Problem 32 - Basic Statistics for Business and Economics

Question

The times patrons at the Grande Dunes Hotel in the Bahamas spend waiting for an elevator follows a uniform distribution between 0 and 3.5 minutes. a. Show that the area under the curve is 1.00 . b. How long does the typical patron wait for elevator service? c. What is the standard deviation of the waiting time? d. What percent of the patrons wait for less than a minute? e. What percent of the patrons wait more than 2 minutes?

Step-by-Step Solution

Answer

a. The area is 1. b. The typical wait is 1.75 minutes. c. The standard deviation is 1.01 minutes. d. 28.6% wait less than 1 minute. e. 42.9% wait more than 2 minutes.

Step 1: Define the Uniform Distribution

A uniform distribution for waiting times between 0 and 3.5 minutes implies that any waiting time has an equal probability. The probability density function (pdf) for a uniform distribution is a constant value between the minimum (0) and maximum (3.5) values.

Step 2: Verify the Area under the Curve

For a uniform distribution, the pdf is given by the constant \[ f(x) = \frac{1}{b-a} = \frac{1}{3.5-0} = \frac{1}{3.5} \].The area under the curve over the interval [0, 3.5] is calculated by integrating the pdf over this range:\[ \text{Area} = \int_{0}^{3.5} \frac{1}{3.5} \, dx = \frac{1}{3.5} \times (3.5 - 0) = 1. \] This confirms the total area under the curve is 1.

Step 3: Calculate the Mean (Typical Wait Time)

The mean of a uniform distribution is given by \( \mu = \frac{a + b}{2} \). For this distribution: \( \mu = \frac{0 + 3.5}{2} = 1.75 \) minutes. This is the typical wait time for a patron.

Step 4: Calculate the Standard Deviation

The standard deviation of a uniform distribution is given by \( \sigma = \frac{b-a}{\sqrt{12}} \). Here this becomes \( \sigma = \frac{3.5 - 0}{\sqrt{12}} \approx 1.01 \) minutes.

Step 5: Calculate Percentage of Patrons Waiting Less Than 1 Minute

To find the percentage of patrons waiting less than 1 minute, calculate the area under the pdf from 0 to 1:\[ \text{Percentage} = \int_{0}^{1} \frac{1}{3.5} \, dx = \frac{1}{3.5} \times (1 - 0) = \frac{1}{3.5} \approx 0.286. \]Convert this to a percentage giving approximately 28.6%.

Step 6: Calculate Percentage of Patrons Waiting More Than 2 Minutes

To find the percentage of patrons waiting more than 2 minutes, calculate the area under the pdf from 2 to 3.5:\[ \text{Percentage} = \int_{2}^{3.5} \frac{1}{3.5} \, dx = \frac{1}{3.5} \times (3.5 - 2) = \frac{1.5}{3.5} \approx 0.429. \]Convert this to a percentage giving approximately 42.9%.

Key Concepts

Probability Density Function

The Probability Density Function (PDF) is a crucial concept in continuous probability distributions, especially for the uniform distribution. In the context of a uniform distribution, the PDF is a constant value indicating that all outcomes within a specified range are equally likely. This means if a person is waiting for an elevator at the Grande Dunes Hotel, any time between 0 and 3.5 minutes has the same chance of happening.
Here, the PDF is given by:

• \( f(x) = \frac{1}{b-a} = \frac{1}{3.5-0} = \frac{1}{3.5} \)

This constant PDF means the graph of a uniform distribution is a horizontal line between the values 0 and 3.5. The total area under this line must sum to 1, representing the whole probability of 100%.
By integrating the PDF from the start point to the endpoint:

• \[ \text{Area} = \int_{0}^{3.5} \frac{1}{3.5} \, dx = \frac{1}{3.5} \times 3.5 = 1 \]

This calculation ensures the entire probability covers the interval from 0 to 3.5 minutes, confirming the function's validity as a probability distribution.

Mean and Standard Deviation

In any probability distribution, understanding the mean and standard deviation is essential in interpreting data characteristics. For a uniform distribution, these metrics offer insights into the average wait time and the data's variability.
### MeanThe mean of a uniform distribution can be calculated simply as the midpoint between the minimum and maximum values:

• \( \mu = \frac{a + b}{2} = \frac{0 + 3.5}{2} = 1.75 \) minutes.

This value of 1.75 minutes represents the average or typical waiting time an elevator patron can expect.
### Standard DeviationMeanwhile, the standard deviation measures the spread or dispersion of wait times around this mean. For a uniform distribution, the formula is:

• \( \sigma = \frac{b-a}{\sqrt{12}} = \frac{3.5 - 0}{\sqrt{12}} \approx 1.01 \) minutes.

This indicates that, while the average wait is 1.75 minutes, there is a moderate amount of variability around this average.

Uniform Distribution Properties

Uniform distribution stands out for its simplicity and symmetry. From the perspective of its properties, here's what we should know:
- **Equally Likely Outcomes**: All wait times between 0 and 3.5 minutes are equally probable, making it a straightforward model when outcomes have no preference. - **Symmetric Graph**: The probability density function is a flat, horizontal line over the interval of interest, showing identical probabilities. - **Defined by Two Parameters**: A uniform distribution is characterized by a minimum and maximum value, here it’s 0 and 3.5 minutes. - **No 'tails'**: Unlike other distributions like normal or exponential, uniform distribution has no tails, everything within the defined limits is possible.
These properties highlight why the uniform distribution is often used for modeling scenarios where each potential outcome within an interval happens with equal frequency. For students, comprehending these properties helps in knowing when to apply this distribution model in different probability contexts.