Problem 30 - Calculus Concepts

Question

Lead-210 Decay Lead 210 decays at a continuous rate of 0.1163 grams per year. If \(x\) is the time, in hours, for one of the lead 210 atoms to decay, the probability density function for \(x\) is given by \(L(x)=0.1163 e^{-0.1163 x}\) for \(x>0 .\) a. Calculate the mean time for one of the lead 210 atoms to decay. b. Calculate the standard deviation of decay times. c. What is the probability that a lead 210 atom will decay between 5 and 9 hours from now?

Step-by-Step Solution

Answer

Mean time to decay is 8.6 hours, standard deviation is 0.344 hours, and probability of decay between 5 and 9 hours is 0.1685.

Step 1: Understanding the Mean of a Continuous Random Variable

For a continuous random variable with probability density function (PDF) \(L(x)\), the mean (expected value) is computed using the integral:\[ E(X) = \int_{0}^{\infty} x \cdot f(x) \, dx \].Here, \(f(x) = 0.1163 e^{-0.1163 x}\).

Step 2: Find the Mean Time

We can find the mean by evaluating the integral:\[ E(X) = \int_{0}^{\infty} x \cdot 0.1163 e^{-0.1163 x} \, dx \].Using integration by parts, let \(u = x\) and \(dv = 0.1163 e^{-0.1163 x} \, dx\). Then, \(du = dx\) and \(v = -e^{-0.1163 x}\).Applying integration by parts:\[ \int u \, dv = uv - \int v \, du \].After solving, we get:\[ E(X) = \frac{1}{0.1163} = 8.6 \text{ hours} \].

Step 3: Standard Deviation Calculation

The standard deviation \(\sigma\) is the square root of the variance. First, compute the variance by finding the second moment:\[ E(X^2) = \int_{0}^{\infty} x^2 \cdot 0.1163 e^{-0.1163 x} \, dx \].After evaluating using integration by parts twice, find the variance:\( \text{Var}(X) = E(X^2) - (E(X))^2 \).Where \(E(X) = 8.6\), we get \( \text{Var}(X) = 74.0784 - 73.96 = 0.1184 \) .Therefore, the standard deviation is \( \sigma = \sqrt{0.1184} \approx 0.344 \text{ hours} \).

Step 4: Probability of Decay between 5 and 9 Hours

To find the probability that decay occurs between 5 and 9 hours, calculate the definite integral:\[ P(5 < X < 9) = \int_{5}^{9} 0.1163 e^{-0.1163 x} \, dx \].Solving this integral, we evaluate:\[ P(5 < X < 9) = \left[-e^{-0.1163 x}\right]_{5}^{9} = e^{-0.1163 \times 5} - e^{-0.1163 \times 9}\].Calculate the result: \( P(5 < X < 9) \approx 0.1685 \).

Key Concepts

Probability Density Function

A Probability Density Function (PDF) provides us with information about how probabilities are distributed over the values of a random variable. In simpler terms, if you want to know how likely a certain outcome is within a set of possible outcomes, the PDF is your friend. In the context of exponential decay, we often encounter PDFs that describe the time it takes for a particular process to complete, like the decay of lead-210.
For exponential decay, the PDF is often given by a function like \( L(x) = ext{constant} imes e^{- ext{constant} imes x} \). Here, the 'constant' often denotes the rate at which decay occurs, and 'x' is the random variable, such as time. In our lead-210 decay problem, the function is \( L(x) = 0.1163 e^{-0.1163 x} \), highlighting that decay is not uniform over time, but instead decreases exponentially.
It's essential to understand that even though the PDF gives us probabilities over a continuous range, the actual value of the PDF at any given point is not probability but rather probability per unit. Integrating the PDF over a specific interval gives us the probability that the random variable falls within that range.

Mean and Standard Deviation

When looking at a continuous random variable, the mean provides us the weighted average value where the variable is expected to 'center' around. In terms of exponential decay, you usually want to know the expected time until an atom decays. The mean \(E(X)\) of a function \(f(x)\) is found by integrating the product of \(x\) (the random variable) and \(f(x)\) over all possible values of \(x\). This process, called integration by parts, simplifies evaluating the integral.
In our case, with \( f(x) = 0.1163 e^{-0.1163 x} \), we end up with a mean decay time of 8.6 hours after doing the math.

• The formula used is \( E(X) = \int_{0}^{\infty} x \cdot 0.1163 e^{-0.1163 x} \, dx \).

Standard deviation \(\sigma\), on the other hand, provides a measure of how spread out the decay times are around the mean. We calculate the variance first, which is the expected value of the squared deviations from the mean. For our lead decay problem, the variance is approximately 0.1184, leading to a standard deviation of about 0.344 hours.

• The formula for variance is \( \text{Var}(X) = E(X^2) - (E(X))^2 \), followed by \( \sigma = \sqrt{\text{Var}(X)} \).

Integration by Parts

Integration by parts is a technique stemming from the product rule of differentiation. It helps us solve integrals where the product of two functions is involved. When you face a complex integral like the ones you see with PDFs, this technique can simplify your work.
The basic formula is \( \int u \, dv = uv - \int v \, du \), where \(u\) and \(dv\) are components of your integral. For solving PDF-related integrals, like finding the mean or variance for exponential decay scenarios, integration by parts turns daunting terms into manageable calculations.
In exponential decay problems, you commonly select \(u = x\) and \(dv = \text{function of } x\) so that differentiation/simplifications bring forth easier integrations. For instance, in solving for the mean, set \(u = x\) and \(dv = 0.1163 e^{-0.1163 x} \, dx\), resulting in the equation after differentiation and integration steps as \(uv - \int v \, du\).

• This method is vital in reducing complex integrals into simpler terms and is extensively used in calculus to manage PDF calculations efficiently.