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The following times were recorded by the quarter-mile and mile runners of a university track team (times are in minutes). \\[\begin{array}{lrrrrr} \text {Quarter-Mile Times:} & .92 & .98 & 1.04 & .90 & .99 \\ \text {Mile Times:} & 4.52 & 4.35 & 4.60 & 4.70 & 4.50 \end{array}\\] After viewing this sample of running times, one of the coaches commented that the quartermilers turned in the more consistent times. Use the standard deviation and the coefficient of variation to summarize the variability in the data. Does the use of the coefficient of variation indicate that the coach's statement should be qualified?

Short Answer

Expert verified
The coefficient of variation shows that the mile times are more consistent than the quarter-mile times.

Step by step solution

01

Calculate Mean Times

First, find the mean of the quarter-mile times and the mile times. The mean is calculated by adding all the times together and dividing by the number of observations.\[\text{Mean}_{\text{quarter}} = \frac{0.92 + 0.98 + 1.04 + 0.90 + 0.99}{5} = 0.966 \text{ minutes}\] \[\text{Mean}_{\text{mile}} = \frac{4.52 + 4.35 + 4.60 + 4.70 + 4.50}{5} = 4.534 \text{ minutes}\]
02

Calculate Standard Deviation for Each Set

The standard deviation measures the amount of variation in the data set. Use the formula: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\). For the quarter-mile: \[ s_{\text{quarter}} = \sqrt{\frac{(0.92-0.966)^2 + (0.98-0.966)^2 + (1.04-0.966)^2 + (0.90-0.966)^2 + (0.99-0.966)^2}{4}} = 0.056 \] For the mile: \[ s_{\text{mile}} = \sqrt{\frac{(4.52-4.534)^2 + (4.35-4.534)^2 + (4.60-4.534)^2 + (4.70-4.534)^2 + (4.50-4.534)^2}{4}} = 0.120 \]
03

Calculate Coefficient of Variation for Each Set

The coefficient of variation (CV) is calculated as the ratio of the standard deviation to the mean, expressed as a percentage. \[ \text{CV}_{\text{quarter}} = \left(\frac{0.056}{0.966}\right) \times 100\% = 5.8\% \] \[ \text{CV}_{\text{mile}} = \left(\frac{0.120}{4.534}\right) \times 100\% = 2.6\% \]
04

Interpret Results

The standard deviation suggests greater variability in the mile times than in the quarter-mile times. However, the coefficient of variation reveals that the mile times, when relative to their mean, are actually more consistent than the quarter-mile times (2.6% for mile times vs. 5.8% for quarter-mile times).
05

Conclusion

The coach's statement about the quarter-mile times being more consistent should be re-evaluated. Although absolute variability as measured by the standard deviation is less in quarter-mile times, the relative measure of the coefficient of variation shows that mile times are actually less variable in relation to their mean.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
The standard deviation is a statistical concept that provides insight into the amount of variation or spread in a set of data values. It tells us how much the individual data points differ from the mean (average) of the data set. A smaller standard deviation indicates that the data points are closer to the mean, which means less variability. Conversely, a larger standard deviation signifies greater variation among the data points.

To calculate standard deviation, we use the formula: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\). Here, \(x_i\) represents each individual data point, \(\bar{x}\) is the mean of the data, and \(n\) is the number of data points.

From the exercise, the quarter-mile times have a standard deviation of 0.056 minutes. This shows that these times are tightly clustered around the average time of 0.966 minutes. Meanwhile, the mile times have a standard deviation of 0.120 minutes, indicating slightly more spread or variability around their mean of 4.534 minutes.
Coefficient of Variation
The coefficient of variation (CV) is a useful statistical tool for comparing the relative variability of different data sets. It is defined as the standard deviation divided by the mean, expressed as a percentage. By using the CV, we can determine which data set has more variability relative to its mean.

The CV is calculated using the formula: \( \text{CV} = \left(\frac{s}{\bar{x}}\right) \times 100\% \), where \(s\) is the standard deviation and \(\bar{x}\) is the mean. This ratio tells us how much variability exists relative to the average value of the data set.

In our data, the quarter-mile times have a CV of 5.8%, while the mile times have a CV of 2.6%. This indicates that, in relative terms, the mile times demonstrate less variability compared to the quarter-mile times. Even though the absolute variability is larger for the mile times, when considering their average value, they show greater consistency.
Data Variability
Data variability refers to how much the data points in a set differ from each other and from the mean. It is an important aspect to understand because it affects how we interpret data.

When assessing variability, the standard deviation reveals the overall spread of the data, providing a straightforward way to see differences within a single data set. However, the coefficient of variation adjusts this spread for the scale or magnitude of the data, offering a normalized measure that allows for comparisons across different data sets or units.

In the context of the running times in the exercise, the variability is assessed using both standard deviation and the coefficient of variation. While the standard deviation gives us a sense of absolute differences, the CV helps us understand variability in relation to the mean. By employing multiple variability measures, we gain a more comprehensive view of the data, which is crucial for making informed conclusions like whether quarter-mile or mile times are genuinely more consistent.

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