Problem 57 - Applied Statistics and Probability for Engineers

Question

A show is scheduled to start at 9: 00 a.M., 9: 30 A.M. and 10: 00 A.M. Once the show starts, the gate will be closed. A visitor will arrive at the gate at a time uniformly distributed between 8: 30 a.m. and 10: 00 A.M. Determine the following: (a) Cumulative distribution function of the time (in minutes) between arrival and 8: 30 A.M. (b) Mean and variance of the distribution in the previous part (c) Probability that a visitor waits less than 10 minutes for a show (d) Probability that a visitor waits more than 20 minutes for a show

Step-by-Step Solution

Answer

(a) CDF: \( F(t) = \frac{t}{90} \) (b) Mean = 45, Variance = 675 (c) Probability <10 minutes = \( \frac{1}{3} \) (d) Probability >20 minutes = \( \frac{5}{9} \).

Step 1: Understanding the time distribution

The visitor arrives at a uniformly random time between 8:30 AM (0 minutes) and 10:00 AM (90 minutes). Thus, the time variable, T, is uniformly distributed over the interval [0, 90].

Step 2: Deriving the CDF

The cumulative distribution function (CDF) for a uniform distribution is given by \( F(t) = \frac{t - a}{b - a} \) for \( a \leq t \leq b \). Here \( a = 0 \) and \( b = 90 \). So, \( F(t) = \frac{t}{90} \) for \( 0 \leq t \leq 90 \).

Step 3: Calculating Mean and Variance

For a uniform distribution over \([a, b]\), the mean is \( \mu = \frac{a + b}{2} \) and the variance is \( \sigma^2 = \frac{(b - a)^2}{12} \). For our interval \([0, 90]\), we have \( \mu = 45 \) minutes and \( \sigma^2 = \frac{90^2}{12} = 675 \).

Step 4: Finding Probability a Visitor Waits less than 10 Minutes

A visitor waits less than 10 minutes if they arrive between 8:30 AM and 8:40 AM, or between 9:00 AM and 9:10 AM, or between 9:30 AM and 9:40 AM. Each period lasts for 10 minutes, thus the probability is \( P(T < 10) = \frac{10}{90} + \frac{10}{90} + \frac{10}{90} = \frac{30}{90} = \frac{1}{3} \).

Step 5: Finding Probability a Visitor Waits more than 20 Minutes

The visitor waits over 20 minutes if they arrive more than 20 minutes before the next show. This is possible if they arrive before 8:40 AM (wait 20 minutes for the 9:00 show), between 9:00 AM and 9:10 AM (wait 20 minutes for the 9:30 show), or 9:30 AM and 9:40 AM (wait 20 minutes for the 10:00 show). Probability of arriving during these times is \( P(T > 20) = \frac{10}{90} + \frac{20}{90} + \frac{20}{90} = \frac{50}{90} = \frac{5}{9} \).

Key Concepts

Cumulative Distribution Function

The cumulative distribution function (CDF) for a uniform distribution describes the probability that a random variable is less than or equal to a certain value. For a uniformly distributed random variable across an interval \(a, b\), the function is defined by the formula \(F(t) = \frac{t - a}{b - a}\), provided that \(a \leq t \leq b\). This CDF helps us understand how probabilities accumulate over the interval.

In the context of our exercise, the time between 8:30 AM (0 minutes) and 10:00 AM (90 minutes) is uniformly distributed. Therefore, the CDF is given by \(F(t) = \frac{t}{90}\) for \(0 \leq t \leq 90\). As time increases within this interval, the likelihood of the visitor having arrived increases smoothly from 0 to 1.

Mean and Variance

The mean and variance of a uniform distribution provide insight into its central tendency and spread. Calculating these can help emphasize the characteristics of the distribution.

• Mean (\( \mu \)): The mean for a uniform distribution over \[a, b\] is given by \( \mu = \frac{a + b}{2} \). It represents the average expected value. In our situation, the time interval is \[0, 90\] minutes, leading to a mean of \( \mu = 45 \) minutes.
• Variance (\( \sigma^2 \)): Variance measures the spread of the distribution and is calculated using \( \sigma^2 = \frac{(b - a)^2}{12} \). For the interval \[0, 90\], the variance becomes \( \sigma^2 = \frac{90^2}{12} = 675 \), indicating how dispersed the arrival times are around the mean.

Probability Calculations

Probability calculations involve determining the likelihood of various events within a distribution. In this exercise, we focus on determining the chances that a visitor waits for a specific duration.

1. **Waiting less than 10 minutes:** To find this probability, identify the periods when the wait is under 10 minutes. These are when visitors arrive within 0 to 10 minutes of each show time start at 9:00 AM, 9:30 AM, and 10:00 AM, respectively. The probability is the sum of these intervals: \( \frac{10}{90} + \frac{10}{90} + \frac{10}{90} = \frac{30}{90} = \frac{1}{3} \).
2. **Waiting more than 20 minutes:** Calculate probability for intervals when a visitor arrives more than 20 minutes early for a show. This includes arrivals before 8:40 AM or shortly after show times start (9:00 AM and 9:30 AM). The probability is: \( \frac{10}{90} + \frac{20}{90} + \frac{20}{90} = \frac{50}{90} = \frac{5}{9} \). These calculations highlight how time is a key variable in understanding the likelihood of different waiting periods.

Uniformly Distributed Random Variable

A uniformly distributed random variable is one where every outcome in a given range is equally likely. Understanding this concept is key in problems where situations are consistent and lack bias towards particular results.

In the context of our problem, the visitor's arrival time is a uniformly distributed random variable between 8:30 AM and 10:00 AM.

• This means any minute within this interval is just as likely as another.
• The uniform nature of this distribution means we don’t favor one segment over another within the 90-minute window.

This is why simple formulas apply when deriving measures such as the cumulative distribution function, mean, and variance. The uniform distribution ensures a predictable and even spread of probabilities over the specified interval.