Problem 163 - An Introduction to Mathematical Statistics and Its Applications

Question

A certain brand of fluorescent bulbs will last, on the average, one thousand hours. Suppose that four of these bulbs are installed in an office. What is the probability that all four are still functioning after one thousand fifty hours? If \(X_{i}\) denotes the \(i\) th bulb's life, assume that $$ f_{X_{1}, X_{2}, X_{3}, X_{4}}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\prod_{i=1}^{4}\left(\frac{1}{1000}\right) e^{-x / 1000} $$

Step-by-Step Solution

Answer

The probability that all four bulbs are still functioning after 1050 hours is \(e^{-4.2}\).

Step 1: Identify the Exponential Distribution Function

The given function \(f_{X_{1}, X_{2}, X_{3}, X_{4}}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\prod_{i=1}^{4}\left(\frac{1}{1000}\right) e^{-x / 1000}\) is the joint probability density function of four independent exponential random variables each with mean 1000. We need the distribution function, which is the integral of the density function.

Step 2: Find the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) gives the probability that a random variable is less than or equal to a certain value. For an exponential random variable with mean \(u\), the CDF is given by \(F(x) = 1 - e^{-x/u}\). For x > 0, here \(u\) = 1000 hours.

Step 3: Calculation

We want to find the probability that all four bulbs are still functioning after 1050 hours. That means each bulb lives more than 1050 hours. Because the bulbs' lifetimes are independent, this is equal to the product of the individual probabilities. To find the probability that one bulb lives more than 1050 hours, we subtract the CDF at 1050 from 1. That is \(1 - F(1050)\). Substituting \(F(x)\) from Step 2, we get \(1 - (1 - e^{-1050/1000})\). Simplifying, this gives \(e^{-1.05}\). Because the bulbs' lifetimes are independent, the probability that all four bulbs live more than 1050 hours is \(e^{-1.05}^4 = e^{-4.2}\).

Key Concepts

Joint Probability Density Function

The joint probability density function (PDF) describes the behavior of multiple random variables simultaneously. In our exercise, we're dealing with four independent exponential random variables, which represent the lifespan of fluorescent bulbs. The joint PDF provided in the exercise is:

• \[f_{X_{1}, X_{2}, X_{3}, X_{4}}(x_{1}, x_{2}, x_{3}, x_{4}) = \prod_{i=1}^{4}\left(\frac{1}{1000}\right)e^{-x/1000}\]

This formula means that each bulb's lifespan is modeled by an exponential distribution with a mean of 1000 hours.
The product notation implies that all four bulbs' lifespans are independent of each other. In other terms, the probability of each bulb lasting a certain amount of time does not affect the others.
By understanding the joint PDF, we grasp the concept of calculating probabilities for multiple events happening together.

Cumulative Distribution Function

The cumulative distribution function (CDF) helps us understand the probability that a random variable takes a value less than or equal to a certain threshold.
For an exponential distribution, which we are looking at in this exercise, the CDF is expressed as:

• \[F(x) = 1 - e^{-x/u}\]

Here, \(u\) is the mean of the distribution. In our case, it's 1000 hours, which represents the average bulb lifespan.
This CDF tells us about the likelihood of a bulb burning out before a specified time. To find out how long the bulb lasts beyond this time, we can simply use the complement property, i.e., \(1 - F(x)\).
Understanding the CDF is crucial for making predictions about the duration a specific bulb can function before failing.

Independent Random Variables

Understanding independent random variables is key in probability theory, especially when computing probabilities for events involving multiple random outcomes. In our problem, we consider the lifetimes of four bulbs as independent events because the failure of one does not affect the others. This is reflected in our joint probability density function which is a product of individual PDFs.

• Independence means the probability of one event does not affect the probability of another.
• The joint probability for multiple independent events is the product of individual probabilities.

In our context, since each bulb independently has a lifespan modeled by an exponential distribution, the calculation for all lasting beyond a certain time is the product of their individual probabilities. Explain this carefully to understand how combinations of independent variables work.

Probability Calculation

To determine the probability of all four bulbs lasting beyond 1050 hours, we need to use the complement of the cumulative distribution function. For one bulb, the probability that it functions more than 1050 hours is given by:

• \(1 - F(1050) = e^{-1050/1000}\)

Substitute the exponential form of the CDF to get \(e^{-1.05}\). Since the events are independent, meaning one bulb's lifespan does not affect another, we multiply this probability for all four bulbs:

• \(e^{-1.05}^4 = e^{-4.2}\)

This formula represents the combined probability of all four bulbs operating beyond 1050 hours. Each step involves simple calculations built on a clear understanding of the exponential probability distribution. Remember, the keys are using the correct complement of the CDF and acknowledging the independence of the variables.