Math Problem Statement

A Providence Hospital experiment involves two different waiting line configurations for patients arriving for admission. The waiting times ( in seconds ) are recorded with a single line configuration that feeds four stations and another configuration with individual lines at the four stations. Find the coefficient of variation for each of the two samples, and then compare the variation. What is the coefficient of variation for single line if this is the data set : 388, 395, 401, 408, 427, 439, 444, 460, 460, 460

Solution

To calculate the coefficient of variation (CV) for the single line configuration, we first need to find the mean and standard deviation of the data set. The coefficient of variation is given by:

Coefficient of Variation (CV)=Standard DeviationMean×100%\text{Coefficient of Variation (CV)} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%

Step 1: Calculate the Mean

The mean (μ\mu) is calculated by summing all the data points and then dividing by the number of data points.

Given dataTo calculate the coefficient of variation (CV) for the single line configuration, we first need to find the mean and standard deviation of the data set. The coefficient of variation is given by:

Coefficient of Variation (CV)=Standard DeviationMean×100%\text{Coefficient of Variation (CV)} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100\%

Step 1: Calculate the Mean

The mean (μ\mu) is calculated by summing all the data points and then dividing by the number of data points.

Given 388,395,401,408,427,439,444,460,460,460388, 395, 401, 408, 427, 439, 444, 460, 460, 460

μ=388+395+401+408+427+439+444+460+460+46010\mu = \frac{388 + 395 + 401 + 408 + 427 + 439 + 444 + 460 + 460 + 460}{10}

μ=434210=434.2 seconds\mu = \frac{4342}{10} = 434.2 \text{ seconds}

Step 2: Calculate the Standard Deviation

The standard deviation (σ\sigma) is calculated using the formula:

σ=i=1n(xiμ)2n\sigma = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n}}

Where:

  • xix_i represents each data point
  • μ\mu is the mean
  • nn is the number of data points

Let's calculate the variance first, which is the sum of squared differences from the mean divided by the number of data points.

σ2=(388434.2)2+(395434.2)2++(460434.2)210\sigma^2 = \frac{(388 - 434.2)^2 + (395 - 434.2)^2 + \ldots + (460 - 434.2)^2}{10}

Let me calculate this.### Results:

  • Mean (μ\mu) = 428.2 seconds
  • Standard Deviation (σ\sigma) = 26.96 seconds (rounded to two decimal places)
  • Coefficient of Variation (CV) = 6.30% (rounded to two decimal places)

Interpretation:

The coefficient of variation for the single line configuration is approximately 6.30%. This metric indicates the relative variability of the waiting times in comparison to the mean waiting time. A lower CV suggests that the waiting times are relatively consistent.

If you have the data for the individual line configuration, you can calculate its CV similarly and compare the variability between the two configurations.

Would you like to proceed with any further details, or have any questions? Here are five related questions you might find interesting:

  1. How does the coefficient of variation differ when applied to different types of data?
  2. What are the implications of a high coefficient of variation in a real-world scenario?
  3. How does the choice between a population and sample standard deviation affect the CV calculation?
  4. What could be potential sources of variability in hospital waiting times?
  5. How does the CV help in comparing two different distributions with different units or scales?

Tip: The coefficient of variation is particularly useful when you want to compare the degree of variation between different datasets, especially when the means of the datasets are different.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics

Formulas

Mean (μ) = Σx / n
Standard Deviation (σ) = √(Σ(x - μ)² / n)
Coefficient of Variation (CV) = (σ / μ) × 100%

Theorems

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Suitable Grade Level

College/University