/*! 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 91 The operator of a pumping statio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 91

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). a. Find the probability that the demand will exceed 200 cfs during the early afternoon on a randomly selected day. b. What water-pumping capacity should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only \(.01 ?\)

Short Answer

Expert verified
(a) 13.53%; (b) Maintain capacity at 460.5 cfs.

Step by step solution

01

Understand the Given Information

The problem provides that the demand for water has an exponential distribution with a mean of 100 cubic feet per second (cfs). This means the rate \( \lambda \) of the distribution can be found using the formula for mean of an exponential distribution, which is \( \mu = \frac{1}{\lambda} \). Thus, \( \lambda = \frac{1}{100} = 0.01 \).
02

Calculate Probability for Part (a)

To find the probability that the demand exceeds 200 cfs, we use the formula for the exponential distribution: \( P(X > x) = e^{-\lambda x} \). Substitute \( \lambda = 0.01 \) and \( x = 200 \) to get \( P(X > 200) = e^{-0.01 \times 200} = e^{-2} \). Evaluate \( e^{-2} \) to get the probability.
03

Evaluate Exponential Function

Calculate \( e^{-2} \) using a calculator or exponential function tables. \( e^{-2} \approx 0.1353 \). This gives us the probability that the demand will exceed 200 cfs.
04

Interpret the Result for Part (a)

The probability that the demand will exceed 200 cfs is approximately 0.1353, or 13.53%.
05

Setup the Equation for Part (b)

For part (b), we need to find the capacity \( C \) such that \( P(X > C) = 0.01 \). Using the exponential probability formula \( P(X > C) = e^{-\lambda C} \), set it equal to 0.01: \( e^{-0.01 C} = 0.01 \).
06

Solve the Equation for Capacity

Take the natural logarithm of both sides to solve for \( C \): \( -0.01 C = \ln(0.01) \). This simplifies to \( C = -\frac{\ln(0.01)}{0.01} \).
07

Calculate Capacity

Evaluate \( \ln(0.01) \), which is approximately -4.605. Thus, \( C = -\frac{-4.605}{0.01} = 460.5 \).
08

Interpret the Result for Part (b)

The water-pumping capacity should be maintained at 460.5 cfs to ensure that the probability demand exceeds capacity is only 1%.

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.

Probability Calculation
When working with exponential distributions, probability calculations are a pivotal concept. Exponential distribution is used to model the time between events in a Poisson process. Here, we're looking to calculate the probability that water demand will exceed a certain threshold. For an exponential distribution with a rate parameter \( \lambda \), the probability that a random variable \( X \) will exceed a value \( x \) is given by the formula:
  • \( P(X > x) = e^{-\lambda x} \)
In this exercise, we are given \( \lambda = 0.01 \) and \( x = 200 \). Plug these into the formula to find the probability:
  • \( P(X > 200) = e^{-0.01 \times 200} = e^{-2} \approx 0.1353 \)
This result tells us that there is about a 13.53% chance that on a randomly selected day, the water demand will exceed 200 cubic feet per second.
Mean of Exponential Distribution
The mean of an exponential distribution is a key concept in understanding how events, such as water demand, are spread over time. For an exponential distribution, the mean \( \mu \) is the reciprocal of the rate \( \lambda \), represented by the formula:
  • \( \mu = \frac{1}{\lambda} \)
In the given exercise, we are told that the mean is 100 cubic feet per second. This means:
  • \( \lambda = \frac{1}{100} = 0.01 \)
The mean provides a central measure, indicating that on average, the water demand during early afternoon hours is expected to be around 100 cubic feet per second. Understanding this helps in making informed decisions regarding infrastructure and resource allocation based on expected demand.
Water Demand Modeling
Water demand modeling involves using mathematical tools to predict future water needs. This exercise uses exponential distribution to understand how water demand fluctuates during specific times. The objective is to ensure that sufficient resources are available, while also mitigating risks of over-demand.For instance, in part (b) of the exercise, determining the right pumping capacity involves setting a acceptable risk level, as stated:
  • Maintain a probability of only 0.01 for demand exceeding capacity.
This involves solving \( P(X > C) = 0.01 \) using the exponential distribution model:
  • \( e^{-0.01 C} = 0.01 \)
To find the necessary capacity \( C \), solve the equation to yield:
  • \( C = -\frac{\ln(0.01)}{0.01} = 460.5 \)
In this case, maintaining a pumping capacity of 460.5 cfs ensures that most of the time, the water station is prepared for the demand during early afternoon hours. Such modeling helps operators make data-driven decisions to optimize service while controlling the likelihood of excessive demand.

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

Explosive devices used in mining operations produce nearly circular craters when detonated. The radii of these craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas produced by these explosive devices.

A normally distributed random variable has density function $$f(y)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(y-\mu)^{2} /\left(2 \sigma^{2}\right)}, \quad-\infty0\).

Use the applet Comparison of Beta Density Functions to compare beta density functions with \((\alpha=9, \beta=7),(\alpha=10, \beta=7),\) and \((\alpha=12, \beta=7)\). a. Are these densities symmetric? Skewed left? Skewed right? b. What do you observe as the value of \(\alpha\) gets closer to \(12 ?\) c. Graph some more beta densities with \(\alpha>1, \beta>1,\) and \(\alpha>\beta .\) What do you conjecture about the shape of beta densities with \(\alpha>\beta\) and both \(\alpha>1\) and \(\beta>1 ?\)

The duration \(Y\) of long-distance telephone calls (in minutes) monitored by a station is a random variable with the properties that $$P(Y=3)=.2 \quad \text { and } \quad P(Y=6)=.1$$ Otherwise, \(Y\) has a continuous density function given by $$f(y)=\left\\{\begin{array}{ll} (1 / 4) y e^{-y / 2}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ The discrete points at 3 and 6 are due to the fact that the length of the call is announced to the caller in three-minute intervals and the caller must pay for three minutes even if he talks less than three minutes. Find the expected duration of a randomly selected long-distance call.

A random variable \(Y\) is said to have a log-normal distribution if \(X=\ln (Y)\) has a normal distribution. (The symbol In denotes natural logarithm.) In this case \(Y\) must be non negative. The shape of the log-normal probability density function is similar to that of the gamma distribution, with a long tail to the right. The equation of the log-normal density function is given by $$ f(y)=\left\\{\begin{array}{ll} \left.\frac{1}{\sigma y \sqrt{2 \pi}} e^{-(\ln (y)-\mu)^{2} / 2 \sigma^{2}}\right), & y>0 \\ 0, & \text { elsewhere } \end{array}\right. $$ Because \(\ln (y)\) is a monotonic function of \(y\) $$ P(Y \leq y)=P[\ln (Y) \leq \ln (y)]=P[X \leq \ln (y)] $$ where \(X\) has a normal distribution with mean \(\mu\) and variance \(\sigma^{2} .\) Thus, probabilities for random variables with a log-normal distribution can be found by transforming them into probabilities that can be computed using the ordinary normal distribution. If \(Y\) has a log-normal distribution with \(\mu=4\) and \(\sigma^{2}=1,\) find a. \(P(Y \leq 4)\) b. \(P(Y>8)\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

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.