Compute the mean, median, mode and range of a data set, find a missing value given a target mean, and interpret which measure of centre best describes data (Statistics and Probability).

What this topic is asking

The ACT tests the four basic summary statistics: the mean (average), median (middle), mode (most frequent) and range (spread). Beyond computing them, it asks you to find a missing value for a target mean and to judge which measure best describes a data set, especially when an outlier is present.

The four measures

Each summarises the data differently.

Computing the median carefully

The median requires sorting first.

Finding a missing value for a target mean

A frequent ACT item gives several values plus a target average and asks for the missing value. Use total $=$ mean $\times$ count. If five numbers must average 20, their total is $5 \times 20 = 100$; subtract the known values to find the missing one. This same relationship answers "what score do you need on the final to get an 85 average", which is one of the most common applied mean questions.

Mean versus median with outliers

The mean uses every value, so a single extreme value (an outlier) drags it toward the extreme. The median depends only on the middle, so it barely moves. Consider salaries $30{,}000$, $32{,}000$, $35{,}000$, $38{,}000$, $500{,}000$: the mean is over \127{,}000$, which describes none of the first four people, while the median is$\$35{,}000$, a far better summary. So for skewed data or data with outliers, the median is usually the more representative measure of centre, a point the ACT likes to test in interpretation questions.

Effect of adding or changing a value

The ACT often asks how a statistic changes when the data changes. Adding a value equal to the current mean leaves the mean unchanged; adding a value above it raises the mean, below it lowers it. The median can jump or stay put depending on where the new value lands relative to the middle. Multiplying every value by a constant multiplies the mean, median and range by that constant; adding a constant to every value shifts the mean and median by it but leaves the range unchanged (since both ends move equally). Reasoning about these effects, rather than recomputing from scratch, answers many "what happens to the average if..." questions quickly.

Mean of a frequency table

When data come in a frequency table (each value with how many times it occurs), the mean is a weighted sum: multiply each value by its frequency, add, and divide by the total count. For scores 5 (appearing 3 times), 8 (appearing 2 times) and 10 (once), the mean is $\frac{5(3) + 8(2) + 10(1)}{3 + 2 + 1} = \frac{15 + 16 + 10}{6} = \frac{41}{6} \approx 6.83$. Reading a frequency table as "value times how many", not just averaging the distinct values, is the key, and it connects this topic to weighted averages.

Why ordering and the total trick matter

Two habits prevent most errors here. First, always sort before reading a median; the middle of an unsorted list is meaningless. Second, for any mean question that gives a target, reach for total $=$ mean $\times$ count rather than guessing. Recognising that the mean is sensitive to outliers while the median is robust lets you answer the "which measure best describes the data" questions quickly and correctly.

Try this

Q1. Find the mean of 12, 15, 9, 14. [1 point]

• Cue. Sum $= 50$, count $= 4$, mean $= \frac{50}{4} = 12.5$.

Q2. Find the range and mode of 4, 7, 7, 2, 9. [1 point]

• Cue. Range $= 9 - 2 = 7$; mode $= 7$ (appears twice).