Problem 37 - Calculus and its applications

Question

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.

Step-by-Step Solution

Answer

The probability is approximately 0.9502.

Step 1: Understand the Probability Density Function

We are given the probability density function (PDF) of learning time for a rat as \( f(t) = 0.02 e^{-0.02 t} \). This function describes the rate of change of the probability that the time \( t \) takes certain values, for \( t \geq 0 \).

Step 2: Set Up the Probability Expression

The task is to find the probability that a rat will learn the maze in 150 seconds or less. This means we need to calculate \( P(T \leq 150) \), which involves integrating the PDF from 0 to 150.

Step 3: Integrate the PDF Over the Given Interval

The probability \( P(T \leq 150) \) is given by the integral \[ P(T \leq 150) = \int_0^{150} 0.02 e^{-0.02 t} \, dt \].

Step 4: Solve the Integral

To solve the integral \( \int 0.02 e^{-0.02 t} \, dt \), use the antiderivative formula for exponential functions. The antiderivative of \( e^{at} \) is \( \frac{1}{a} e^{at} \), so \[ \int 0.02 e^{-0.02 t} \, dt = -e^{-0.02 t} + C \].

Step 5: Evaluate the Definite Integral

Substitute the limits 0 and 150 into the antiderivative to find the definite integral:\[ -e^{-0.02 \times 150} + e^{-0.02 \times 0} = -e^{-3} + 1 \].

Step 6: Calculate the Final Probability

Compute \( e^{-3} \) using a calculator to find the final probability: \[ P(T \leq 150) = 1 - e^{-3} \approx 1 - 0.0498 = 0.9502 \].

Key Concepts

Probability Density Function

The concept of a Probability Density Function (PDF) is central in statistics and helps describe the likelihood of a random variable taking on various values. In the context of the exponential distribution, which is often used to model time until an event occurs, the PDF provides us with a continuous measure of probability over a range of outcomes.
For the given problem, the PDF is given by the function \( f(t) = 0.02 e^{-0.02 t} \), which models the time it takes a rat to learn a maze. Here, \( e \) is the base of the natural logarithm, and \( t \) is the time in seconds. The function describes how probability is distributed over time. When dealing with PDFs:

• The function never goes negative, as probabilities can't be less than zero.
• The area under the curve of the PDF over its entire range is always 1, representing the total probability.
• It tells us nothing about specific outcomes but rather the density, or the rate, at which the probabilities change for different times \( t \).

The exponential PDF in this scenario reflects a decreasing probability rate as time increases, meaning the more time passes without solving the maze, the less likely it becomes that the rat will do so shortly.

Probability Calculation

To calculate the probability that an event occurs within a specific time frame using a PDF, we integrate it over the desired interval. This process is akin to finding the area under the PDF curve between two given points. For our exercise, we're interested in the probability that the rat learns the maze in 150 seconds or less.
Therefore, we compute the probability \( P(T \leq 150) \) by integrating the PDF from 0 to 150. This operation is expressed as:
\[ P(T \leq 150) = \int_0^{150} 0.02 e^{-0.02 t} \, dt \]
This integral represents the sum of all infinitesimal probabilities over the range from 0 to 150 seconds, thus giving the cumulative probability up to that time.
Using integration in this context not only helps pinpoint probabilities for specific intervals but also showcases the accumulation of probabilities as time progresses, aligning perfectly with how we perceive events' likelihoods in continuous domains.

Antiderivative

In solving problems involving exponential distributions, finding the antiderivative, or the integral, of the function is crucial. This is because the integral of the PDF over the specified interval directly provides the cumulative probability.
For exponential functions like \( 0.02 e^{-0.02 t} \), the antiderivative can be determined using the formula for integrals of exponential terms. Given a function of the form \( a e^{bt} \), its antiderivative is:
\[ \int a e^{bt} \, dt = \frac{a}{b} e^{bt} + C \]
In our case, integrating \( 0.02 e^{-0.02 t} \) results in:
\[ \int 0.02 e^{-0.02 t} \, dt = -e^{-0.02 t} + C \]
here \( C \) is the constant of integration typically disregarded in definite integrals.
Once the antiderivative is calculated, apply the limits of integration (150 and 0) to find the definite integral. It's evaluated as:
\[ -e^{-0.02 \times 150} + e^{-0.02 \times 0} = -e^{-3} + 1 \]
This calculation simplifies to \( 1 - e^{-3} \), approximating to 0.9502, reflecting a 95% probability that the rat will learn the maze in 150 seconds or less. Understanding antiderivatives in this context helps bridge the gap between calculus and probability, providing a clear path to solving real-world problems involving time-based events.