequency table is a list written in shorthand. Each row says, “This value shows up this many times.” So the table below is the list $4,\ 4,\ 7,\ 7,\ 7,\ 7,\ 7,\ 10,\ 10,\ 10$ .

Value	Frequency	Adds to the total
$4$	$2$	$4(2)=8$
$7$	$5$	$7(5)=35$
$10$	$3$	$10(3)=30$

Add what each row contributes to get the weighted total:

4(2)+7(5)+10(3)=73.

Add the frequencies to count the values. This sum is the total frequency:

2+5+3=10.

So the weighted mean is

\frac{73}{10}=7.3.

In general,

\boxed{ \text{weighted mean} = \frac{\sum(\text{value}\times\text{frequency})} {\sum \text{frequencies}} }.

Weighted means some values count more times than others. The $7$ shows up five times, so it pulls on the total harder than the $4$ , which shows up only twice.

Check your understanding:

In the table above, why do you divide by $10$ and not by $3$ ?

Common mistake:

Working out $(4+7+10)/3$ gives $7$ , not $7.3$ , because it counts each value once. The table doesn’t: it has two $4$ s, five $7$ s and three $10$ s. Multiply each value by its frequency, add those products, and divide by the total frequency.

Example: Recover a missing value

If you know the mean and how many values there are, you know the total they have to reach. Compare it with what the values you can see add up to.

Worked example

The mean of six measurements is $18.5$ centimeters. Five of the measurements are

$14,\ 17,\ 19,\ 20,\ 22$ .

What is the missing measurement, in centimeters?

A
$17$
B
$18$
C
$19$
D
$20$

Step 1

Find the total all six must reach

Six measurements with a mean of $18.5$ have to add up to

18.5(6)=111\text{ centimeters}.

Step 2

Add the five you know

Add the five measurements you’re given:

14+17+19+20+22=92\text{ centimeters}.

Step 3

The gap is the missing value

The sixth measurement has to make up the difference:

111-92=19\text{ centimeters}.

Choice C is correct.

To check, put $19$ back in:

\frac{92+19}{6}=\frac{111}{6}=18.5.

Check your understanding:

Why does subtracting the known sum from $111$ give you exactly one missing measurement?

Common mistake:

Once you subtract, the difference is already the one missing value. Means involve dividing, so it’s tempting to divide by $6$ again, but that shrinks $19$ to about $3.2$ , far too small for a set with a mean of $18.5$ . Check your answer by putting it back in and recomputing the mean.

Combine groups with unequal sizes

Say $18$ volunteers read a mean of $42$ pages, and $12$ other volunteers read a mean of $51$ pages. What’s the mean for all $30$ ? The two means don’t count equally, because the groups aren’t the same size.

Rebuild each group’s total:

\begin{aligned} \text{first-group total}&=18(42)=756,\\[1.4em] \text{second-group total}&=12(51)=612. \end{aligned}

Then add the totals, add the counts, and divide:

\text{combined mean} = \frac{756+612}{18+12} = \frac{1368}{30} =45.6\text{ pages}.

The answer $45.6$ lands between $42$ and $51$ , but closer to $42$ , because the first group is bigger.

For any two groups,

\boxed{ \text{combined mean} = \frac{(\text{mean}_1)(\text{count}_1)+(\text{mean}_2)(\text{count}_2)} {\text{count}_1+\text{count}_2} }.

It’s the frequency-table idea again. Each group’s mean plays the part of a value, and each group’s size plays the part of its frequency.

Check your understanding:

Two groups have means of $20$ and $30$ . Is their combined mean $25$ ?

Common mistake:

Working out $(42+51)/2=46.5$ treats the two groups as if they were the same size, but the group of $18$ should count for more. Write each group’s size next to its mean, turn both means into totals, and divide by the combined number of observations.

Reverse the structure for an unknown frequency

Sometimes what’s missing is a frequency, not a value. Call it $x$ , and write both the total and the count using $x$ .

Value	Frequency
$3$	$2$
$5$	$x$
$9$	$4$

Say the mean is $6$ . The weighted total is

3(2)+5x+9(4)=42+5x,

and the total frequency is

2+x+4=6+x.

Now use total equals mean times count:

\begin{aligned} 42+5x&=6(6+x)\\[1.4em] 42+5x&=36+6x\\[1.4em] x&=6. \end{aligned}

So the missing frequency is $6$ . A frequency counts things, so make sure it’s a whole number that isn’t negative. Then check that the rebuilt mean comes out right:

\frac{3(2)+5(6)+9(4)}{2+6+4} = \frac{72}{12} =6.

Check your understanding:

In the equation $42+5x=6(6+x)$ , what does each side stand for?

Choose hand work or a calculator table

Whatever tool you use, set up the math first. Write total = mean × count or mean = weighted total ÷ total count, then decide where to do the arithmetic.

Short problems go faster by hand:

One missing value takes one multiplication, one sum and one subtraction, like $18.5(6)-92$ .
Two groups take two multiplications, like $18(42)$ and $12(51)$ , then two additions and one division.
A table with only a few rows, like the bus-trip table, takes three products, two sums and one division.
A short equation for an unknown frequency, like $42+5x=6(6+x)$ , takes two lines to solve.

A longer frequency table is where Desmos helps. A Desmos table keeps each value lined up with its frequency and does the multiplying for you:

Enter the values in x_1.
Enter the matching frequencies in y_1.
Enter total(x_1*y_1) to get the weighted total.
Enter total(y_1) to get the count.
Divide with total(x_1*y_1)/total(y_1).

The calculator holds the $4$ , $7$ , $10$ table from earlier. It shows the same weighted total of $73$ , count of $10$ and mean of $7.3$ . Change one frequency and watch all three update. Reset brings back the original table.

Check your understanding:

Why does each frequency have to stay in the same row as its value?

Calculator loads as you approach

Each value sits beside its frequency. The last line divides the weighted total by the total frequency.

Questions that add, remove, correct or replace values and ask how the statistics change come later, in Analyze changed data and outliers.

Practice problems

Each problem starts the same way: turn what you’re given into a total. They get harder as you go, ending with an unknown group size.

Weight every row

Practice problem

The table summarizes the masses, in kilograms, of $10$ packages.

Mass (kilograms)	Frequency
$2$	$2$
$5$	$5$
$8$	$3$

What is the mean mass, in kilograms, of the packages?

Your answer

Calculator loads as you approach

Three rows are quick by hand. Use this to check your work if it helps.

Find one missing value

Practice problem

The mean of eight recorded temperatures is $14.5^\circ\text{C}$ . Seven of the temperatures are

10,\ 12,\ 13,\ 14,\ 15,\ 16,\ 18.

What is the eighth temperature?

Calculator loads as you approach

Rebuild the total first. Use this for the arithmetic if it helps.

Find an unknown group size

Practice problem

Data set A contains $24$ values and has a mean of $68$ . Data set B has a mean of $82$ . When the two data sets are combined, the $24$ values from data set A and all the values from data set B have a mean of $74$ .

How many values are in data set B?

Your answer

Calculator loads as you approach

Write the equation first. Then solve it or check it here if it helps.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Mean, count and total are linked: $\text{mean}=\frac{\text{total}}{\text{count}}$ and $\text{total}=\text{mean}\times\text{count}$ .
A frequency table is a list in shorthand. Each value repeats as many times as its frequency.
For a weighted mean, find $\sum(\text{value}\times\text{frequency})$ and divide by the total frequency.
For one missing value, subtract the known sum from the total the mean requires. That difference is the answer.
To combine groups of different sizes, turn each mean into a total before you add.
For an unknown frequency or group size, write both the total and the count with a variable, then use total equals mean times count.
Short problems go faster by hand. Use a Desmos table when a long table has many products to multiply and rows to keep lined up.

Next lesson

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Move from center to spread by measuring and comparing how widely data values vary.

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