Averages in Context

Averages in Context

The average (mean) of a set of numbers is their sum divided by how many there are. In Part V you used that definition on its own. Here it appears inside word problems about grades, classes, and trips, and the key is knowing which version of "average" the situation calls for. There are three tools: working with totals, weighting, and average speed.

Totals and missing values

The definition rearranges into a more useful form: sum = average times count. When a problem gives the average and all but one value, multiply the average by the count to get the total, then subtract the known values.

Four numbers have an average of $15$, and three of them are $10$, $14$, and $19$. The total is $15\times 4=60$, and the known values add to $43$, so the missing value is $60-43=17$.

The same idea handles a value being added to a set. If five quiz scores average $82$, their total is $410$. Adding a sixth score of $94$ gives a new total of $504$, and the new average is $\frac{504}{6}=84$. In every case, move from average to total, adjust the total, then move back.

Weighted averages

A weighted average is not the same as the simple average of two numbers. When groups contribute unequally, because of credit hours, class sizes, or exam weights, multiply each value by its weight before adding, and divide by the total weight. Averaging the raw numbers and ignoring the weights is one of the most common errors on this topic.

Worked example. A final grade is weighted $30\%$ on a midterm and $70\%$ on a final exam. A student scores $80$ on the midterm and $95$ on the final. Find the final grade.

Multiply each score by its weight and add: $0.30(80)+0.70(95)=24+66.5=90.5$. The simple average would be $87.5$, which understates the grade because it gives the lower midterm too much influence.

When the weights are counts instead of percents, the same method applies. A class of $25$ students averaging $75$ and a class of $15$ students averaging $91$ have a combined average of $\frac{25(75)+15(91)}{40}=\frac{3{,}240}{40}=81$. A reasonableness check: the combined average must fall between $75$ and $91$, and it must lie closer to $75$ because the larger group has the lower average. Grade point averages work identically, with credit hours as the weights: $3$ credit hours at $4.0$ and $2$ credit hours at $3.0$ give $\frac{3(4.0)+2(3.0)}{5}=3.6$.

Average speed for a full trip

Average speed is total distance divided by total time. It is almost never the simple average of the individual speeds. On a trip with equal distances at two different speeds, you spend more time at the slower speed, which pulls the true average below the midpoint.

Worked example. A drive covers $120$ miles at $40$ miles per hour, then $120$ miles at $60$ miles per hour. Find the average speed for the entire trip.

Compute each time: $\frac{120}{40}=3$ hours and $\frac{120}{60}=2$ hours. Then the average speed is $\frac{240}{5}=48$ miles per hour, not $50$.

The same approach works when the distances differ. A trip of $240$ miles in $4$ hours followed by $180$ miles in $4$ hours has an average speed of $\frac{420}{8}=52.5$ miles per hour. Always find the time for each leg using time = distance divided by speed, then add.

More on averages and reasonableness

Averages can also include negative numbers. To find the average of $-5$, $-2$, $8$, $15$, and $-1$, add carefully, keeping track of signs: $-5-2+8+15-1=15$, and $\frac{15}{5}=3$. Slow, deliberate addition matters here, since a sign slip produces a plausible-looking wrong choice.

Weighted averages appear in several disguises. A grade point average weights each course grade by its credit hours: $3$ credit hours at $3.0$ and $5$ credit hours at $3.8$ give $\frac{3(3.0)+5(3.8)}{8}=\frac{28}{8}=3.5$. The simple average of $3.0$ and $3.8$ is $3.4$, and the weighted value is higher because the course with the higher grade carries more credit. Whenever the weights are equal, the weighted average and the simple average coincide, as with two exams that count $50\%$ each, so notice when a problem is truly unequal.

For a trip with uneven distances, say $90$ miles at $45$ miles per hour and then $150$ miles at $50$ miles per hour, the times are $2$ and $3$ hours, so the average speed is $\frac{240}{5}=48$ miles per hour. Because $48$ is between $45$ and $50$, and closer to $50$ (the leg that took more time), the answer passes the reasonableness test.

Strategy

• If a problem hides a missing value, turn the average into a total first.
• If groups are unequal in size, credit, or weight, use a weighted average. Ask: "Does each value count the same?"
• For speed, find total distance and total time. Never average the speeds.
• Use the "between the values" check: a weighted average always lies between the smallest and largest value, closer to the one with more weight.
• When rounding is requested (for example, to the nearest hundredth), do the exact computation first and round only at the end.