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The lifetime, measure in hours, of the ACME super light bulb is a random variable \(T\) with density function \(f_{T}(t)=\lambda^{2} t e^{-\lambda t},\) where \(\lambda=.05 .\) What is the expected lifetime of this light bulb? What is its variance?

Short Answer

Expert verified
The expected lifetime is 40 hours, and the variance is 800 hours squared.

Step by step solution

01

Identify the given density function

The problem provides the probability density function (pdf) of a random variable \( T \) which represents the lifetime of the light bulb. The pdf is given by \( f_T(t) = \lambda^2 t e^{-\lambda t} \), and \( \lambda = 0.05 \). This is a form of the gamma distribution with shape parameter \( k = 2 \) and rate parameter \( \lambda = 0.05 \).
02

Expected Value for Gamma Distribution

For a gamma distribution with parameters \( k \) and \( \lambda \), the expected value is given by \( E(T) = \frac{k}{\lambda} \). Substitute \( k = 2 \) and \( \lambda = 0.05 \) to find the expected lifetime: \[ E(T) = \frac{2}{0.05} = 40 \] hours.
03

Variance for Gamma Distribution

The variance of a gamma distribution with parameters \( k \) and \( \lambda \) is given by \( \text{Var}(T) = \frac{k}{\lambda^2} \). Use \( k = 2 \) and \( \lambda = 0.05 \) to calculate the variance: \[ \text{Var}(T) = \frac{2}{(0.05)^2} = 800 \] hours squared.

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Key Concepts

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

Gamma Distribution
The gamma distribution is a continuous probability distribution often used to model phenomena where the waiting times between events are considered. It's particularly useful in fields like reliability engineering and queuing models. One of the key properties of a gamma distribution is its parameters: the shape parameter \( k \) and the rate parameter \( \lambda \). In this context, the shape parameter \( k \) often represents the number of events or stages, while the rate parameter \( \lambda \) indicates the occurrence frequency of the events.

In our exercise, the lifetime of the ACME super light bulb follows a gamma distribution with a shape parameter \( k = 2 \) and a rate parameter \( \lambda = 0.05 \). This tells us that we are dealing with a process that involves multiple events and the bulb's lifetime is modeled as a waiting time for these events to occur.

Some vital characteristics of the gamma distribution include:
  • It is defined for \( t > 0 \)
  • The probability density function (pdf) involves \( t^{k-1} e^{-\lambda t} \)
  • It is generalized enough to simplify to other distributions like the exponential distribution when \( k = 1 \)
Understanding gamma distribution helps in predicting the lifetime of products, thus improving reliability testing and inventory planning.
Expected Value
The expected value is one of the main concepts in probability and statistics. It represents the long-run average or the mean of the random variable. For distributions like the gamma distribution, the expected value can help provide insights into what the average outcome may be over numerous trials or instances.

In the case of the gamma distribution, the expected value \( E(T) \) is calculated using the formula \( \frac{k}{\lambda} \). This formula uses the shape \( k \) and rate \( \lambda \) parameters of the distribution. For our problem involving the ACME light bulb, substituting the values \( k = 2 \) and \( \lambda = 0.05 \) gives us:

\[ E(T) = \frac{2}{0.05} = 40 \] hours.

This expected value of 40 hours indicates that, on average, the bulbs are expected to last 40 hours before they burn out. Knowing this helps manufacturers and consumers understand the average performance standards of the light bulbs.
Variance
Variance is a measure of how spread out or dispersed the values of a random variable are, around its expected value. Specifically, for the gamma distribution, variance provides information about the variability or uncertainty in the lifetime of the ACME super light bulb.

For a gamma distribution, the variance \( \text{Var}(T) \) is calculated using the formula \( \frac{k}{\lambda^2} \). Using the specified parameters from the problem, \( k = 2 \) and \( \lambda = 0.05 \), the variance is calculated as:

\[ \text{Var}(T) = \frac{2}{(0.05)^2} = 800 \] hours squared.

This variance value of 800 hours squared indicates that there is a significant amount of spread or variability in the lifetimes of light bulbs around the mean value of 40 hours. Understanding variance in this context is crucial as it helps in assessing the consistency and reliability of the product's lifespan predictions.

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Most popular questions from this chapter

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