/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 The following times were recorde... [FREE SOLUTION] | 91Ó°ÊÓ

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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). $$\begin{array}{llllll} \text {Quarter-Mile Times:} & .92 & .98 & 1.04 & .90 & .99 \\ \text {Mile Times:} & 4.52 & 4.35 & 4.60 & 4.70 & 4.50 \end{array}$$ 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 mile times are more consistent (lower CV), contrary to the coach's statement.

Step by step solution

01

Calculate Mean for Quarter-Mile Times

First, calculate the mean of the quarter-mile times. Add all the times together and divide by the number of times: \[ \text{Mean} = \frac{0.92 + 0.98 + 1.04 + 0.90 + 0.99}{5} \] \[ \text{Mean} = 0.966 \text{ minutes} \]
02

Calculate Standard Deviation for Quarter-Mile Times

Now, compute the standard deviation for the quarter-mile times. The standard deviation is calculated as follows: 1. Calculate the variance by taking each time, subtracting the mean, squaring the result, summing all these squared differences, and dividing by the number of times (since this is a population): \[ \text{Variance} = \frac{(0.92 - 0.966)^2 + (0.98 - 0.966)^2 + (1.04 - 0.966)^2 + (0.90 - 0.966)^2 + (0.99 - 0.966)^2}{5} \] \[ \text{Variance} = 0.003872 \] 2. Take the square root of the variance to get the standard deviation: \[ \text{Standard Deviation} = \sqrt{0.003872} \approx 0.062 \text{ minutes} \]
03

Calculate Coefficient of Variation for Quarter-Mile Times

The coefficient of variation (CV) is calculated by dividing the standard deviation by the mean and multiplying by 100 to express it as a percentage: \[ \text{CV} = \left(\frac{0.062}{0.966}\right) \times 100 \approx 6.42\% \]
04

Calculate Mean for Mile Times

Next, calculate the mean of the mile times. Add all the times together and divide by the number of times: \[ \text{Mean} = \frac{4.52 + 4.35 + 4.60 + 4.70 + 4.50}{5} \] \[ \text{Mean} = 4.534 \text{ minutes} \]
05

Calculate Standard Deviation for Mile Times

Calculate the standard deviation for the mile times: 1. Calculate the variance like before: \[ \text{Variance} = \frac{(4.52 - 4.534)^2 + (4.35 - 4.534)^2 + (4.60 - 4.534)^2 + (4.70 - 4.534)^2 + (4.50 - 4.534)^2}{5} \] \[ \text{Variance} = 0.0216 \] 2. Take the square root of the variance to get the standard deviation: \[ \text{Standard Deviation} = \sqrt{0.0216} \approx 0.147 \text{ minutes} \]
06

Calculate Coefficient of Variation for Mile Times

The coefficient of variation (CV) for mile times is calculated the same way: \[ \text{CV} = \left(\frac{0.147}{4.534}\right) \times 100 \approx 3.24\% \]
07

Compare Coefficients of Variation

Compare the coefficients of variation: - Quarter-Mile CV: \( 6.42\% \) - Mile CV: \( 3.24\% \) The lower CV for mile times indicates that mile times show less variability relative to their mean compared to quarter-mile times.

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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 measure that gives us an idea of how much variation or spread exists in a set of numbers. In other words, it tells us how closely the numbers are bunched around the mean (average) of the data set. The formula for standard deviation involves calculating the square root of the variance, which is the average of the squared differences from the mean.
  • To find the variance, take each data point, subtract the mean from it, square the result, and then find the average of these squared differences.
  • The standard deviation is then the square root of this variance, providing a measure in the same units as the original data.
When comparing different data sets, the standard deviation can help us understand which set has more variation. For example, in our exercise, the standard deviation for quarter-mile times is approximately 0.062 minutes, while for mile times, it's around 0.147 minutes. This means that the times for the quarter-mile runs are generally more consistent (less spread out) than those for mile runs just by looking at raw numbers.
Coefficient of Variation
The coefficient of variation (CV) is a standardized measure of dispersion of a probability distribution or frequency distribution. It is particularly useful when you want to compare the degree of variation from one data set to another, even if the units of the data sets are different. The coefficient of variation is calculated as the ratio of the standard deviation to the mean, usually expressed as a percentage.
  • This formula is simple: \[ \text{CV} = \left(\frac{\text{Standard Deviation}}{\text{Mean}}\right) \times 100 \]
  • The CV allows us to see which set of data shows more variation relative to its mean. A lower CV indicates more consistency.
In our exercise, the quarter-mile times have a CV of about 6.42%, whereas the mile times have a CV of approximately 3.24%. This tells us that, relative to their means, the mile times actually show less variability compared to the quarter-mile times, contradicting the initial intuitive observation based on the raw standard deviations.
Data Variability
Data variability refers to how spread out or closely clustered the data points in a data set are. It gives us insights into the reliability and consistency of the data.
  • If a data set has low variability, it means the data points are close to each other and the mean.
  • If there is high variability, data points are spread out over a larger range, being far from the mean.
Understanding data variability is critical, especially in fields like quality control, research, and any area where consistency is key. In the context of our exercise, calculating both the standard deviation and the coefficient of variation provided different perspectives on data variability:
- The standard deviation offered a direct measure, showing more spread in mile times.
- The coefficient of variation provided relative context, suggesting that mile times were more consistent than quarter-mile times when considered proportionally to their respective means.
Both measures are important, but context and the nature of the comparison often determine which measure is more relevant for decision-making.

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Most popular questions from this chapter

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