/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 A show is scheduled to start at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
(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 by step solution

01

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].
02

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 \( 0 \leq t \leq 90 \).
03

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 \( \sigma^2 = \frac{90^2}{12} = 675 \).
04

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 \( P(T < 10) = \frac{10}{90} + \frac{10}{90} + \frac{10}{90} = \frac{30}{90} = \frac{1}{3} \).
05

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 \( P(T > 20) = \frac{10}{90} + \frac{20}{90} + \frac{20}{90} = \frac{50}{90} = \frac{5}{9} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When a bus service reduces fares, a particular trip from New York City to Albany, New York, is very popular. A small bus can carry four passengers. The time between calls for tickets is exponentially distributed with a mean of 30 minutes. Assume that each caller orders one ticket. What is the probability that the bus is filled in less than three hours from the time of the fare reduction?

Calls to the help line of a large computer distributor follow a Poisson distribution with a mean of 20 calls per minute. Determine the following: (a) Mean time until the one-hundredth call (b) Mean time between call numbers 50 and 80 (c) Probability that three or more calls occur within 15 seconds

Assume that the flaws along a magnetic tape follow a Poisson distribution with a mean of 0.2 flaw per meter. Let \(X\) denote the distance between two successive flaws. (a) What is the mean of \(X ?\) (b) What is the probability that there are no flaws in \(10 \mathrm{con}-\) secutive meters of tape? (c) Does your answer to part (b) change if the 10 meters are not consecutive? (d) How many meters of tape need to be inspected so that the probability that at least one flaw is found is \(90 \% ?\) (e) What is the probability that the first time the distance between two flaws exceeds eight meters is at the fifth flaw? (f) What is the mean number of flaws before a distance between two flaws exceeds eight meters?

The life (in hours) of a computer processing unit (CPU) is modeled by a Weibull distribution with parameters \(\beta=3\) and \(\delta=900\) hours. Determine (a) and (b): (a) Mean life of the CPU. (b) Variance of the life of the CPU. (c) What is the probability that the CPU fails before 500 hours?

The time between failures of a laser in a cytogenics machine is exponentially distributed with a mean of 25,000 hours. (a) What is the expected time until the second failure? (b) What is the probability that the time until the third failure exceeds 50,000 hours?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.