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Chapter 11: Problem 31

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?

Short Answer

Expert verified
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 by step solution

01

Understanding the Distribution

The exponential distribution of the age at which the fish dies is given by \[f(t) = \begin{cases}\lambda e^{-\lambda x} & \text{for } t \geq 0 \ 0 & \text{otherwise}\end{cases}\]
02

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.
03

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: \[P(10 \leq X < 20) = \int_{10}^{20} \lambda e^{-\lambda x} dx\]\[= \left[-e^{-\lambda x}\right]_{10}^{20} = -(e^{-20\lambda} - e^{-10\lambda})\].\[= e^{-10\lambda} - e^{-20\lambda}\]
04

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,\[ \lambda = \frac{\ln 3}{10}\]
05

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 \[1 - e^{-\frac{\ln 3}{2}}\].
06

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.

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Key Concepts

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

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 \text{ } \ \ \text{otherwise}\text{otherwise} \text{}\text{}\text{}\text{} \ \ \text{} & \text{} \text{} & \text{}\text{} & \text{} \ \ \text{}\t \ \ \text{} \ \ \text{} \ \ \text{} \ \

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Most popular questions from this chapter

In Exercises 1 and 2, the outcomes and corresponding probability assignments for a discrete random variable \(X\) are listed. Draw the histogram for \(X\). Then find the expected value \(E(X)\), the variance \(\operatorname{Var}(X)\), and the standard deviation \(\sigma(X)\). $$ \begin{array}{l|c|c|c|c|c} \hline \text { Outcomes for } X & 0 & 2 & 4 & 6 & 8 \\ \hline \text { Probability } & \frac{1}{8} & \frac{3}{8} & \frac{1}{4} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$

TIME MANAGEMENT A shuttle tram arrives at a tram stop at a randomly selected time \(X\) within a 1-hour period, and a tourist independently arrives at the same stop also at a randomly selected time \(Y\) within the same hour. The tourist has the patience to wait for the tram for up to 20 minutes before calling a taxi. The joint probability density function for \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}1 & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 & \text { otherwise }\end{cases} $$ a. What is the probability that the tram takes longer than 20 minutes to arrive? b. What is the probability that the tourist arrives after the tram? c. What is the probability that the tourist connects with the tram? [Hint: The event of this occurring has the form $$ Y+a \leq X \leq Y+b $$ for suitable numbers \(a\) and \(b\).]

TRAFFIC CONTROL Suppose the time (in minutes) between the arrivals of successive cars at a toll booth is measured by the random variable \(X\) with probability density function $$ f(t)= \begin{cases}0.5 e^{-0.5 t} & \text { if } t \geq 0 \\ 0 & \text { otherwise }\end{cases} $$ a. Find the probability that a randomly selected pair of successive cars will arrive at the toll booth at least 6 minutes apart. b. Find the average time between the arrivals of successive cars at the toll booth.

If the random variable \(X\) is normally distributed with mean \(\mu=7\) and standard deviation \(\sigma=2\), what is \(P(X \geq 9)\) ?

QUALITY CONTROL An automobile manufacturer claims that its new cars get an average of 30 miles per gallon in city driving. Assume the manufacturer's claim is correct and that gas mileage is normally distributed, with standard deviation of 2 miles per gallon. a. Find the probability that a randomly selected car will get less than 25 miles per gallon. b. If you test two cars, what is the probability that both get less than 25 miles per gallon?
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