Statistics & Data

What you need to know

• Mean $=\dfrac{\text{sum}}{\text{count}}$, so sum $=$ mean $\times$ count. Almost every problem about a mean is really about the total. Example: five numbers with mean $12$ add up to $60$.
• Median: put the numbers in order first. With an odd count, the median is the middle number. With an even count, it is the average of the two middle numbers. Example: $8,3,5,9,1$ sorts to $1,3,5,8,9$, with median $5$. And $2,7,4,10$ sorts to $2,4,7,10$, with median $\frac{4+7}2=5.5$.
• Mode: the value that appears most often. "Unique mode" means it appears strictly more often than every other value; a tie is not allowed. Example: $2,3,3,3,5,5$ has unique mode $3$, but $2,3,3,5,5,7$ has no unique mode, since $3$ and $5$ tie.
• Range $=$ largest $-$ smallest. Example: $2,3,3,5,5,7$ has range $7-2=5$.
• Combining two groups (a weighted average). If a group of $a$ people has mean $m_1$ and a group of $b$ people has mean $m_2$, the mean of everyone is $\dfrac{am_1+bm_2}{a+b}$. Example: $20$ students averaging $70$ and $30$ averaging $80$ give $\frac{1400+2400}{50}=76$.

How AMC 10 tests it

• Problems 1–8: find one statistic from another. The hard part is not mixing up the mean and the median.
• "After the next test my average will be $k$": change the averages into totals and subtract. Example: to go from an average of $85$ on $4$ tests to $87$ on $5$, you need $5\cdot87-4\cdot85=95$.
• Middle-of-the-paper problems where the mean, median, mode, and range are all given and you find the largest possible number in the list. Solve them with a bound plus a list that reaches it.

Standard approaches

1. Rewrite every mean as a sum, and every median as a fact about a certain position in the sorted list.
2. When numbers are added or removed, keep track of totals, not averages.
3. For largest/smallest problems, first place the values that are forced (the median spot, the repeated mode). Then push the free values as far as they can go.
4. Bound first, then build: show the larger answer is impossible, then write down a list that reaches the next one.

Worked example

A list of five positive integers has mean $12$, median $13$, and unique mode $15$. What is the largest possible range of the list?

(A) $8$ (B) $9$ (C) $10$ (D) $11$ (E) $12$

Solution. Write the list in order as $a\le b\le c\le d\le e$.

Use the mean and median. The mean is $12$, so the sum is $5\cdot12=60$. The median is the middle number, so $c=13$.

Use the mode. The mode $15$ must appear at least twice. Only $d$ and $e$ can be $15$, since $a$, $b$, $c$ are at most $13$. So $d=e=15$, and $a+b=60-13-15-15=17$.

Make the range big. The range is $15-a$, so we want $a$ as small as possible, which means $b$ as large as possible. Since $15$ appears exactly twice and is the unique mode, no other value may appear twice: $a\ne b$ and $b\ne13$. As $b\le c=13$, this leaves $b\le12$, so $a=17-b\ge5$.

Check that it works. The list $5,12,13,15,15$ has sum $60$, median $13$, and unique mode $15$. Its range is $15-5=10$. The answer is $\boxed{\textbf{(C)}\ 10}$.

Pitfalls

• Giving the mean when the sum is asked, or the median when the mean is asked.
• Forgetting to sort before finding the median. Example: the middle of $9,2,7,4,3$ as written is $7$, but the median is $4$.
• Ignoring "unique" in "unique mode." Example: $4,13,13,15,15$ has sum $60$ and median $13$, but $13$ ties with $15$, so its range of $11$ is not allowed.
• Averaging group averages without weighting by group size. Example: above, $\frac{70+80}{2}=75$ is wrong; the true mean is $76$.

Traps that recur

• Treating the median as 5.2 or one of the given numbers; with nine values the fifth must be an integer, so one of x, y, z has to sit in the middle. (2024 AMC 10B #15)
• Using five 8s to fill the bottom, which forces the other values up and yields 24; fewer 8s lets 1s in and gives a bigger maximum. (2014 AMC 10B #18)
• Ignoring that the mode must be unique, or forgetting the mean fixes the total at 64. (2002 AMC 10A #21)
• Assuming the mean is always the largest of the three statistics; when the median is x you must test all three orderings, and when the median is 4 you must discard solutions that leave the case's range. (2000 AMC 10 #23)