/*! 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 10 During a certain part of the day... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 10

During a certain part of the day, the time between arrivals of automobiles at the tollgate on a turnpike is an exponential random variable with expected value 20 seconds. Find the probability that the time between successive arrivals is greater than 10 seconds and less than 30 seconds.

Short Answer

Expert verified
The probability that the time between arrivals is between 10 and 30 seconds is 0.3834.

Step by step solution

01

Identify the parameter of the exponential distribution

The time between arrivals is described by an exponential distribution with an expected value of 20 seconds. The mean (expected value, \(E[X]\)) of an exponential distribution is \(\frac{1}{\lambda}\), where \(\lambda\) is the rate parameter. Therefore, \(\lambda = \frac{1}{20} = 0.05\).
02

Define the cumulative distribution function (CDF) of the exponential distribution

For an exponential distribution, the CDF is given by \(F(t) = 1 - e^{-\lambda t}\).
03

Calculate the probability for less than 30 seconds

Using the CDF, \(P(X < 30)\) can be calculated as \(F(30) = 1 - e^{-0.05 \cdot 30} = 1 - e^{-1.5} \approx 0.7769\).
04

Calculate the probability for less than 10 seconds

Similarly, calculate \(P(X < 10)\) as \(F(10) = 1 - e^{-0.05 \cdot 10} = 1 - e^{-0.5} \approx 0.3935\).
05

Calculate the probability between 10 and 30 seconds

The probability that the time between successive arrivals is between 10 and 30 seconds is \(P(10 < X < 30) = F(30) - F(10)\). Using the values from Steps 3 and 4, \(P(10 < X < 30) = 0.7769 - 0.3935 = 0.3834\).

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 91影视!

Key Concepts

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

Understanding Probability in Exponential Distribution
Probability is a fundamental concept in statistics and denotes the likelihood of an event occurring. In the context of exponential distribution, probability helps us determine the likelihood of specific time intervals between events. For our tollgate example, the probability that the time between successive arrivals is between 10 and 30 seconds is calculated using the exponential distribution function. This involves subtracting the cumulative probabilities for upper and lower bounds (30 and 10 seconds, respectively). This gives us
\(P(10 < X < 30) = F(30) - F(10) \) where \((X)\) is the time between arrivals.
Exploring the Cumulative Distribution Function (CDF)
To understand the exponential distribution, we must be familiar with its cumulative distribution function (CDF). The CDF represents the probability that a random variable takes on a value less than or equal to a certain number. For an exponential random variable with rate parameter \((\lambda)\), the CDF is given by\(F(t) = 1 - e^{-\lambda t}\). This CDF helps in calculating the probabilities for different ranges. For instance, the probability that the time between arrivals is less than 30 seconds is computed as \( F(30) = 1 - e^{-0.05 \) \( 30}\) which indicates how we find cumulative probabilities effectively.
By comparing cumulative probabilities at different time points, we can find the probability of the time interval lying between two given values.
The Role of the Rate Parameter (\lambda)
The rate parameter \(\lambda\) is crucial in exponential distribution. It signifies the rate at which events occur. In our context, it is the rate at which cars arrive at the tollgate. \(\lambda\) is the reciprocal of the mean (expected value) of the distribution. For example, if the expected arrival time is 20 seconds, \(\lambda\) would be \(1/20 = 0.05\). This means, on average, a car arrives every 20 seconds.
Understanding \(\lambda\) is essential as it shapes the entire distribution curriculum and it helps in calculating both the Probability Density Function (PDF) and Cumulative Distribution Function (CDF). By knowing \(\lambda\), we can find the probability of any time-related event modelled by the exponential distribution.

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

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{3 \sqrt{x}}{16}, 0 \leq x \leq 4\)

The number of accidents occurring each month at a certain intersection is Poisson distributed with \(\lambda=4.8\). (a) During a particular month, are five accidents more likely to occur than four accidents? (b) What is the probability that more than eight accidents will occur during a particular month?

In a certain cell population, cells divide every 10 days, and the age of a cell selected at random is a random variable \(X\) with the density function \(f(x)=2 k e^{-k x}, 0 \leq x \leq 10, k=(\ln 2) / 10\). Let \(X\) be a continuous random variable with values between \(A=1\) and \(B=\infty\), and with the density function \(f(x)=4 x^{-5}\). (a) Verify that \(f(x)\) is a probability density function for \(x \geq 1\) (b) Find the corresponding cumulative distribution function \(F(x)\). (c) Use \(F(x)\) to compute \(\operatorname{Pr}(1 \leq X \leq 2)\) and \(\operatorname{Pr}(2 \leq X)\).

If \(X\) is a random variable with density function \(f(x)\) on \(A \leq x \leq B\), the median of \(X\) is that number \(M\) such that $$ \int_{A}^{M} f(x) d x=\frac{1}{2} $$ In other words, \(\operatorname{Pr}(X \leq M)=\frac{1}{2}\). Find the median of the random variable whose density function is \(f(x)=2(x-1), 1 \leq x \leq 2\).

Decision Making Based on Expected Value A citrus grower anticipates a profit of \(\$ 100,000\) this year if the nightly temperatures remain mild. Unfortunately, the weather forecast indicates a \(25 \%\) chance that the temperatures will drop below freezing during the next week. Such freezing weather will destroy \(40 \%\) of the crop and reduce the profit to \(\$ 60,000 .\) However, the grower can protect the citrus fruit against the possible freezing (using smudge pots, electric fans, and so on) at a cost of \(\$ 5000\). Should the grower spend the \(\$ 5000\) and thereby reduce the profit to \(\$ 95,000 ?[\) Hint: Compute \(\mathrm{E}(X)\), where \(X\) is the profit the grower will get if he does nothing to protect the fruit.]
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
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.