One-Variable Data: Distributions, Centre, Spread

One-variable data is a list of values of a single quantity: test scores, heights, numbers of pets. The SAT asks you to find the center (mean and median), to read data from frequency tables and dot plots, to predict how outliers affect each measure, and to compare spread (range and standard deviation) without heavy calculation.

• The mean is the total divided by the count. Many questions are easier if you think in totals: $\text{total} = \text{mean} \times \text{count}$.
• The median is the middle value after sorting. With an even count, it is the average of the two middle values.
• A frequency table is a compressed list: the row "value $3$, frequency $4$" means $3, 3, 3, 3$.
• An outlier pulls the mean toward it a lot but barely moves the median. A data set with a high outlier usually has mean greater than median.
• Standard deviation measures how far values typically are from the mean. Tightly clustered data has a small standard deviation; widely spread data has a large one. Adding the same number to every value does not change the spread.

$\text{mean} = \frac{\text{sum of values}}{\text{number of values}} \qquad \text{mean from a table} = \frac{\sum (\text{value} \times \text{frequency})}{\sum \text{frequency}}$

• Combined mean of two groups: $\frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$. For $12$ students averaging $70$ and $18$ averaging $80$: $\frac{840 + 1440}{30} = 76$.
• Median position in a sorted list of $n$ values: the $\frac{n + 1}{2}$th value.

Find the mean and median of $3, 8, 5, 12, 5, 9$.

1. Mean: the sum is $3 + 8 + 5 + 12 + 5 + 9 = 42$, and $42 \div 6 = 7$.
2. Median: sort to get $3, 5, 5, 8, 9, 12$. The middle two values are $5$ and $8$, so the median is $\frac{5 + 8}{2} = 6.5$.

Check: the mean is a little above the median because $12$ stretches the upper end.

The table shows how many books each of $20$ students read over the summer. Find the mean and the median.

1. Total books: $0(4) + 1(7) + 2(5) + 3(3) + 4(1) = 0 + 7 + 10 + 9 + 4 = 30$.
2. Mean: $30 \div 20 = 1.5$.
3. Median: with $20$ values, average the $10$th and $11$th. The first $4$ values are $0$; values $5$ through $11$ are $1$. Both middle values are $1$, so the median is $1$.

Check: the frequencies add to $4 + 7 + 5 + 3 + 1 = 20$. The mean exceeds the median, which fits the long tail toward $3$ and $4$ books.

• Averaging the values column in a table. The mean is not $\frac{0 + 1 + 2 + 3 + 4}{5} = 2$; each value must be weighted by its frequency.
• Forgetting to sort before finding the median.
• Averaging two averages. When groups have different sizes, the combined mean is not the midpoint of the two means; use totals.
• Assuming the median changes when an extreme value changes. Making the largest value larger, or the smallest value smaller, leaves the median of an odd sized list unchanged. It does, however, increase the range and the standard deviation.
• Confusing spread with center. The lists $40, 45, 50, 55, 60$ and $48, 49, 50, 51, 52$ have the same mean and median, but the first has a much larger standard deviation.