The Most Misunderstood Average

Most uninformed students, when asked to calculate the average speed for a round trip with a “going” average speed of 30 miles per hour and a “returning” average speed of 60 miles per hour, would think that their average speed for the entire trip is 45 miles per hour [calculated as (30+60)/2 = 45]. The first task is to convince the students that this is the wrong answer.

For starters, you might ask the students if they believe it is fair to consider the two speeds with equal “weight.” Some may realize that the two speeds were achieved for different lengths of time and therefore cannot get the same weight. This might lead someone to offer that the trip at the slower speed, 30 mph, took twice as long and, therefore, ought to get twice the weight in the calculation of the average round-trip speed. This would then bring the calculation to the following:

(30+30+60)/3 = 40, which happens to be the correct average speed.

For those not convinced by this argument, try something a bit closer to “home.” A question can be posed about the grade a student deserves who scored 100% on nine of ten tests in a semester and on one test scored only 50%. Would it be fair to assume that this student’s performance for the term was 75% (i.e., 100+50/2)? The reaction to this suggestion will tend toward applying appropriate weight to the two scores in consideration. The 100% was achieved nine times as often as the 50% and therefore ought to get the appropriate weight. Thus, a proper calculation of the student’s average ought to be

(9×100)+50/10 = 95

This clearly appears more just!

An astute student may now ask, “What happens if the rates to be averaged are not multiples of one another?” For the speed problem above, one could find the time “going” and the time “returning” to get the total time, and then, with the total distance, calculate the “total rate,” which is, in fact, the average rate.

There is a more efficient way; a concept called the harmonic mean, which is the mean of a harmonic sequence. This frequently misunderstood mean (or average) usually causes confusion, but to avoid this, once we identify that we are to find the average of rates (i.e., the harmonic mean), then we have a lovely formula for calculating the harmonic mean for rates over the same base. In the above situation, the rates were for the same distance (round-trip legs).

The harmonic mean for two rates, a and b, is 2ab/(a+b); and for three rates, a, b, and c, the harmonic mean is 3abc/(ab+bc+ac).

You can see the pattern evolving, so that for 4 rates the harmonic mean is

4abcd/(abc+abd+acd+bcd).

Applying this to the above speed problem gives us

(2×30×60)/(30+60) = 3600/90 = 40

Begin by posing the following problem:

On Monday, a plane makes a round-trip flight from New York City to Washington with an average speed of 300 miles per hour. The next day, Tuesday, there is a wind of constant speed (50 miles per hour) and direction (blowing from New York City to Washington). With the same speed setting as on Monday, this same plane makes the same round trip on Tuesday. Will the Tuesday trip require more time, less time, or the same time as the Monday trip?

This problem should be slowly and carefully posed, so that students notice that the only thing that has changed is the “help and hindrance of the wind.” All other controllable factors are the same: distances, speed regulation, airplane’s conditions, etc. An expected response is that the two round-trip flights ought to be the same, especially since the same wind is helping and hindering two equal legs of a round-trip flight.

Realization that the two legs of the “wind trip” require different amounts of time should lead to the notion that the two speeds of this trip cannot be weighted equally as they were done for different lengths of time. Therefore, the time for each leg should be calculated and then appropriately apportioned to the related speeds.

We can use the harmonic mean formula to find the average speed for the “windy trip.” The harmonic mean is 2(350)(250)/(250+350) = 291.667, which is slower than the no-wind trip.

What a surprise!