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Harmonic Mean
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Objective

Here you will learn how and when to calculate the harmonic mean of a set.

Concept

Can you define the difference between the arithmetic, geometric, and harmonic means? Can you think of a situation when each might be appropriately used?

By the end of this lesson you will!

Guidance

The harmonic mean is a ‘middle number’ you can use when you are worried that the arithmetic mean of your set would be skewed by a few very large values (as in figure 1), and/or when particularly small values are disproportionately important. The harmonic mean is also useful for calculations of average rates, or finding other sorts of weighted averages .

The harmonic mean can be a much better average number than the arithmetic mean.

Of the three mean value calculations we discuss in this chapter, the harmonic mean is always the lowest value if calculated using the same data.

Calculating the harmonic mean is a bit more complex than the geometric mean, but it quickly gets easier with practice:

  • Count the number of values in your data set. This number becomes your numerator.
  • Calculate the sum of the inverses of the data values, this sum becomes your denominator.
  • Divide the numerator by the denominator, the resulting quotient is the harmonic mean of the values.

\frac{number \ of \ values}{\left(\frac{1}{value \ 1}\right) + \left(\frac{1}{value \ 2}\right) + \left(\frac{1}{value \ 3}\right) \cdots \left(\frac{1}{value \ n}\right)}

For example, to find the harmonic mean of the set {1, 2, 3}:

  • First count the values, there are 3. The number 3 becomes the numerator of your final calculation.
  • Sum the inverses of the values: \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6}\frac{11}{6} becomes the denominator of your final calculation.
  • Final calculation:
    • Divide the numerator (3) by the denominator \left(\frac{11}{6}\right) :

\frac{3}{\frac{11}{6}} &= \frac{18}{11} = 1.636 \\1.636 \ \text{is the } & \text{harmonic mean of} \ \{1, 2, 3\}

This type of harmonic mean assumes un-weighted values, such as the same distance travelled at different rates. To calculate the harmonic mean of differently weighted values ( the weighted harmonic mean ), such as when calculating the average speed of a trip made of segments with different distances and different rates, (see Example C), the process is nearly the same:

  • First calculate the sum of the weights of your values (as opposed to just counting the values, as with the basic harmonic mean), this number becomes your numerator.
  • Second, find the sum of the values (weights divided by amounts – NOT the inverses); this number becomes your denominator.
  • Divide the numerator by the denominator, the resulting quotient is the weighted harmonic mean of the values.

As a formula, this looks like: WHM =\frac{(\sum w)}{\left(\sum \frac{w}{a}\right)}   Where  w is the weight (the numerator if your values are fractional) and  a is the amount (the denominator if your values are fractional).

Example A

Calculate the harmonic mean of x .

x=\{4, 15, 17, 5, 22\}

Solution: Follow the steps above for the un-weighted harmonic mean:

  • The number of values is 6, this is the numerator
  • The sum of the inverses is \frac{1}{4} + \frac{1}{15} + \frac{1}{17} + \frac{1}{5} + \frac{1}{22} =.621 , this is the denominator
  • The harmonic mean is \frac{6}{.621} = 9.66

Example B

Kiera decided to take a quick tour around town on her bike, if the information below describes the speeds she travelled over each equal-length segment, what was her average speed for the trip?

Segment #1 8 mph 3 mi
Segment #2 12 mph 3 mi
Segment #3 14 mph 3 mi
Segment #4 7 mph 3 mi

Solution: There are two ways to solve this:

1. We could find the average speed for the whole trip by dividing the entire distance by the entire time:

  1. Sum the Distances: 3 \ mi + 3 \ mi + 3 \ mi + 3 \ mi = 12 \ mi\mathbf{ \ total \ distance}
  2. Sum the times:
    1. \text{Segment} \ 1 \ \frac{3 \ mi}{8 \ mph}=.375 \ hr
    2. \text{Segment} \ 2 \ \frac{3 \ mi}{12 \ mph}=.25 \ hr
    3. \text{Segment} \ 3 \ \frac{3 \ mi}{14 \ mph}=.214 \ hr
    4. \text{Segment} \ 4 \ \frac{3 \ mi}{7 \ mph}=.429 \ hr
    5. TOTAL \ TIME: .375 + .25 + .214 + .429 = 1.268 \ hrs\mathbf{ \ total \ time}
  3. Average \ rate: \frac{total \ distance}{total \ time} = \frac{12 \ mi}{1.268 \ hrs} = 9.46 \ mph

2. The other method is to find the harmonic mean of the four rates:

  1. HM = \frac{4}{\frac{1}{8} + \frac{1}{12} + \frac{1}{14} + \frac{1}{7}} = \frac{4}{.423} = 9.46 \ mph

Obviously, the second method is more efficient!

Example C

Use the weighted harmonic mean to find the average rate of a traveler who records the following itinerary:

1 st Segment 230 km 66 kph
2 nd Segment 310 km 83 kph
3 rd Segment 199 km 72 kph
4 th Segment 210 km 77 kph
5 th Segment 240 km 91 kph

Solution: Because this problem includes segments with varying speeds and varying distances, we will need to use a weighted harmonic mean to calculate the average rate.

Use the formula:  WHM= \frac{(\sum w)}{\left(\sum \frac{w}{a}\right)}  Where  w (the weight) is the distance travelled at each rate, and  a (the amount) is the rate for each segment.

  • First, calculate the sum of the weights (distances):

230 + 310 + 199 + 210 + 240 = 1189

  • Next, calculate the sum of the distances over the rates (remember that  time = \frac{distance}{rate} ):

\frac{230 \ km}{66 \ kph} + \frac{310 \ km}{83 \ kph} + \frac{199 \ km}{72 \ kph} + \frac{210 \ km}{77 \ kph} + \frac{240 \ km}{91 \ kph} = 3.48 + 3.735 + 2.764 + 2.727 + 2.637 = 15.343

  • Finally, divide the sum of the weights by the sum of the times.
  • The weighted harmonic mean is \frac{1189}{15.343} = 77.5 \ kph

Concept Problem Revisited

Can you define the difference between the arithmetic, geometric, and harmonic means? Can you think of a situation when each might be appropriately used?

The arithmetic mean is the simple average of a set of values, the sum of the values divided by the number of values.  It is appropriate for situations such as the average rate for a journey composed of segments of equal time and distance.

The geometric rate is the  n^{th} root of the product of the values, and is appropriate for situations such as deriving a single value to represent scores from multiple scales. The single value could then be used to compare an overall ranking of scores.

The harmonic mean is the reciprocal of the arithmetic means of the reciprocals of a set of values. It is useful for calculations such as the average rate for a journey composed of segments of differing times or distances. A weighted harmonic mean can be used to calculate the average rate of a journey composed of differing times and distances.

Vocabulary

A harmonic mean is defined as the reciprocal of the mean of the reciprocals of the values in a set.

A weighted harmonic mean is a harmonic mean of values with varying frequencies or weights.

A weighted average is a common term for a central value meant to account for values of different weights in the same set.

Guided Practice

1. Find the harmonic mean of the set: \{13, 17, 22, 29, 39, 45, 50\}

2. Find the weighted harmonic mean of the set: \left \{ \frac{1}{3}, \frac{2}{5}, \frac{2}{7}, \frac{1}{2}, \frac{3}{11}\right \}

3. Use the weighted harmonic mean to find the average rate of a traveler who records the following itinerary:

1 st Segment 110 km 23 kph
2 nd Segment 230 km 56 kph
3 rd Segment 259 km 42 kph
4 th Segment 300 km 102 kph
5 th Segment 330 km 71 kph

4. Compare the a) basic and b) weighted, harmonic means of the set: \left \{ \frac{1}{3}, 5, 3, \frac{3}{5}, 6, \frac{5}{11} \right \}

Solutions:

1. First, count the values: 7

Second, sum the inverses of the values: \frac{1}{13} + \frac{1}{17} + \frac{1}{22} + \frac{1}{29} + \frac{1}{39} + \frac{1}{45} + \frac{1}{50} = \frac{4497703}{15862275} = .283

Finally, divide the count by the sum of the inverses: \frac{7}{.283} = 24. 74

2. First, sum the numerators: 1 + 2 + 2 + 1 + 3 = 9

Second, sum the values: \frac{1}{3} + \frac{2}{5} + \frac{2}{7} + \frac{1}{2} + \frac{3}{11} = \frac{4, 139}{2,310} = 1.792

Finally, divide the sum of the numerators by the sum of the values: \frac{9}{1.792}=5.022

3. First, find the sum of the distances (the weights of the values):

110 \ km + 230 \ km + 259 \ km + 300 \ km + 330 \ km = 1,229

Next, find the sum of the times \left(\frac{\text{distance}}{\text{rate}} = \text{time}\right) :

\frac{110 \ km}{23 \ kph} + \frac{230 \ km}{56 \ kph} + \frac{259 \ km}{42 \ kph} + \frac{300 \ km}{102 \ kph} + \frac{330 \ km}{71 \ kph} = 4.78 + 4.10 + 6.17 + 2.94 + 4.65 =22.64

Finally, divide the distance by the time: \frac{1,229}{22.65} = 54.26 \ kph

4. a) Basic harmonic mean:

  1. First, count the values: there are 6
  2. Second, sum the inverses of the values: \frac{3}{1} + \frac{1}{5} + \frac{1}{3} + \frac{5}{3} + \frac{1}{6} + \frac{11}{5} = \frac{227}{30} \ or \ 7.57
  3. Finally, divide the count by the sum of the inverses: \frac{6}{7.57} = 0.792

b) Weighted harmonic mean:

  1. First, find the sum of the numerators: 1 + 5 + 3 + 3 + 6 + 5 = 23
  2. Second, find the sum of the values: \frac{1}{3} + \frac{5}{1} + \frac{3}{1} + \frac{3}{5} + \frac{6}{1} + \frac{5}{11} = 15.39
  3. Finally, divide the sum of the numerators by the sum of the values: \frac{23}{15.39} = 1.494

Practice

For questions 1 - 10, calculate the harmonic mean.

1. \{5, 8, 9, 7, 6, 8, 3, 4\}

2. \{12, 16, 18, 13, 14, 16\}

3. \{23, 24, 26, 28\}

4. \{33, 37, 38, 36, 35, 35, 36, 34\}

5. \left \{ \frac{5}{8}, \frac{13}{21}, \frac{7}{9}, \frac{3}{5}, \frac{11}{23} \right \}

6. \left \{ \frac{6}{7}, \frac{17}{23}, \frac{17}{19}, \frac{6}{11}, \frac{5}{13} \right \}

7. \left \{\frac{2}{5}, \frac{5}{7} \ 2 \frac{3}{4}, \frac{15}{3}, \frac{9}{11}, \frac{14}{17} \right \}

8. \left \{ 3 \frac{3}{4}, 2 \frac{2}{3}, 5 \frac{7}{8}, 3.25, 1 \frac{6}{5} \right \}

9. \{5, 8, 9, 7, 6, 8, 3, 4\}

10. \{12, 16, 18, 13, 14, 16\}

11. Brian records the following trip data, use it and the basic harmonic mean to find his average rate for the complete trip:

Segment #1 18 mph 4.2 mi
Segment #2 21 mph 4.2 mi
Segment #3 24 mph 4.2 mi
Segment #4 17 mph 4.2 mi

12. Use the weighted harmonic mean to find the average rate of a traveler who records the following itinerary:

1 st Segment 21.1 km 23 kph
2 nd Segment 32.0 km 15.6 kph
3 rd Segment 29.5 km 14.2 kph
4 th Segment 30.5 km 10.2 kph

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