Problem 57 - Applied Calculus

Question

BUSINESS: Product Reliability The proportion of light bulbs that last longer than \(t\) hours is predicted to be \(\int_{t}^{\infty} 0.001 e^{-0.001 s} d s .\) Use this formula to find the proportion of light bulbs that will last longer than 1200 hours.

Step-by-Step Solution

Answer

Approximately 30.12% of light bulbs last longer than 1200 hours.

Step 1: Understand the Problem

We are given an integral representing the proportion of light bulbs lasting longer than a certain time \( t \), specifically \( \int_{t}^{\infty} 0.001 e^{-0.001 s} d s \). We are asked to find what this proportion is when \( t = 1200 \) hours.

Step 2: Recognize the Integral Type

The integral \( \int_{t}^{\infty} 0.001 e^{-0.001 s} d s \) is a form of the exponential function integral, which is often evaluated using its association with the cumulative distribution function of the exponential distribution. The exponential distribution with a rate parameter \( \lambda = 0.001 \) uses this exact integrand.

Step 3: Evaluate the Integral

The integral \( \int_{1200}^{\infty} 0.001 e^{-0.001 s} d s \) can be computed using the antiderivative of \( 0.001 e^{-0.001 s} \), which is \( -e^{-0.001 s} \). We evaluate this from 1200 to infinity: \[ \left[-e^{-0.001 s}\right]_{1200}^{\infty} = 0 - (-e^{-0.001 \times 1200}) = e^{-1.2}. \]

Step 4: Simplify the Result

The expression \( e^{-1.2} \) is calculated as approximately 0.3012. Thus, the proportion of light bulbs that last longer than 1200 hours is about 0.3012, or 30.12%.

Key Concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution used to model the time between events in a process that occurs continuously and independently at a constant average rate. It is often used in reliability analysis, such as in predicting the lifespan of products.

**Key Features of Exponential Distribution:**

• It is characterized by its rate parameter \( \lambda \). The rate is the average number of occurrences in a given time period.
• The probability density function (PDF) is given by \( f(s) = \lambda e^{-\lambda s} \), where \( s \) is the time.
• This distribution is memoryless, meaning that the probability of an event occurring in the future is independent of how much time has already elapsed.

In the given problem, the rate parameter \( \lambda \) is 0.001, meaning that the average lifespan between events (or failures) is 1,000 hours. This distribution helps to predict the proportion of light bulbs that last longer than a specific time, such as 1200 hours.

Integral Calculus

Integral calculus is a branch of mathematics focusing on the accumulation of quantities and the areas under and between curves. In the context of probability distributions, integrals help to calculate probabilities by summing up tiny segments to find areas corresponding to probabilities.

**Why Integrals Matter in Probability:**

• The integral of a probability density function (PDF) gives the probability that a random variable falls within a certain range.
• For continuous random variables, the cumulative distribution function (CDF) is found by integrating the PDF from \(-\infty\) to a given value.

In the problem, the integral \( \int_{t}^{\infty} 0.001 e^{-0.001 s} d s \) is used to calculate the probability or proportion of light bulbs lasting longer than \( t \) hours. Calculating this integral from 1200 to infinity gives the proportion of bulbs lasting longer than 1200 hours.

Antiderivative Evaluation

Antiderivative evaluation is the process of finding the original function given its derivative, which is essential in solving integrals, especially indefinite ones.

**Understanding Antiderivatives:**

• The antiderivative of a function \( f'(x) \) yields the original function \( f(x) \), plus an arbitrary constant \( C \).
• For exponential functions like \( e^{-\lambda s} \), the antiderivative involves dividing by \(-\lambda\) and retaining the exponential term.

In this exercise, the function \( 0.001 e^{-0.001 s} \) has an antiderivative \( -e^{-0.001 s} \). The integral is evaluated from 1200 to infinity:- At \( s = \infty \), \( e^{-0.001s} \) approaches zero.- At \( s = 1200 \), the value becomes \( e^{-0.001 \times 1200} = e^{-1.2} \).Hence, by subtracting these values, we obtain that \( 0 - (-e^{-1.2}) = e^{-1.2} \), providing the required probability that a light bulb lasts longer than 1200 hours.