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Chapter 3: Problem 78

According to the article "'Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt.s 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

Short Answer

Expert verified
Probability of exactly 3 intervals: 0.144; At most 3: 0.739. Probability of exceeding \( \mu + \sigma \) not directly computed but involves complement and cumulative distribution.

Step by step solution

01

Understand the Geometric Distribution

In this exercise, we are dealing with a geometric distribution. The geometric distribution models the number of trials until the first success in a sequence of independent and identically distributed Bernoulli trials. The probability mass function for a geometric distribution with parameter \( p \) is given by:\[ P(Y = k) = (1 - p)^{k-1} p \]where \( k \) is the number of intervals (or trials) needed to get the first success.
02

Calculate Probability of Drought Lasting Exactly 3 Intervals

Using the probability mass function of the geometric distribution, calculate the probability that the drought lasts exactly 3 intervals:\[ P(Y = 3) = (1 - 0.409)^{3-1} \cdot 0.409 \]\[ P(Y = 3) = (0.591)^{2} \cdot 0.409 \]Calculate this expression to find the probability.
03

Calculate Probability of Drought Lasting At Most 3 Intervals

The probability that the drought lasts at most 3 intervals is the cumulative probability for 1, 2, or 3 intervals:\[ P(Y \leq 3) = P(Y = 1) + P(Y = 2) + P(Y = 3) \]Calculate each as:\[ P(Y = 1) = (0.591)^{0} \cdot 0.409 = 0.409 \]\[ P(Y = 2) = (0.591)^{1} \cdot 0.409 \]Add the results.
04

Calculate Mean and Standard Deviation of the Geometric Distribution

For a geometric distribution, the mean \( \mu \) and standard deviation \( \sigma \) are calculated as follows:\[ \mu = \frac{1}{p} \]\[ \sigma = \sqrt{\frac{1-p}{p^2}} \]Compute these values using \( p = 0.409 \) to find \( \mu \) and \( \sigma \).
05

Probability of Drought Exceeding Mean by At Least One Standard Deviation

Calculate the value of \( \mu + \sigma\) using the computations from Step 4. Then find the probability that the drought length \( Y \) exceeds this value. This involves calculating the cumulative distribution function and using the complement rule to find \( P(Y > \mu + \sigma) \).

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) for the geometric distribution gives us the probability of the first occurrence of a "success" after a certain number of "failures".
In the context of this problem, each period in which the water supply is below the critical value (deficit) can be considered a "failure" and the first period with a surplus is the "success."
The formula for the PMF of the geometric distribution is:
  • \( P(Y = k) = (1 - p)^{k-1} p \)
where \( k \) is the number of periods (or trials) until the first success.
Here, \( p \) represents the probability of success (in our case, exiting the drought period). If we want to calculate the probability that a drought lasts exactly 3 intervals, we apply the formula using \( k = 3 \) and \( p = 0.409 \).
So, \( P(Y = 3) = (0.591)^2 \times 0.409 \), where \( 0.591 = 1 - 0.409 \).
It's crucial to understand how this function simplifies the prediction of drought intervals based on past data and statistical patterns.
Bernoulli Trials
In probability theory, a Bernoulli trial is a random experiment where there are only two possible outcomes, often termed "success" and "failure.鈥�
Each Bernoulli trial is independent of the others, meaning the outcome of one does not affect the others.
For the geometric distribution used in this exercise, each trial represents a single period that is analyzed to see if the drought condition persists or if the drought ends.
  • A "success" in this context means the drought ends (water supply is above a critical level).
  • A "failure" means the drought continues.
Bernoulli trials are fundamental to understanding not just geometric distributions, but also other probability distributions that model binary processes, such as the binomial distribution.
In this problem, each interval of water measurement is like flipping a coin where the outcome tells if the drought will persist or end. The geometric distribution then models how these "coin flips" forecast drought persistence.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) for a geometric distribution helps us determine the probability that a random variable is less than or equal to a certain value.
It aggregates the probabilities from the PMF for all values up to a certain point, providing a more comprehensive view of the probability over a range.
  • For example, to find the probability that a drought lasts at most 3 intervals (\( P(Y \leq 3) \)), we sum the probabilities \( P(Y = 1) + P(Y = 2) + P(Y = 3) \).
Calculating these probabilities individually using the PMF and then summing them gives the CDF.
This cumulative approach allows researchers to understand the likelihood of different drought lengths, offering insights into planning for water resource management during drought conditions.
Using the CDF is especially useful for questions that ask for probabilities over ranges rather than specific outcomes, as it quickly conveys aggregate probabilities.

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Most popular questions from this chapter

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then \(X\), the number of failures, has a Poisson distribution with \(\mu=1\). a. Obtain \(P(X \leq 5)\) by using Appendix Table A.2. b. Determine \(P(X=2)\) first from the pmf formula and then from Appendix Table A.2. c. Determine \(P(2 \leq X \leq 4)\). d. What is the probability that \(X\) exceeds its mean value by more than one standard deviation?

Customers at a gas station pay with a credit card (A), debit card \((B)\), or cash \((C)\). Assume that successive customers make independent choices, with \(P(A)=.5\), \(P(B)=.2\), and \(P(C)=.3\). a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\rho\) will have an infinite set of possible values? If yes, say why. If no, give an example.

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable \(Y\) as the number of ticketed passengers who actually show up for the flight. The probability mass function of \(Y\) appears in the accompanying table. \begin{tabular}{l|ccccccccccc} \(y\) & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\ \hline\(p(y)\) & \(.05\) & \(.10\) & \(.12\) & \(.14\) & \(.25\) & \(.17\) & \(.06\) & \(.05\) & \(.03\) & \(.02\) & \(.01\) \end{tabular} a. What is the probability that the flight will accommodate all ticketed passengers who show up? b. What is the probability that not all ticketed passengers who show up can be accommodated? c. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.
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