Four summaries, four different blind spots
For a list of $n$ numbers:
Mean $=\dfrac{\text{sum of all values}}{n}$.
Median $=$ the middle value after sorting (average of the two middle values when $n$ is even).
Mode $=$ the value that occurs most often (a list may have none, one, or several).
Range $=$ largest value $-$ smallest value.
Each summary throws away different information, and every SAT question at this level is built on top of what a particular summary cannot see.
- The mean uses every value, so one extreme value drags it.
- The median uses only position, so it barely notices an extreme value.
- The mode ignores size completely — a mode of $2$ tells you nothing about whether the other values are $3$ or $3000$.
- The range uses only two values, the two least typical ones.
Đây là chỗ đề ra bẫy, không phải chỗ tính toán. Ví dụ: một danh sách có trung vị $8$ — điều đó không cho biết giá trị lớn nhất là bao nhiêu, cũng không cho biết có bao nhiêu giá trị bằng $8$. Mỗi khi gặp câu “which must be true”, hãy hỏi ngược: đại lượng đề cho nhìn thấy được gì của dữ liệu, và không nhìn thấy được gì.
Frequency tables
A frequency table is the standard way the SAT hides a data set of $30$ or $40$ values inside five short rows. The mean is a weighted mean: \[ \text{mean}=\frac{\sum (\text{value}\times\text{frequency})}{\sum \text{frequency}} . \] The median is found by counting positions, not by looking at the middle row.
The table records the number of pets owned by each of $30$ students.
| Pets | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Students | 6 | 9 | 7 | 5 | 3 |
Mean. Total pets $=0(6)+1(9)+2(7)+3(5)+4(3)=0+9+14+15+12=50$, so the mean is $\dfrac{50}{30}=\dfrac53\approx 1.67$.
Median. With $n=30$ the median is the average of the $15$th and $16$th values. Running totals: through “0” we have $6$ students, through “1” we have $15$, through “2” we have $22$. So the $15$th value is $1$ and the $16$th value is $2$, giving a median of $\boxed{1.5}$ — a number that does not appear in the data at all.
Mode $=1$ (highest frequency, $9$). Range $=4-0=4$.
Cộng năm số ở hàng “Pets” rồi chia cho $5$: $(0+1+2+3+4)/5=2$. Đó là trung bình của các loại giá trị, không phải của học sinh. Cũng sai không kém: lấy $30\div 5=6$. Trung bình luôn là tổng tất cả giá trị chia cho tổng tần số.
Trung vị trong bảng tần số phải đếm vị trí. Với $n$ giá trị: $n$ lẻ thì trung vị nằm ở vị trí $\frac{n+1}{2}$; $n$ chẵn thì lấy trung bình hai vị trí $\frac n2$ và $\frac n2+1$. Cộng dồn tần số từ trái sang cho tới khi vượt qua vị trí cần tìm — đừng chỉ nhìn cột ở giữa bảng.
Working backwards from a mean
Every “find the missing value” question is the same identity read in reverse: \[ \text{sum}=\text{mean}\times n . \] Convert each mean into a sum first; only then compare.
The mean of $9$ numbers is $24$. One number is removed and the mean of the remaining $8$ numbers is $25$. What number was removed?
Original sum $=24\times 9=216$. Remaining sum $=25\times 8=200$. The removed number is $216-200=\boxed{16}$.
Five samples have masses $1.2$ kg, $850$ g, $1.05$ kg, $720$ g and $1.18$ kg. What is the mean mass, in grams?
Put everything in grams first: $1200,\;850,\;1050,\;720,\;1180$. Sum $=1200+850+1050+720+1180=5000$, so the mean is $\dfrac{5000}{5}=\boxed{1000\text{ g}}$.
Trộn đơn vị: cộng thẳng $1.2+850+1.05+720+1.18$ ra một con số vô nghĩa. Ở nhóm Xử lý số liệu, đổi đơn vị trước khi tính là bước bắt buộc, và đề luôn đặt sẵn một phương án ứng với việc quên đổi.
Combining two groups
Two groups of sizes $n_1$ and $n_2$ with means $\bar x_1$ and $\bar x_2$ combine to \[ \bar x=\frac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2}, \] which equals $\dfrac{\bar x_1+\bar x_2}{2}$ only when $n_1=n_2$.
Class A has $18$ students with a mean score of $76$; class B has $12$ students with a mean score of $86$. What is the mean score of all $30$ students?
Sums: $18\times 76=1368$ and $12\times 86=1032$, so the combined sum is $2400$ and the combined mean is $\dfrac{2400}{30}=\boxed{80}$.
The naive answer $\frac{76+86}{2}=81$ is too high because the larger class is the lower-scoring one, so it pulls the combined mean down.
Trung bình gộp luôn nằm giữa hai trung bình thành phần, và lệch về phía nhóm đông hơn. Đó là cách kiểm tra nhanh: nếu nhóm đông hơn có trung bình thấp hơn mà đáp số của bạn lại lớn hơn $\frac{\bar x_1+\bar x_2}{2}$ thì chắc chắn sai.
What moves when the data moves
- Add a constant $c$ to every value: mean, median and mode all shift by $c$; the range does not change.
- Multiply every value by $k>0$: mean, median, mode and range are all multiplied by $k$.
- Change one value only: the mean always moves; the median moves only if the change crosses the middle position; the range moves only when an extreme is involved — either the value that changed was the largest or the smallest, or the replacement lands outside the old extremes.
In the list $3,\;5,\;7,\;9,\;41$ the value $41$ is replaced by $91$. What happens to the mean, the median and the range?
Mean: from $\dfrac{65}{5}=13$ to $\dfrac{115}{5}=23$ — it rises by $10$.
Median: the sorted order is unchanged and the third value is still $7$ — no change.
Range: from $41-3=38$ to $91-3=88$ — it rises by $50$.
The list $4,\;6,\;8,\;10,\;12$ has mean $8$, median $8$ and range $8$. A sixth value is added. Compare the effect of adding $8$ with the effect of adding $30$.
Adding $8$: the sum becomes $48$ over $6$ values, so the mean stays $8$; the two middle values are now $8$ and $8$, so the median stays $8$; the range stays $8$.
Adding $30$: the sum becomes $70$ over $6$ values, so the mean rises to $\dfrac{70}{6}=\dfrac{35}{3}\approx 11.67$; the two middle values are $8$ and $10$, so the median rises to $9$; the range becomes $30-4=26$.
Câu hỏi “thêm một giá trị bằng đúng trung bình hiện tại thì trung bình có đổi không” xuất hiện rất nhiều. Câu trả lời là không đổi — và đó cũng là mẹo kiểm tra: nếu giá trị mới lớn hơn trung bình cũ thì trung bình mới tăng, nhỏ hơn thì giảm. Không cần tính lại tổng.
Reading the shape from the summaries
When the mean sits noticeably above the median, a few unusually large values are pulling the sum up while leaving the middle position where it was; when the mean sits below the median, a few unusually small values are doing the opposite.
A city reports that the mean household income is $\$92{,}000$ while the median household income is $\$61{,}000$. What does the gap tell us?
The middle household earns $\$61{,}000$, so at least half of all households earn no more than that. The mean is far higher only because a small number of very large incomes inflate the total. The gap says nothing about how many households there are, and nothing about the smallest income — it says the upper tail is long.
A list of five positive integers has a mean of $10$ and a median of $12$. What is the greatest possible value in the list?
The sum is $10\times 5=50$, and sorting the list as $a\le b\le c\le d\le e$ forces $c=12$. To make $e$ as large as possible, make everything else as small as possible: $a=b=1$ (positive integers) and $d=12$ (it cannot be below the median). Then $e=50-1-1-12-12=\boxed{24}$, and the list $1,1,12,12,24$ does have mean $10$ and median $12$.
Đọc “median $=12$” rồi cho luôn $d$ nhỏ hơn $12$. Sau khi sắp xếp, mọi giá trị đứng sau trung vị đều phải $\ge$ trung vị. Ép $d=12$ là giá trị nhỏ nhất hợp lệ, không phải $1$.