Problem 6 - Basic Statistics for Business and Economics

Question

Customers experiencing technical difficulty with their Internet cable hookup may call an 800 number for technical support. It takes the technician between 30 seconds to 10 minutes to resolve the problem. The distribution of this support time follows the uniform distribution. a. What are the values for \(a\) and \(b\) in minutes? b. What is the mean time to resolve the problem? What is the standard deviation of the time? c. What percent of the problems take more than 5 minutes to resolve. d. Suppose we wish to find the middle 50 percent of the problem-solving times. What are the end points of these two times?

Step-by-Step Solution

Answer

a: 0.5, 10 minutes; b: Mean: 5.25 min, SD: 2.74 min. c: 52.63%. d: 2.875 to 7.625 minutes.

Step 1: Identify the Range for Uniform Distribution

The range of the uniform distribution is defined by values \(a\) and \(b\), where \(a\) is the smallest value and \(b\) is the largest value of the distribution. Since the support time takes between 30 seconds and 10 minutes, we convert these to minutes: \(a = 0.5\) minutes and \(b = 10\) minutes.

Step 2: Calculate the Mean of the Uniform Distribution

The mean of a uniform distribution is given by \(\mu = \frac{a+b}{2}\). Substituting the values, we have \(\mu = \frac{0.5+10}{2} = 5.25\) minutes.

Step 3: Calculate the Standard Deviation of the Uniform Distribution

The standard deviation \(\sigma\) of a uniform distribution is calculated as \(\sigma = \frac{b-a}{\sqrt{12}}\). Therefore, \(\sigma = \frac{10-0.5}{\sqrt{12}} \approx 2.74\) minutes.

Step 4: Find Percent of Problems Taking More than 5 Minutes

To find what percent of the problems take more than 5 minutes, we calculate the probability \(P(X > 5)\). For a uniform distribution, this is \(\frac{b-x}{b-a}\). Substituting the values, \(P(X > 5) = \frac{10-5}{10-0.5} = \frac{5}{9.5} \approx 0.5263\), which is 52.63%.

Step 5: Find the Middle 50 Percent of Problem-Solving Times

The middle 50% refers to the interval between the 25th and 75th percentiles of the distribution. The 25th percentile (\(Q1\)) is \(a + 0.25(b-a)\) and the 75th percentile (\(Q3\)) is \(a + 0.75(b-a)\). So:\(Q1 = 0.5 + 0.25(10-0.5) = 2.875\) minutes\(Q3 = 0.5 + 0.75(10-0.5) = 7.625\) minutes. The times range between 2.875 and 7.625 minutes.

Key Concepts

Mean Calculation

The mean, also known as the average, gives a central value around which all other data points are distributed. In the case of a uniform distribution, which is characterized by every outcome being equally likely between the minimum (\(a\)) and maximum (\(b\)) values, the calculation of the mean is straightforward.

To find the mean of a uniform distribution, simply sum up the minimum and the maximum value and divide by 2. This formula can be written as: \[ \mu = \frac{a+b}{2} \] Using the problem's values of \(a = 0.5\) (minutes) and \(b = 10\) (minutes), we compute:

\[ \mu = \frac{0.5+10}{2} = 5.25 \] minutes.

This result indicates that on average, it takes about 5.25 minutes to resolve a technical support issue under this uniform distribution.

Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of values. It shows how much individual values deviate, on average, from the mean.

In the context of a uniform distribution, standard deviation can be calculated using the formula: \[ \sigma = \frac{b-a}{\sqrt{12}} \] Here, \(a\) is the lowest value, and \(b\) is the highest value in the range.

Substituting our example's values, \(a = 0.5\) and \(b = 10\): \[ \sigma = \frac{10-0.5}{\sqrt{12}} \approx 2.74 \] minutes.

This calculation reflects that the time to resolve issues can vary within about ±2.74 minutes from the mean of 5.25 minutes. Most issues will likely be solved within 5.25 ± 2.74 minutes, capturing this spread.

Percentile Calculation

A percentile provides a way to understand the relative standing of a particular value within a dataset. In our scenario, it's important to also focus on the time measurements within the dataset.

To compute the percentage of problems taking longer than a certain time (e.g., 5 minutes), convert it into a fractional representation of how much of the distribution is beyond that point.
The probability that the time exceeds 5 minutes in a uniform distribution is: \[ P(X > 5) = \frac{b-x}{b-a} \] Plugging in the values (\(x = 5\), \(a = 0.5\), \(b = 10\)): \[ P(X > 5) = \frac{10-5}{10-0.5} = \frac{5}{9.5} \approx 0.5263 \]
Therefore, approximately 52.63% of technical support issues take more than 5 minutes to resolve.

This percentile calculation helps in understanding what proportion of calls need more time.

Probability Calculation

Probability in the context of a uniform distribution allows us to predict how likely a particular outcome is, given all options are equally probable.

Finding the middle 50% of problem-solving times demands calculating specific percentiles—specifically the 25th and 75th. These endpoints encapsulate the middle range where half of the observations fall.

The 25th percentile (\(Q1\)) and the 75th percentile (\(Q3\)) are calculated using these formulas:

• \( Q1 = a + 0.25(b-a) \)
• \( Q3 = a + 0.75(b-a) \)

Substituting the exercise's values:

• \( Q1 = 0.5 + 0.25(10-0.5) = 2.875 \) minutes
• \( Q3 = 0.5 + 0.75(10-0.5) = 7.625 \) minutes

Hence, the middle 50% of the data falls between 2.875 and 7.625 minutes. This means that half of the customers' issues are resolved within this time frame, providing a critical insight into service performance.