Mean, Median and Weighted Averages, SAT math lesson

Two classes take the same quiz. One class of 20 averages 80 points and the other class of 30 averages 70. It is tempting to say the combined average is 75, right in the middle. But the larger class pulls harder, and the true combined average is 74. Getting that right comes down to one habit, which is to think in totals and counts rather than in averages alone.

Key terms

Mean

Definition The mean of $n$ numbers $x_1, x_2, \dots, x_n$ is their sum divided by how many there are,

\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}.

It is also called the arithmetic mean or the average.

In plain words Add everything up and share it out equally. The mean is the amount each observation would have if the total were split evenly.

Example For the five values $3, 7, 8, 10, 12$ the sum is $40$ , so the mean is $40/5 = 8$ .

Total

Definition The total of a data set is the sum of all its observations. From the definition of the mean, for $n$ observations,

\text{total} = n \cdot \bar{x}.

This turns any statement about a mean into a statement about a total, which is often the easier quantity to work with.

In plain words If you know the average and the count, you know the sum. When values are added or removed, update the total and the count, then divide again.

Example A student’s five quiz scores have mean $82$ , so their total is $5 \cdot 82 = 410$ . What sixth score would raise the mean to $84$ ? Six scores with mean $84$ need a total of $6 \cdot 84 = 504$ , so the sixth score must be

504 - 410 = 94.

Median

Definition The median of a data set is found by sorting the $n$ values from least to greatest. If $n$ is odd, the median is the value in position $\frac{n+1}{2}$ . If $n$ is even, it is the mean of the values in positions $\frac{n}{2}$ and $\frac{n}{2} + 1$ . At least half of the values are less than or equal to the median and at least half are greater than or equal to it.

In plain words Line the values up in order and take the middle one, or the average of the middle two when there is no single middle.

Example The values $12, 3, 9, 7, 15, 4$ must be sorted first, giving $3, 4, 7, 9, 12, 15$ . There are six values, so the median is the mean of the third and fourth,

\frac{7 + 9}{2} = 8.

Taking the middle of the unsorted list would give $9$ or $7$ , both wrong.

Frequency tables

Definition In a frequency table each distinct value $x_i$ appears once with its frequency $f_i$ , the number of observations equal to $x_i$ . The number of observations is $n = f_1 + f_2 + \dots + f_k$ , and the mean is

\bar{x} = \frac{f_1 x_1 + f_2 x_2 + \dots + f_k x_k}{f_1 + f_2 + \dots + f_k}.

The median is found by counting through the frequencies to the middle position.

In plain words Each value counts as many times as it occurs. Multiply each value by its frequency to get the total, and divide by the number of observations, not by the number of rows.

Example Twenty people rate a film from $1$ to $4$ .

Rating	$1$	$2$	$3$	$4$
Frequency	$4$	$6$	$7$	$3$

The total is $4 + 12 + 21 + 12 = 49$ , so the mean is $49/20 = 2.45$ . For the median, positions $1$ to $4$ are ratings of $1$ , positions $5$ to $10$ are $2$ and positions $11$ to $17$ are $3$ . The tenth value is $2$ and the eleventh is $3$ , so the median is $2.5$ .

Weighted mean

Definition If numbers $x_1, \dots, x_k$ have positive weights $w_1, \dots, w_k$ , their weighted mean is

\frac{w_1 x_1 + w_2 x_2 + \dots + w_k x_k}{w_1 + w_2 + \dots + w_k}.

When groups of sizes $n_1, \dots, n_k$ have means $\bar{x}_1, \dots, \bar{x}_k$ , the mean of all the observations together is the weighted mean of the group means with the group sizes as weights. The plain average of the group means is correct only when every group has the same size.

In plain words Bigger groups count for more. Rebuild each group’s total, add the totals, and divide by everyone.

Example A class of $20$ has mean $80$ and a class of $30$ has mean $70$ . The totals are $1600$ and $2100$ , so the combined mean is

\frac{1600 + 2100}{20 + 30} = \frac{3700}{50} = 74.

The unweighted guess of $75$ treats the two classes as if they were the same size. The true answer sits closer to $70$ because more students scored around $70$ .

Outlier

Definition An outlier is an observation that lies unusually far from the rest of the data. There is no single universal cutoff. One common rule calls a value an outlier when it is below $Q_1 - 1.5 \cdot \text{IQR}$ or above $Q_3 + 1.5 \cdot \text{IQR}$ , where $Q_1$ and $Q_3$ are the quartiles and $\text{IQR} = Q_3 - Q_1$ . Because the mean uses the size of every value, an outlier can move it a lot. The median depends only on which values sit in the middle positions, so changing an extreme value does not change the median at all as long as that value stays on the same side of the middle values.

In plain words One very large or very small value drags the mean toward it, while the median barely notices. That is why the median is often preferred for data such as incomes or home prices.

Example The eight values $10, 12, 13, 13, 14, 15, 16, 17$ have total $110$ , mean $13.75$ and median $13.5$ . Replace $17$ with $60$ . The total becomes $153$ and the mean becomes $153/8 = 19.125$ , higher than seven of the eight values. The middle two values are still $13$ and $14$ , so the median stays $13.5$ . Taking the quartiles as the medians of the lower four and upper four values of the new data gives $Q_1 = 12.5$ and $Q_3 = 15.5$ , so the upper cutoff is $15.5 + 1.5 \cdot 3 = 20$ and $60$ is an outlier by the common rule.

Check yourself

A survey of $20$ households found $5$ households owning $0$ pets, $8$ owning $1$ , $4$ owning $2$ and $3$ owning $3$ . What is the mean number of pets per household? (A) $1$ (B) $1.25$ (C) $1.5$ (D) $5$

Show the answer

The correct choice is (B). The total number of pets is $0 \cdot 5 + 1 \cdot 8 + 2 \cdot 4 + 3 \cdot 3 = 25$ , and

\frac{25}{20} = 1.25.

Choice (A) is the median, (C) averages the values $0$ to $3$ without their frequencies, and (D) averages the frequencies.

The mean of six numbers is $15$ . Five of the numbers are $12$ , $18$ , $9$ , $20$ and $14$ . What is the sixth number?

Show the answer

The answer is $17$ . Six numbers with mean $15$ have total $6 \cdot 15 = 90$ . The five known numbers add to $73$ , so the sixth is $90 - 73 = 17$ .

In a survey, $12$ adults slept a mean of $6$ hours and $18$ teenagers slept a mean of $9$ hours. What is the mean number of hours for all $30$ people? (A) $7.2$ (B) $7.5$ (C) $7.8$ (D) $9$

Show the answer

The correct choice is (C). The totals are $12 \cdot 6 = 72$ and $18 \cdot 9 = 162$ , so the combined mean is

\frac{72 + 162}{30} = 7.8.

Choice (B) ignores the group sizes, and (A) swaps them, giving the larger weight to the adults. Choice (D) is the mean of the teenagers alone, which leaves out the adults. The answer must lie between $6$ and $9$ , closer to $9$ because there are more teenagers.

Mean, Median and Weighted Averages

Theory, part 77

Practice, part 78

Key terms

Mean

Total

Median

Frequency tables

Weighted mean

Outlier

Check yourself

How this lesson connects

Builds on

Used later in