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Chapter 5: Problem 37

In a psychology experiment, the time \(t\) in seconds, that it takes a rat to learn its way through a maze is an exponentially distributed random variable with the probability density function $$f(t)=0.02 e^{-0.02 t}, \quad 0 \leq t<\infty$$. Find the probability that a rat will learn its way through a maze in 150 sec or less.

Short Answer

Expert verified
The probability is approximately 0.9502, or 95.02%.

Step by step solution

01

Identify the Problem

Given an exponentially distributed random variable with the probability density function (PDF) given by \( f(t) = 0.02 e^{-0.02 t} \), we need to find the probability that the rat will learn its way through the maze in 150 seconds or less.
02

Set Up the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) for an exponentially distributed random variable is found by integrating the PDF. So, the CDF is: \( F(t) = \int_{0}^{t} 0.02 e^{-0.02 x} \, dx \).
03

Integrate the PDF

To find the CDF, we first integrate the PDF: \( \int_{0}^{t} 0.02 e^{-0.02 x} \, dx = -e^{-0.02 x} \bigg|_0^t = 1 - e^{-0.02 t} \). Thus, \( F(t) = 1 - e^{-0.02 t} \).
04

Substitute the Given Time into the CDF

We need to find the probability that the rat will learn the maze in 150 seconds or less. Substituting \( t = 150 \) seconds into the CDF: \( F(150) = 1 - e^{-0.02 \times 150} \).
05

Calculate the Probability

Now, calculate the value: \( F(150) = 1 - e^{-3} \approx 1 - 0.0498 = 0.9502 \).

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Key Concepts

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

Exponential Distribution
The exponential distribution is a continuous probability distribution. It describes the time between events in a Poisson process, where the events occur continuously and independently at a constant average rate. In our context, it models the time a rat takes to complete a maze. One key property of the exponential distribution is its memoryless property. This means that the probability of an event occurring in the next time interval is independent of how much time has already passed.
Probability Density Function (PDF)
The probability density function (PDF) gives us the likelihood of a random variable to take a specific value. For the exponential distribution, the PDF is expressed as:
\( f(t) = \beta e^{-\beta t} \).
Here, \( \beta \) is the rate parameter, indicating how frequently events occur. In our exercise, the PDF given is \( f(t) = 0.02 e^{-0.02 t} \), where \( \beta = 0.02 \), meaning that on average events occur approximately every 50 seconds \( (\frac{1}{0.02}) \). The PDF is non-negative, and its total area under the curve is 1, ensuring all possibilities are accounted for.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) indicates the probability that a random variable takes on a value less than or equal to a particular value. For the exponential distribution, we compute the CDF by integrating the PDF. For example, the CDF, \( F(t) \), for the given PDF, \( f(t) = 0.02 e^{-0.02 t} \), is calculated by integrating:
\( F(t) = \int_{0}^{t} 0.02 e^{-0.02 x} \, dx = 1 - e^{-0.02 t} \).
This integral shows the probability that the rat completes the maze within a time \( t \) seconds or less.
Integration
Integration is the mathematical process of finding the area under a curve. In probability, we often use integration to find the cumulative distribution function (CDF) from the probability density function (PDF). For our example, we need to integrate the PDF \( f(t) = 0.02 e^{-0.02 t} \) to find the CDF.
The process involves:
  • Setting up the integral: \( \int_{0}^{t} 0.02 e^{-0.02 x} \, dx \).
  • Solving the integral: \( -e^{-0.02 x} \bigg|_{0}^{t} \).
  • Evaluating the limits: \( 1 - e^{-0.02 t} \).
Integration helps in determining probabilities over an interval rather than a single point, providing a fuller understanding of the distribution.

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Most popular questions from this chapter

Find the accumulated future value of cach continuous income stream at rate \(R(t),\) for the given time \(T\) and interest rate \(k\) compounded continuously. Round to the nearest \(\$ 10 .\) $$R(t)=550,000, \quad T=22 y r, \quad k=7 \%$$

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