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The average number of days that construction workers miss per year is \(11 .\) The standard deviation is \(2.3 .\) The average number of days that factory workers miss per year is 8 with a standard deviation of \(1.8 .\) Which class is more variable in terms of days missed?

Short Answer

Expert verified
Factory workers are more variable in terms of days missed.

Step by step solution

01

Understand the Concept of Variability

Variability refers to how spread out the values are in a data set. One common measure of variability is the standard deviation. A larger standard deviation indicates greater variability.
02

Identify Given Data for Both Groups

The construction workers have an average (mean) of 11 days missed with a standard deviation of 2.3. Factory workers have an average of 8 days missed with a standard deviation of 1.8.
03

Understand the Coefficient of Variation (CV)

Since the datasets have different means, it is helpful to use the coefficient of variation (CV) to compare their variability. The coefficient of variation is defined as the ratio of the standard deviation to the mean, expressed as a percentage: \[CV = \left( \frac{\text{Standard Deviation}}{\text{Mean}} \right) \times 100\% \]
04

Calculate the CV for Construction Workers

For construction workers:\[CV = \left( \frac{2.3}{11} \right) \times 100 \% \approx 20.91\% \]
05

Calculate the CV for Factory Workers

For factory workers:\[CV = \left( \frac{1.8}{8} \right) \times 100 \% \approx 22.50\% \]
06

Compare the Coefficient of Variation

Comparing the CVs, construction workers have a CV of approximately 20.91%, and factory workers have a CV of 22.50%. A higher CV indicates more variability.

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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 key measure used to quantify the amount of variation or dispersion in a set of data values. It's a statistic that indicates how each number in a dataset deviates from the mean, or average, value of the dataset.
Calculating the standard deviation involves the following steps:
  • Find the mean (average) of the dataset.
  • Subtract the mean from each data point and square the result.
  • Calculate the mean of these squared differences.
  • Take the square root of this mean. This result is the standard deviation.
The larger the standard deviation, the more spread out the values in the dataset are. For example, in the exercise, construction workers have a standard deviation of 2.3, meaning their number of missed days varies more than the factory workers, who have a standard deviation of 1.8. However, when comparing variability between datasets with different means, it's better to use another measure called the coefficient of variation.
Coefficient of Variation
The coefficient of variation (CV) is a statistical measure that helps to understand the degree of variability in data compared to its mean. It is especially useful when the means of the datasets are significantly different. This is because CV normalizes variability by expressing it as a percentage of the mean, allowing for a direct comparison of variability across datasets.
To calculate the CV, use the formula:
  • CV = (Standard Deviation / Mean) × 100%
In the exercise, the CV for the construction workers is calculated as approximately 20.91%, while the CV for the factory workers is about 22.50%.
A higher CV value signifies greater relative variability. In this case, factory workers have a slightly higher CV, indicating that, relative to their mean, their attendance variability is higher than that of the construction workers.
Mean Comparison
When comparing datasets, the mean provides a central value that reflects the average of the dataset. Although it is not a direct measure of variability, it is crucial when using methods like the coefficient of variation, where the mean is part of the calculation.
In the exercise, the mean for construction workers is 11 days missed per year, while for factory workers it is 8 days missed per year. These different means show that, on average, construction workers miss more days. However, just comparing average days missed doesn't tell us about the relative variability of those days. That's why CV is used to take the mean into account when assessing variability. In conclusion, while the mean provides helpful information about the central tendency of the data, it should be used alongside other statistical measures to fully understand data variability.

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Most popular questions from this chapter

Harmonic Mean The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2 . The average is found as shown. Since Time \(=\) distance \(\div\) rate then Time \(1=\frac{100}{40}=2.5\) hours to make the trip Time \(2=\frac{100}{50}=2\) hours to return Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

An employment evaluation exam has a variance of \(250 .\) Two particular exams with raw scores of 142 and 165 have \(z\) scores of -0.5 and \(0.955,\) respectively. Find the mean of the distribution.

The average weekly earnings in dollars for various industries are listed below. Find the percentile rank of each value. $$ \begin{array}{lllllllll} 804 & 736 & 659 & 489 & 777 & 623 & 597 & 524 & 228 \end{array} $$ For the same data, what value corresponds to the 40 th percentile?

What types of symbols are used to represent sample statistics? Give an example. What types of symbols are used to represent population parameters? Give an example.

The average full time faculty member in a post secondary degree-granting institution works an average of 53 hours per week. a. If we assume the standard deviation is 2.8 hours, what percentage of faculty members work more than 58.6 hours a week? b. If we assume a bell-shaped distribution, what percentage of faculty members work more than 58.6 hours a week?

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