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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2.725 hours is a good model for rainfall duration (Urban Stormwater Management Planning with Analytical Probabilistic Models, 2000, p. 69). a. What is the probability that the duration of a particular rainfall event at this location is at least 2 hours? At most 3 hours? Between 2 and 3 hours? b. What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations? What is the probability that it is less than the mean value by more than one standard deviation?

Short Answer

Expert verified
a. At least 2 hrs: ~0.485; At most 3 hrs: ~0.656; Between 2 and 3 hrs: ~0.171. b. Exceeds by more than 2 std. dev.: ~0.048; Less than mean by more than 1 std. dev.: ~0.

Step by step solution

01

Understanding Exponential Distribution

The probability density function (PDF) for an exponential distribution is given by \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \), where \( \lambda = \frac{1}{\text{mean}} \). Here, the mean is 2.725 hours, so \( \lambda = \frac{1}{2.725} \approx 0.367 \).
02

Probability at Least 2 Hours

We need to find \( P(X \geq 2) \). This is calculated using the cumulative distribution function (CDF): \( P(X \geq 2) = 1 - P(X < 2) = 1 - (1 - e^{-\lambda \cdot 2}) = e^{-0.367 \cdot 2} \approx 0.485 \).
03

Probability at Most 3 Hours

Find \( P(X \leq 3) \) using the CDF: \( P(X \leq 3) = 1 - e^{-\lambda \cdot 3} = 1 - e^{-0.367 \cdot 3} \approx 0.656 \).
04

Probability Between 2 and 3 Hours

Find \( P(2 \leq X \leq 3) \): \( P(2 \leq X \leq 3) = P(X \leq 3) - P(X < 2) = (1 - e^{-0.367 \cdot 3}) - (1 - e^{-0.367 \cdot 2}) = e^{-0.367 \cdot 2} - e^{-0.367 \cdot 3} \approx 0.656 - 0.485 = 0.171 \).
05

Compute Standard Deviation

The standard deviation for an exponential distribution is the same as the mean, which is 2.725 hours.
06

Probability Exceeds Mean by More Than 2 Standard Deviations

Calculate \( P(X > 2.725 + 2 \times 2.725) = P(X > 8.175) \): \( P(X > 8.175) = e^{-\lambda \cdot 8.175} = e^{-0.367 \cdot 8.175} \approx 0.048 \).
07

Probability Less Than Mean by More Than One Standard Deviation

Calculate \( P(X < 2.725 - 2.725) = P(X < 0) \). Since exponential distributions are only defined for \( x \geq 0 \), this probability is 0. Hence, compute \( P(X < 0) \) possibly refers to more than one standard deviation as only possible in context internally. Need reevaluation or adjust standards to find \( P(X < 2.725) \approx 0.632 \).

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Key Concepts

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

Probability Density Function (PDF)
In the realm of probability and statistics, the Probability Density Function, often abbreviated as PDF, plays a crucial role in understanding distributions. The PDF provides the likelihood of a continuous random variable falling within a particular range of values. When dealing with exponential distributions, the PDF is defined by the formula: \[ f(x; \lambda) = \lambda e^{-\lambda x} \]where \( x \geq 0 \) and \( \lambda \) represents the rate parameter. This parameter is the reciprocal of the mean of the distribution, \( \lambda = \frac{1}{\text{mean}} \).
In practical scenarios, like rainfall duration at Toronto Pearson International Airport, the mean is set at 2.725 hours, which sets \( \lambda \) to approximately 0.367. The PDF helps model how likely it is for the rain to last a specific duration, providing a foundation for calculating probabilities of different events using this exponential distribution model.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function, known as CDF, is an integral tool that helps compute the probability that a random variable \( X \) is less than or equal to a certain value \( x \). For an exponential distribution, the CDF is represented as: \[ F(x; \lambda) = 1 - e^{-\lambda x} \]Utilizing the CDF, you can easily determine probabilities over intervals. For example:
  • The probability that rainfall lasts at most 3 hours at Toronto Pearson is given by \( P(X \leq 3) = 1 - e^{-0.367 \times 3} \).
  • Alternatively, the chance that the rain lasts at least 2 hours is computed using \( 1 - P(X < 2) = e^{-0.367 \times 2} \).
The CDF thus simplifies calculating probabilities of different events, enabling more straightforward decision-making under uncertainty.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion of a set of values. In the context of exponential distributions, a unique characteristic is that the standard deviation is equal to the mean. So here, the standard deviation of the exponential distribution describing rainfall duration at the airport is 2.725 hours.
This feature branches from the definition of the exponential distribution, which is naturally linked to its mean. Analyzing how values deviate using the standard deviation can illuminate how likely it is for occurrences to stray from the average duration, assisting in better forecasting and planning.
Mean
The mean of a distribution is the average, a central value that summarizes the bulk of the data points. For the exponential distribution, the mean is a pivotal parameter since it not only acts as the average duration but also determines the rate parameter \( \lambda \) of the PDF. For example, in analyzing rainfall duration, if the mean is 2.725 hours, it implies that rain on average lasts for this duration.
The relationship between the mean and \( \lambda \) is given by \( \lambda = \frac{1}{\text{mean}} \). This average helps predict that about 63.2% of the rainfall events will last less than or equal to the mean duration. Such insights assist in harnessing analytical models for predicting weather patterns and preparing accordingly.

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