What is the difference between mean and median?

The mean is the average — add all values and divide by the count. The median is the middle value of the ordered data. The mean is pulled by outliers, while the median stays near the center, so the median often better represents skewed data.

How do you find a missing number when you know the average?

Multiply the average by how many values there are to get the total sum. Then subtract the values you already know. The difference is the missing number.

How do you calculate a weighted mean or weighted grade?

Multiply each value by its weight, add those products together, and divide by the total of the weights. If the weights are percentages that add to 100%, the weighted sum is already the answer.

Can you average two averages together on the ACT?

Only if the two groups are the same size. Otherwise you must rebuild each group total by multiplying its average by its count, add the totals, and divide by the combined number of items.

CramClub • Mean, Median, Mode, and Range

$\text{mean} = \dfrac{\text{sum of values}}{\text{number of values}}$

Middle value of ordered data (average the two middles if the count is even).

$\text{range} = \text{maximum} - \text{minimum}$

$\dfrac{\sum (\text{value} \times \text{weight})}{\sum \text{weights}}$

Describing a Data Set with a Single Number

When the ACT hands you a list of values, it usually wants you to summarize them. Four measures do most of the work: three describe the center of the data (mean, median, mode) and one describes its spread (range). Knowing which measure a question asks for — and how each responds to extreme values — is what separates a quick point from a careless miss.

Mean (the Average)

The mean is the ordinary average: add every value and divide by how many there are.

$\text{mean} = \frac{\text{sum of values}}{\text{number of values}}$

For the scores $8, 12, 15, 20, 10$ , the mean is $\tfrac{8 + 12 + 15 + 20 + 10}{5} = \tfrac{65}{5} = 13$ . The mean uses every value, which makes it powerful but also sensitive: a single very large or very small number pulls it noticeably.

Median (the Middle)

The median is the middle value once the data is written in order. The ordering step is not optional — a list given out of order will give a wrong median if you skip it.

With an odd count, the median is the single middle value.
With an even count, the median is the average of the two middle values.

For $3, 7, 9, 12, 15, 20$ (six values, even count), the two middle numbers are $9$ and $12$ , so the median is $\tfrac{9 + 12}{2} = 10.5$ .

Mode (the Most Frequent)

The mode is the value that appears most often. A set can have no mode (all values distinct), one mode, or several. The mode is the only one of these measures that also works for non-numerical data such as favorite colors.

Range (the Spread)

The range measures how spread out the data is:

$\text{range} = \text{maximum} - \text{minimum}$

For $3, 7, 9, 12, 15, 20$ the range is $20 - 3 = 17$ . Range is quick but fragile — it depends only on the two most extreme values.

Mean vs. Median: The Outlier Question

The ACT loves to ask how an outlier (an unusually large or small value) affects these measures.

Measure	Sensitive to outliers?	Why
Mean	Very	It uses every value in the sum
Median	Barely	It depends only on position, not magnitude
Mode	No	It depends only on frequency
Range	Very	It is built from the two extremes

If a data set of typical salaries includes one executive earning far more than everyone else, the mean salary jumps upward while the median barely moves. That is exactly why the median is often the "fairer" summary of skewed data.

The Weighted Mean

Sometimes values do not count equally. A weighted mean multiplies each value by its weight, sums those products, and divides by the total weight:

$\text{weighted mean} = \frac{\sum (\text{value} \times \text{weight})}{\sum \text{weights}}$

A course grade is the classic example. Suppose homework counts $20\%$ , the midterm $30\%$ , and the final $50\%$ , with scores $90$ , $80$ , and $85$ :

$0.20(90) + 0.30(80) + 0.50(85) = 18 + 24 + 42.5 = 84.5$

Because the weights already add to $1$ , no extra division is needed. When the weights are raw counts instead of percentages, divide by their total.

Working Backward: Missing Values and Combined Groups

Two setups appear again and again.

Missing value from a known mean. If a mean is given, multiply it by the count to recover the total. Four tests averaging $85$ means the four scores sum to $4 \times 85 = 340$ . Subtract the known scores to find the missing one.

Combining two groups. You cannot simply average two averages unless the groups are the same size. Instead, rebuild each group's total. If $20$ students average $78$ and $30$ students average $88$ , the combined mean is

$\frac{20(78) + 30(88)}{50} = \frac{1560 + 2640}{50} = \frac{4200}{50} = 84.$

Notice the answer is closer to $88$ than to $78$ , because the larger group pulls the combined mean toward its own average. Averaging the two averages naively would give $83$ , which is wrong.

A Reliable Game Plan

Identify which measure the question wants. For the mean, think "sum then divide"; for the median, order first; for the mode, count frequencies; for range, subtract the extremes. If the problem gives you an average and asks for a value, convert the average back into a total right away.

Find the mean of the values $12, 15, 20, 8,$ and $10$ .

What is the median of the data set $4, 20, 7, 12, 9, 15$ ?

On four tests a student scored $82, 90,$ and $88$ on the first three. What must the fourth score be for the four-test average to be exactly $85$ ?

A course grade is $20\%$ homework, $30\%$ midterm, and $50\%$ final. A student earns $90$ on homework, $80$ on the midterm, and $85$ on the final. What is the course grade?

Class A has $20$ students with an average score of $78$ . Class B has $30$ students with an average of $88$ . What is the average score of all $50$ students combined?

A student's six quiz scores have a mean of $82$ . After the lowest score is dropped, the mean of the remaining five scores is $85$ . What was the dropped score?

Finding the median without ordering the data, e.g. calling $9$ the median of $4, 20, 7, 12, 9, 15$ because it sits in the middle of the list.

Sort the values from least to greatest first, then locate the middle. For an even count, average the two central numbers.

Averaging two group averages directly, so a class of $20$ (avg $78$ ) and a class of $30$ (avg $88$ ) is reported as $83$ .

Weight by group size: combined mean $= \dfrac{20(78) + 30(88)}{50} = 84$ . The larger group pulls the result toward its own average.

Assuming an outlier changes the median as much as the mean, or claiming it changes the mode.

Remember the sensitivity order: the mean and range react strongly to outliers, the median barely moves, and the mode does not change at all.

Forgetting to convert a given average back into a total when solving for a missing value.

A mean of $85$ over $4$ items means the sum is $4 \times 85 = 340$ . Recover the total first, then subtract the known values.

What is the range of the data set $3, 17, 9, 12, 20,$ and $5$ ?

What is the mode of the data set $6, 9, 4, 9, 12, 6,$ and $9$ ?

The mean of five numbers is $24$ . Four of the numbers are $20, 30, 18,$ and $27$ . What is the fifth number?

A final grade is $40\%$ tests, $35\%$ projects, and $25\%$ participation. A student earns $70$ on tests, $90$ on projects, and $80$ on participation. What is the final grade?

A group of $15$ hikers averaged $12$ miles, and a group of $10$ hikers averaged $17$ miles. What is the average distance for all $25$ hikers?

The set $\{8, 11, x, 19, 24\}$ is listed in increasing order and has a median that is $3$ more than its smallest value. What is the value of $x$ ?

$\text{mean} = \dfrac{\text{sum of values}}{\text{number of values}}$

Middle value of ordered data (average the two middles if the count is even).

$\text{range} = \text{maximum} - \text{minimum}$

$\dfrac{\sum (\text{value} \times \text{weight})}{\sum \text{weights}}$

Describing a Data Set with a Single Number

Mean (the Average)

The mean is the ordinary average: add every value and divide by how many there are.

$\text{mean} = \frac{\text{sum of values}}{\text{number of values}}$

Median (the Middle)

The median is the middle value once the data is written in order. The ordering step is not optional — a list given out of order will give a wrong median if you skip it.

With an odd count, the median is the single middle value.
With an even count, the median is the average of the two middle values.

For $3, 7, 9, 12, 15, 20$ (six values, even count), the two middle numbers are $9$ and $12$ , so the median is $\tfrac{9 + 12}{2} = 10.5$ .

Mode (the Most Frequent)

Range (the Spread)

The range measures how spread out the data is:

$\text{range} = \text{maximum} - \text{minimum}$

For $3, 7, 9, 12, 15, 20$ the range is $20 - 3 = 17$ . Range is quick but fragile — it depends only on the two most extreme values.

Mean vs. Median: The Outlier Question

The ACT loves to ask how an outlier (an unusually large or small value) affects these measures.

Measure	Sensitive to outliers?	Why
Mean	Very	It uses every value in the sum
Median	Barely	It depends only on position, not magnitude
Mode	No	It depends only on frequency
Range	Very	It is built from the two extremes