Problem 31 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

FISHERY MANAGEMENT Brooke, the manager of a fishery, determines that the age \(X\) (in weeks) at which a certain species of fish dies follows an exponential distribution with probability density function $$ f(t)= \begin{cases}\lambda e^{-\lambda x} & \text { for } t \geq 0 \\ 0 & \text { otherwise }\end{cases} $$ Brooke observes that it is twice as likely for a randomly selected fish to die during the first 10 -week period as during the next 10 weeks (from week 10 to week 20 ). a. What is \(\lambda\) ? b. What is the probability that a randomly chosen fish will die within the first 5 weeks? c. How long should Brooke expect a randomly selected fish to live?

Step-by-Step Solution

Answer

The value of \(\lambda\) is \(\frac{\ln 3}{10}\). The probability of a fish dying within the first 5 weeks is \(1 - e^{-\frac{\ln 3}{2}}\). The expected lifespan is \(\frac{10}{\ln 3}\) weeks.

Step 1: Understanding the Distribution

The exponential distribution of the age at which the fish dies is given by \[f(t) = \begin{cases}\lambda e^{-\lambda x} & \text{for } t \geq 0 \ 0 & \text{otherwise}\end{cases}\]

Step 1: Find \lambda\

It is given that the probability that a fish dies in the first 10 weeks is twice the probability it dies in the next 10 weeks.Let's set up the probability equations:\[P(0 \leq X < 10) = \int_0^{10} \lambda e^{-\lambda x} dx\]\[P(10 \leq X < 20) = \int_{10}^{20} \lambda e^{-\lambda x} dx\].Since \(P(0 \leq X < 10) = 2 \, P(10 \leq X < 20)\), we find both integrals and relate them.

Step 2: Solve the Integrals

First integral: \[P(0 \leq X < 10) = \int_0^{10} \lambda e^{-\lambda x} dx\]Using the antiderivative: \[= \left[-e^{-\lambda x}\right]_0^{10} = -(e^{-10\lambda} - 1) = 1 - e^{-10\lambda}\].Second integral: \[P(10 \leq X < 20) = \int_{10}^{20} \lambda e^{-\lambda x} dx\]\[= \left[-e^{-\lambda x}\right]_{10}^{20} = -(e^{-20\lambda} - e^{-10\lambda})\].\[= e^{-10\lambda} - e^{-20\lambda}\]

Step 3: Equate Probabilities

Given: \(P(0 \leq X < 10) = 2 \, P(10 \leq X < 20)\) We have:\[1 - e^{-10\lambda} = 2 (e^{-10\lambda} - e^{-20\lambda})\].Re-organize and solve for \(\lambda\): \(1 - e^{-10\lambda} = 2 e^{-10\lambda} - 2 e^{-20\lambda}\).\[1 + e^{-10\lambda} = 2 e^{-10\lambda} - 2 e^{-20\lambda}\]\[1 = e^{-10\lambda} - 2 e^{-20\lambda}\]\[1 = 3e^{-10\lambda}\]\[e^{-10\lambda} = \frac{1}{3}\]\[ -10\lambda = \ln(\frac{1}{3}) = -\ln(3)\]So,\[ \lambda = \frac{\ln 3}{10}\]

Step 4: Find Probability of Fish Dying Within 5 Weeks

Now calculate the probability for the first 5 weeks:\[P(0 \leq X < 5) = \int_0^5 \lambda e^{-\lambda x} dx\]Use the previously found \(\lambda\):\[\int_0^{5} \frac{\ln 3}{10} e^{-\frac{\ln 3}{10} x} dx\]\[= \left[-e^{-\frac{\ln 3}{10} x}\right]_0^5 = -(e^{-\frac{\ln 3}{2}} - 1)\]\[= 1 - e^{-\frac{\ln 3}{2}}\]Thus, the probability is \[1 - e^{-\frac{\ln 3}{2}}\].

Step 5: Calculate Expected Lifespan

The expected value for an exponential distribution is given by \(E(X) = \frac{1}{\lambda}\).Thus,\[E(X) = \frac{1}{\frac{\ln 3}{10}} = \frac{10}{\ln 3}\].So, the expected lifespan of a fish is \(\frac{10}{\ln 3}\) weeks.

Key Concepts

Exponential Distribution

The exponential distribution is a probability distribution often used in reliability engineering and fishery management. It helps describe the time until an event occurs, such as the age at which a fish dies. It's characterized by its probability density function (PDF): \( f(t) = \begin{cases}\begin{array}{ll}\text{for } t \underline{\phantom{xxx}} \ \ \text{otherwise}\text{otherwise} \text{}\text{}\text{}\text{} \ \ \text{} & \text{} \text{} & \text{}\text{} & \text{} \ \ \text{}\t \ \ \text{} \ \ \text{} \ \ \text{} \ \

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