Understanding Average Ratios and Central Tendency

The question asks for the most appropriate measure of central tendency to calculate an average of ratios or rates, such as 'price per unit', 'work done per hour', or 'kilometres per hour'. These are examples of rates where we want to find an average value under certain conditions.

Measures of Central Tendency Explained

Central tendency measures provide a single value that represents the center or typical value of a dataset. Let's look at the options:

• Arithmetic Mean: This is the most common average, calculated by summing all values and dividing by the number of values. It's suitable for data where the values are directly additive, like heights or scores. Formula: \(\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\).
• Geometric Mean: Used primarily for averaging rates of change or ratios where values are multiplied together, like average investment returns over several years. Formula: \(G = \sqrt[n]{x_1 \times x_2 \times \dots \times x_n}\).
• Harmonic Mean: This is specifically useful for averaging rates or ratios, particularly when the numerator represents a fixed quantity or when dealing with inverse relationships. It's calculated as the reciprocal of the arithmetic mean of the reciprocals of the observations. Formula: \(H = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}\).
• Mode: The mode is the value that appears most frequently in a dataset. It's used for categorical data or to find the most common occurrence, not for calculating an average rate.

Why Harmonic Mean is Suitable for Average Ratios

Ratios like 'price/unit' or 'kilometres/hour' represent rates. When calculating an average of such rates, the harmonic mean is often the correct choice because it appropriately accounts for the underlying structure of these rates.

Consider the example of calculating the average speed for a journey:

Suppose you travel 100 kilometres at 50 km/hr and then another 100 kilometres at 100 km/hr.

• Time for the first part: \(\frac{100 \text{ km}}{50 \text{ km/hr}} = 2 \text{ hours}\).
• Time for the second part: \(\frac{100 \text{ km}}{100 \text{ km/hr}} = 1 \text{ hour}\).
• Total distance = 200 km.
• Total time = 2 + 1 = 3 hours.
• Actual average speed = \(\frac{\text{Total Distance}}{\text{Total Time}} = \frac{200 \text{ km}}{3 \text{ hours}} \approx 66.67 \text{ km/hr}\).

If we incorrectly used the arithmetic mean:

Arithmetic mean of speeds = \(\frac{50 \text{ km/hr} + 100 \text{ km/hr}}{2} = 75 \text{ km/hr}\). This is incorrect because it doesn't account for the fact that you spent more time travelling at the slower speed.

Now, let's use the harmonic mean for the speeds:

Harmonic Mean (H) = \(\frac{2}{\frac{1}{50} + \frac{1}{100}}\)

H = \(\frac{2}{\frac{2}{100} + \frac{1}{100}}\)

H = \(\frac{2}{\frac{3}{100}}\)

H = \(\frac{2 \times 100}{3} = \frac{200}{3} \approx 66.67 \text{ km/hr}\).

This matches the actual average speed. Similarly, for 'price/unit', if you buy different quantities at different prices, the harmonic mean helps find the average cost per unit correctly, especially if considering the total amount spent. For 'work done/hour', it helps find the average rate of work.

Conclusion

Therefore, when the task is to find the average of rates or ratios like price per unit, work done per hour, or kilometres per hour, the harmonic mean is the suitable measure of central tendency.