#### 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.

\begin{align*}\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)}\end{align*}

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: \begin{align*}\frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6}\end{align*}
11+12+13=66+36+26=116 . \begin{align*}\frac{11}{6}\end{align*}116 becomes the denominator of your final calculation. **Final calculation:**- Divide the numerator (3) by the denominator \begin{align*}\left(\frac{11}{6}\right)\end{align*}
(116) :

- Divide the numerator (3) by the denominator \begin{align*}\left(\frac{11}{6}\right)\end{align*}

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

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
of the values.*weighted harmonic mean*

As a formula, this looks like: \begin{align*}WHM =\frac{(\sum w)}{\left(\sum \frac{w}{a}\right)}
\end{align*}

**Example A**

Calculate the harmonic mean of \begin{align*}x\end{align*}

\begin{align*}x=\{4, 15, 17, 5, 22\}\end{align*}

**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 \begin{align*}\frac{1}{4} + \frac{1}{15} + \frac{1}{17} + \frac{1}{5} + \frac{1}{22} =.621\end{align*}
14+115+117+15+122=.621 , this is the denominator - The harmonic mean is \begin{align*}\frac{6}{.621} = 9.66 \end{align*}
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:

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

- \begin{align*}Average \ rate: \frac{total \ distance}{total \ time} = \frac{12 \ mi}{1.268 \ hrs} = 9.46 \ mph\end{align*}

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

- \begin{align*}HM = \frac{4}{\frac{1}{8} + \frac{1}{12} + \frac{1}{14} + \frac{1}{7}} = \frac{4}{.423} = 9.46 \ mph\end{align*}

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

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

\begin{align*}230 + 310 + 199 + 210 + 240 = 1189\end{align*}

- Next, calculate the sum of the distances over the rates (remember that \begin{align*} time = \frac{distance}{rate}\end{align*}):

\begin{align*}\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\end{align*}

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

**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 \begin{align*}n^{th}\end{align*} 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: \begin{align*}\{13, 17, 22, 29, 39, 45, 50\}\end{align*}

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

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: \begin{align*}\left \{ \frac{1}{3}, 5, 3, \frac{3}{5}, 6, \frac{5}{11} \right \}\end{align*}

**Solutions:**

1. First, count the values: 7

Second, sum the inverses of the values: \begin{align*}\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\end{align*}

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

2. First, sum the numerators: \begin{align*}1 + 2 + 2 + 1 + 3 = 9\end{align*}

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

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

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

\begin{align*}110 \ km + 230 \ km + 259 \ km + 300 \ km + 330 \ km = 1,229\end{align*}

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

\begin{align*}\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\end{align*}

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

4. a) **Basic harmonic mean:**

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

b) **Weighted harmonic mean:**

- First, find the sum of the numerators: \begin{align*}1 + 5 + 3 + 3 + 6 + 5 = 23 \end{align*}
- Second, find the sum of the values: \begin{align*}\frac{1}{3} + \frac{5}{1} + \frac{3}{1} + \frac{3}{5} + \frac{6}{1} + \frac{5}{11} = 15.39 \end{align*}
- Finally, divide the sum of the numerators by the sum of the values: \begin{align*}\frac{23}{15.39} = 1.494 \end{align*}

#### Practice

For questions 1 - 10, calculate the harmonic mean.

1. \begin{align*}\{5, 8, 9, 7, 6, 8, 3, 4\} \end{align*}

2. \begin{align*}\{12, 16, 18, 13, 14, 16\} \end{align*}

3. \begin{align*}\{23, 24, 26, 28\} \end{align*}

4. \begin{align*}\{33, 37, 38, 36, 35, 35, 36, 34\} \end{align*}

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

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

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

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

9. \begin{align*}\{5, 8, 9, 7, 6, 8, 3, 4\} \end{align*}

10. \begin{align*}\{12, 16, 18, 13, 14, 16\} \end{align*}

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 |

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.3.