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Chapter 5: Problem 136

Probability The median waiting time (in minutes) for people waiting for service in a convenience store is given by the solution of the equation \(\int_{0}^{x} 0.3 e^{-0.3 t} d t=\frac{1}{2}\) What is the median waiting time?

Short Answer

Expert verified
The median waiting time for people in the convenience store is approximately 2.31 minutes.

Step by step solution

01

Understanding the integral

The integral presented in the equation is of the form \(\int_{a}^{b} e^{f(x)} dx\) which is calculated as \(e^{f(x)}\) evaluated at the limits \(a\) and \(b\). The integral \(\int_{0}^{x} 0.3 e^{-0.3t} dt\) simplifies to \(-e^{-0.3t}\) between the limits \(0\) and \(x\).
02

Evaluate the integral at the limits

Evaluating \(-e^{-0.3t}\) at the lower limit \(0\) gives \(-1\) (since \(e^0 = 1\)) and at the upper limit \(x\) gives \(-e^{-0.3x}\). The result, after subtracting, is \(1-e^{-0.3x}\).
03

Solve the equation

We have the equation \(1-e^{-0.3x} = 1/2\). We must solve for \(x\). First, subtract 1 from both sides and multiply by -1 to get: \(e^{-0.3 x} = 1/2\).
04

Solving for x

To solve this equation, we'll make use of the property of logarithms that states \(-0.3x = ln(1/2)\), which leads to \(x = -ln(1/2) / 0.3\).
05

Calculating the numerical value

Using a calculator, we find that the numerical value of x, or the median waiting time, is approximately 2.31 minutes.

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Key Concepts

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

Integral Calculus
Integral calculus helps us find the area under a curve or function. In this problem, we see an integral used to determine the median waiting time in a convenience store. The integral \[ \int_{0}^{x} 0.3 e^{-0.3 t} dt \]represents the cumulative distribution function (CDF) of a probability density function (PDF). The function describes how the waiting times are distributed.
When solving this integral, we're essentially summing up infinitesimal probabilities (tiny pieces of the curve) from 0 to some time "x". Our goal is to find the value of "x" such that this cumulative probability equals 0.5, as per the equation in the exercise.
Here's how it all fits together:
  • The function inside the integral \(0.3 e^{-0.3 t}\) is an exponential decay function, representing how waiting times diminish over time.
  • Upon solving the integral, the result we obtain symbolizes a probability at a specific point "x".
  • By setting this equal to 0.5, we're determining which "x" corresponds to the median of our data distribution.
Exponential Function
An exponential function is a mathematical function in which an independent variable appears in the exponent. In this case, our function \[ 0.3 e^{-0.3 t} \]is an example of exponential decay.
This means that as time increases, the impact or 'height' of the function decreases exponentially.
  • It models many natural phenomena where something diminishes over time, like radioactive decay or cooling processes.
  • The rate of decrease here is specified by the constant 0.3. This value impacts how quickly the probabilities drop as time goes on.
  • These types of functions are crucial in understanding problems dealing with time, growth, and decay.
In the context of the problem, this function describes how likely it is for a person to wait a certain amount of time in line for service.
The function decreases rapidly initially, suggesting it's more likely for people to wait for a shorter time rather than a longer time.
Logarithms
Logarithms help us solve for variables in exponents, and they are intimately related to exponential functions. In the solution, we reached an equation: \[ e^{-0.3 x} = \frac{1}{2} \]To solve for "x", we use logarithms. A logarithm is essentially the inverse of an exponential function.
  • It allows us to "undo" the effect of raising a number to a power, helping us isolate the variable in the exponent.
  • The natural logarithm, denoted as "ln", is particularly handy when dealing with exponential functions involving the constant "e".
  • In our equation, taking the natural logarithm of both sides gives us: \(-0.3x = ln(\frac{1}{2})\)
By rearranging, we find that "x" can be solved by isolating it: \[ x = \frac{-ln(\frac{1}{2})}{0.3} \]This algebraic manipulation provides us with the precise median waiting time.
Probability Distributions
Understanding probability distributions is key to solving this median waiting time problem. A probability distribution provides a function that describes the likelihood of different outcomes. Here, the exponential function models the waiting times of customers as described by a probability density function (PDF).
  • A probability density function helps us understand the nature of continuous random variables, like waiting times.
  • The median, specifically, is a statistic measuring the central tendency of a probability distribution. It's the value where half of the observations fall below and half above.
  • In practical terms, understanding the median helps us gain insight into what a "typical" waiting time might be in a distribution.
In the exercise, the integral evaluates the cumulative probability up to a certain point, where it becomes equal to 0.5. That's precisely when the median is reached.

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Most popular questions from this chapter

True or False? In Exercises \(83-86\) , determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false. $$\frac{d}{d x}[\arctan (\tan x)]=1$$ for all \(x\) in the domain.

Population Growth A lake is stocked with 500 fish, and the population \(p\) is growing according to the logistic curve \(p(t)=\frac{10,000}{1+19 e^{-t / 5}}\) where \(t\) is measured in months. (a) Use a graphing utility to graph the function. (b) Find the fish populations after 6 months, 12 months, 24 months, 36 months, and 48 months. What is the limiting size of the fish population? (c) Find the rates at which the fish population is changing after 1 month and after 10 months. (d) After how many months is the population increasing most rapidly?

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