Measures of Central Tendency

A measure of central tendency is a single number that describes the “middle” or “typical” value of a data set. The mean, median, and mode each describe the centre in a different way, and they don’t always agree. Knowing which one to use (and when a number is hiding something) is one of the most useful skills in statistics.

Key ideas

Mean

The mean is the sum of the values divided by how many there are. Statisticians use different symbols for a population and a sample:

$$\mu = \frac{\sum x}{N} \qquad\qquad \bar{x} = \frac{\sum x}{n}$$

Symbol	Meaning
$\mu$ (mu)	mean of a whole population
$\bar{x}$ (x-bar)	mean of a sample
$N$	population size
$n$	sample size
$\sum x$	the sum of all the data values

The formula is the same; only the symbols change. The mean uses every value, which makes it sensitive to extreme values.

Median

The median is the middle value when the data are in order.

• If $n$ is odd, the median is the middle value: the one in position $\dfrac{n + 1}{2}$.
• If $n$ is even, the median is the mean of the two middle values.

Mode

The mode is the value that occurs most often. A data set can have one mode, more than one mode, or no mode (if every value occurs once). The mode is the only measure you can use for categorical data, like favourite colours.

Weighted mean

When some values count more than others, use a weighted mean. Multiply each value by its weight, add, and divide by the total weight:

$$\text{weighted mean} = \frac{\sum w x}{\sum w}$$

If the weights are percentages that add to $100\%$, just multiply each value by its weight as a decimal and add.

Mean of grouped data

When data are given in a frequency table with intervals, you don’t know the exact values. Estimate the mean by assuming every value in an interval sits at the interval’s midpoint $m$:

$$\bar{x} \approx \frac{\sum f m}{\sum f}$$

where $f$ is the frequency of each interval. The median interval is the interval that contains the middle value.

Outliers and choosing a measure

An outlier is a value far away from the rest of the data. Outliers pull the mean toward them, but they barely move the median. (A precise rule for spotting outliers is on the quartiles and percentiles page.)

Measure	Best when…	Weakness
Mean	data are numerical and roughly symmetric, with no outliers	pulled by outliers
Median	data are skewed or have outliers (incomes, house prices)	ignores how far the other values are from the middle
Mode	data are categorical, or you want the most common value (shoe sizes to stock)	may not exist, or may not be near the centre

The centre is only half the story. Two data sets can have the same mean but very different spreads. To measure spread, see standard deviation.

On the SAT

In Desmos, type a list and use mean(...) and median(...), or name it first (L = [...], then mean(L)). For a frequency table, you don’t have to type every repeated value: multiply each value by its frequency and divide by the total count, e.g. values $2, 3, 4$ with frequencies $3, 5, 2$ give $\dfrac{6 + 15 + 8}{10} = 2.9$. Many SAT questions ask how the mean or median changes when a value is added or removed, or when there’s an outlier, and those are faster to reason out: an outlier pulls the mean toward it but barely moves the median. See using Desmos on the SAT.

Worked examples

Example 1: Mean, median, and mode

A sample of $8$ students reported how many hours they slept last night:

$$7,\ 8,\ 6,\ 9,\ 7,\ 8,\ 7,\ 10$$

• (a) Find the mean, median, and mode.
• (b) A ninth student reports $2$ hours. How does each measure change?

Solution. (a) The sum is $62$, so

$$\bar{x} = \frac{62}{8} = 7.75 \text{ h}$$

In order: $6, 7, 7, 7, 8, 8, 9, 10$. With $n = 8$ (even), the median is the mean of the 4th and 5th values:

$$\text{median} = \frac{7 + 8}{2} = 7.5 \text{ h}$$

The value $7$ occurs three times, so the mode is $7$ h.

(b) Now $n = 9$ and the sum is $64$, so $\bar{x} = \dfrac{64}{9} \approx 7.11$ h. In order: $2, 6, 7, 7, 7, 8, 8, 9, 10$. The median is the 5th value, $7$ h. The mode is still $7$ h.

The one low value dragged the mean down by about $0.64$ h, but the median only moved by $0.5$ h and the mode didn’t move at all.

Example 2: A weighted course mark

Mei’s final mark is calculated like this: tests are worth $45\%$, assignments $25\%$, and the final exam $30\%$. Her averages are $76\%$ on tests, $88\%$ on assignments, and $70\%$ on the exam. Find her final mark.

Solution. The weights add to $100\%$, so multiply each mark by its weight as a decimal:

$$\begin{aligned} \text{final mark} &= 0.45(76) + 0.25(88) + 0.30(70) \\ &= 34.2 + 22 + 21 \\ &= 77.2 \end{aligned}$$

Her final mark is $77.2\%$. Notice that a simple mean of the three marks, $\dfrac{76 + 88 + 70}{3} \approx 78.0$, gives the wrong answer because it treats the three parts as equally important.

Example 3: Mean of grouped data

A class of $30$ students recorded their commute times to school.

Time (min)	Frequency $f$	Midpoint $m$	$f m$
$0$ to under $10$	$6$	$5$	$30$
$10$ to under $20$	$11$	$15$	$165$
$20$ to under $30$	$8$	$25$	$200$
$30$ to under $40$	$4$	$35$	$140$
$40$ to under $50$	$1$	$45$	$45$
Total	$30$		$580$

Estimate the mean commute time, and find the median interval.

Solution.

$$\bar{x} \approx \frac{\sum f m}{\sum f} = \frac{580}{30} \approx 19.3 \text{ min}$$

With $30$ values, the median is between the 15th and 16th values. The first interval holds the first $6$ values, and the second holds values $7$ to $17$. So the median interval is $10$ to under $20$ minutes.

This is only an estimate of the mean, because we assumed every value is at the midpoint of its interval.

Example 4: Comparing two data sets

Two players on a school basketball team scored these points in their last $7$ games:

• Aisha: $12, 14, 15, 15, 16, 17, 18$
• Maya: $4, 6, 8, 10, 11, 30, 36$

Compare their scoring using the mean and median. Who is the more reliable scorer?

Solution.

	Mean	Median
Aisha	$\dfrac{107}{7} \approx 15.3$	$15$
Maya	$\dfrac{105}{7} = 15$	$10$

Their means are almost the same, so the means alone suggest they’re equally good scorers. The medians tell a different story: in a typical game, Aisha scores about $15$ points but Maya scores about $10$. Maya’s mean is pulled up by two big games ($30$ and $36$).

Aisha is the more reliable scorer: her scores are close together, while Maya’s swing widely. When you compare data sets, compare both a measure of centre and a measure of spread.

Common mistakes

Finding the median without sorting first. The median is the middle of the ordered list. Always put the data in order before you look for the middle.

Taking the middle position as the median value. For $n = 9$, the median is the 5th value, not $5$. The formula $\dfrac{n + 1}{2}$ gives a position, not the answer.

Taking a simple mean when the parts have different weights. Course marks, GPA, and averages of groups of different sizes all need a weighted mean.

Using interval endpoints instead of midpoints for grouped data. Use the midpoint of each interval, and remember that the result is an estimate, not an exact mean.

Reporting the mean when there are outliers. One very large salary can make an “average” salary misleading. For skewed data or data with outliers, the median usually describes a typical value better.

Comparing data sets with only one number. Two sets with the same mean can be very different. Look at the median and the spread too.

Practice

1. (Warm-up) Find the mean, median, and mode: $4, 9, 6, 4, 7, 10, 2$.

Solution

In order: $2, 4, 4, 6, 7, 9, 10$. The sum is $42$, so the mean is $\dfrac{42}{7} = 6$. The median is the 4th value, $6$. The mode is $4$.

2. (Warm-up) Find the mean and median: $13, 18, 11, 22, 15, 16$.

Solution

The sum is $95$, so the mean is $\dfrac{95}{6} \approx 15.8$.

In order: $11, 13, 15, 16, 18, 22$. The median is $\dfrac{15 + 16}{2} = 15.5$.

3. (Warm-up) Which measure of central tendency (mean, median, or mode) is best in each situation? Explain briefly.

• (a) A store decides which shoe size to order the most of.
• (b) A survey asks students for their favourite season.
• (c) A real estate agent describes a typical house price in a town with a few mansions.

Solution

(a) Mode: the store wants the size that sells most often.

(b) Mode: the data are categorical, so the mean and median don’t make sense.

(c) Median: the mansions are outliers that would pull the mean up.

4. (Core) In a course, term work is worth $70\%$ and the final exam is worth $30\%$. Jordan’s term mark is $84\%$. What mark does Jordan need on the exam to finish with at least $80\%$?

Solution

Let $x$ be the exam mark:

$$\begin{aligned} 0.70(84) + 0.30x &\ge 80 \\ 58.8 + 0.30x &\ge 80 \\ 0.30x &\ge 21.2 \\ x &\ge 70.\overline{6} \end{aligned}$$

Jordan needs about $71\%$ on the exam.

5. (Core) Forty students recorded how long they spent on homework last night.

Time (min)	$0$ to under $30$	$30$ to under $60$	$60$ to under $90$	$90$ to under $120$
Frequency	$5$	$12$	$15$	$8$

• (a) Estimate the mean time.
• (b) Which interval contains the median?

Solution

(a) The midpoints are $15, 45, 75, 105$:

$$\bar{x} \approx \frac{5(15) + 12(45) + 15(75) + 8(105)}{40} = \frac{2580}{40} = 64.5 \text{ min}$$

(b) The median is between the 20th and 21st values. The first two intervals hold $5 + 12 = 17$ values, and the third holds values $18$ to $32$. The median interval is $60$ to under $90$ minutes.

6. (Core) A runner’s times (in seconds) for six $100$ m sprints are $13.2, 13.5, 13.8, 14.0, 14.1, 21.6$. In the last sprint, she tripped.

• (a) Find the mean and median of all six times.
• (b) Find the mean and median without the $21.6$.
• (c) Which measure better describes her typical time? Why?

Solution

(a) Mean: $\dfrac{90.2}{6} \approx 15.03$ s. Median: $\dfrac{13.8 + 14.0}{2} = 13.9$ s.

(b) Mean: $\dfrac{68.6}{5} = 13.72$ s. Median: $13.8$ s.

(c) The median. The outlier raised the mean by more than $1.3$ s, but changed the median by only $0.1$ s. A mean of $15.03$ s is slower than five of her six runs.

7. (Core) Two brands of batteries were tested in the same flashlight. Lifetimes in hours:

• Brand A: $48, 50, 51, 52, 52, 53, 54$
• Brand B: $40, 45, 52, 55, 58, 60, 61$

Find the mean and median for each brand. Which brand would you buy, and why?

Solution

Brand A: mean $\dfrac{360}{7} \approx 51.4$ h, median $52$ h.

Brand B: mean $\dfrac{371}{7} = 53$ h, median $55$ h.

Brand B lasts longer on average, but its lifetimes are much more spread out ($40$ to $61$ h, compared with $48$ to $54$ h for A). If you want the longest life, choose B. If you need batteries you can count on (for example, for an emergency kit), A is more predictable. Either answer is fine with a good reason.

8. (Challenge) One class of $24$ students has a mean test mark of $72\%$. Another class of $30$ students has a mean of $81\%$. Find the mean mark of all $54$ students. Why isn’t it $76.5\%$?

Solution

Find each class’s total, then divide by the total number of students:

$$\mu = \frac{24(72) + 30(81)}{54} = \frac{1728 + 2430}{54} = \frac{4158}{54} = 77\%$$

This is a weighted mean with the class sizes as weights. The answer isn’t $\dfrac{72 + 81}{2} = 76.5\%$ because the second class has more students, so it pulls the combined mean toward $81\%$.

9. (Challenge) Make up a set of five positive whole numbers with a mean of $10$, a median of $8$, and a mode of $6$.

Solution

The median is $8$, so the middle (3rd) value is $8$. The mode is $6$, so $6$ must appear at least twice, and it has to be below the median: the list starts $6, 6, 8$. The mean is $10$, so the sum is $50$, and the last two values add to $50 - 20 = 30$. They must both be greater than $8$ and different from each other (so $6$ stays the only mode).

One answer: $6, 6, 8, 12, 18$. Check: mean $\dfrac{50}{5} = 10$, median $8$, mode $6$. ✓ (Other answers, such as $6, 6, 8, 14, 16$, also work.)