Question

Let \(\mathrm{X}\) be a random variable denoting the hours of life in an electric light bulb. Suppose \(\mathrm{X}\) is distributed with density function \(\mathrm{f}(\mathrm{x})=(1 / 1,000) \mathrm{e}^{-\mathrm{x} / 1000} \quad\) for \(\mathrm{x}>0\) Find the expected lifetime of such a bulb.

Short answer

The expected lifetime of such a light bulb is 1000 hours.

Step-by-step solution

Write down the integral formula for the expected value

Recall the formula for the expected value of a continuous random variable: \[E(X) = \int xf(x) dx\]

Substitute the given density function into the integral

Substitute the given density function into the integral formula: \[E(X) = \int x \Big(\frac{1}{1000}e^{-x/1000}\Big) dx\] This integral will be calculated over the range of x > 0, so our limits will be from 0 to +∞.

Solve the integral

To solve the integral, we'll need to use integration by parts. Let: \[u = x\] \[dv = \frac{1}{1000}e^{-x/1000} dx\] Now, differentiate u and then integrate dv: \[du = dx\] \[v = -1000e^{-x/1000}\] By applying integration by parts formula: \(\int u dv = uv - \int v du\), we get: \[E(X) = -x\cdot1000e^{-x/1000} \Big|_0^{+\infty} + \int_0^{+\infty} 1000e^{-x/1000} dx\]

Solve the remaining integral

The remaining integral is much simpler to solve: \[\int_0^{+\infty} 1000e^{-x/1000} dx = -1000^2 e^{-x/1000}\Big|_0^{+\infty}\]

Evaluate the integrals and find the expected value

Now, we simply need to plug in the limits and find the value of the expected lifetime: \[E(X) = \Big[-x\cdot 1000e^{-x/1000} + 1000^2 e^{-x/1000}\Big] \Big|_0^{+\infty}\] Evaluate the limits: \[E(X) = \Big[-(\infty)\cdot0 + 0\Big] - \Big[-(0)\cdot 0 +1000^2(1)\Big] = 1000^2\]

Final Answer

The expected lifetime of such a light bulb is 1000 hours.

Key concepts

Random Variable

A random variable represents a numerical outcome of a random phenomenon, and it serves as a cornerstone concept in probability and statistics. In our light bulb example, the random variable, denoted by \(X\), signifies the hours of life of an electric light bulb. There are primarily two types of random variables: discrete and continuous. Discrete random variables take on a countable number of distinct values, such as the number of coins that show heads when flipping a set number of coins. On the other hand, a continuous random variable can take an infinite number of possible values within a given range, like the lifespan of our light bulb, which can be any positive real number.

When dealing with continuous random variables, we use probability density functions to describe the likelihood of the variable falling within a particular range, instead of calculating probabilities of exact values, which are zero for continuous ranges. By understanding the nature of the random variable in question, we can better approach problems concerning it, such as finding expected values or variances.

Density Function

The density function, often termed as the probability density function (PDF), gives us the probability of a continuous random variable lying within a certain interval. The density function \(f(x)\) associated with our light bulb's lifetime is \((1/1000)e^{-x/1000}\) for \(x > 0\), ensuring that the probabilities are non-negative and that the total probability across all possible values (from zero to infinity) sums up to one.

The density function is a powerful tool in calculus-based probability because it enables us to compute probabilities for intervals and expected values by integrating over the desired range. For instance, to find the probability that a light bulb lasts between 1000 and 2000 hours, one would integrate the given density function over that interval. Clearly explained density functions are key to solving many probabilistic problems.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation and is used to integrate products of functions. It is expressed as \(\textstyle\frac{d}{dx}u(x) \times v(x) = u(x) \times \frac{d}{dx}v(x) + v(x) \times \frac{d}{dx}u(x)\), which after rearranging and integrating, yields the formula \(\textstyle\frac{d}{dx}u \times v = uv - \textstyle\frac{d}{dx}v \times u\). In our example, we set \(u = x\) and \(dv = (1/1000)e^{-x/1000}dx\).

By differentiating \(u\) and integrating \(dv\), we arrive at \(du = dx\) and \(v = -1000e^{-x/1000}\) respectively. Applying the integration by parts formula makes complex integrals more manageable, as evident in our step-by-step solution. This technique often transforms an intractable integral into a simpler form, which is indispensable for solving many integrals, especially those involving exponential functions or products of polynomials and logarithms.

Continuous Random Variable

A continuous random variable differs from its discrete counterpart by taking an infinite number of potential values in a continuum. In the context of our problem, the lifetime of the light bulb denoted by \(X\) is a continuous random variable because it can take on any positive real value. It could be 1234.567 hours, or any other number within the range. This continuity requires us to calculate probabilities and expected values through integration rather than simple summations.

Recognizing a variable as continuous informs us that we should look to methods such as the probability density function (PDF) for analysis. Calculating expected values for continuous random variables, like in our example, involves integrating the product of the variable itself and its density function over its range, which usually spans from minus to plus infinity, or in this case from zero to plus infinity due to the lifespan of a light bulb being a positive quantity. Understanding the concept of a continuous random variable is critical as it guides the mathematical approach for probability calculations and statistical analyses.