/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Let \(X=\) the time between two ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X=\) the time between two successive arrivals at the drive-up window of a local bank. If \(X\) has an exponential distribution with \(\lambda=1\) (which is identical to a standard gamma distribution with \(\alpha=1\) ), compute the following: a. The expected time between two successive arrivals b. The standard deviation of the time between successive arrivals c. \(P(X \leq 4)\) d. \(P(2 \leq X \leq 5)\)

Short Answer

Expert verified
a) 1, b) 1, c) 0.9817, d) 0.1286

Step by step solution

01

Identify the Distribution Parameters

Given that \(X\) follows an exponential distribution, the rate parameter \(\lambda = 1\). This also means that the shape parameter for the equivalent gamma distribution is \(\alpha = 1\).
02

Calculate the Expected Time

For an exponential distribution with rate \(\lambda\), the expected value (mean) is given by the formula \(E(X) = \frac{1}{\lambda}\). With \(\lambda = 1\), it follows that \(E(X) = \frac{1}{1} = 1\).
03

Calculate the Standard Deviation

The standard deviation of an exponential distribution is given by \(\text{SD}(X) = \frac{1}{\lambda}\). Using \(\lambda = 1\), it follows that \(\text{SD}(X) = \frac{1}{1} = 1\).
04

Calculate the Probability \(P(X \leq 4)\)

The cumulative distribution function (CDF) for the exponential distribution is \(F(x) = 1 - e^{-\lambda x}\). For \(P(X \leq 4)\), calculate \(1 - e^{-1 \times 4} = 1 - e^{-4}\). Evaluating this gives approximately \(1 - 0.0183 = 0.9817\).
05

Calculate the Probability \(P(2 \leq X \leq 5)\)

The probability that \(X\) falls within a range is given by \(P(a \leq X \leq b) = F(b) - F(a)\), where \(F(x) = 1 - e^{-\lambda x}\). First, compute \(F(5) = 1 - e^{-5}\) and \(F(2) = 1 - e^{-2}\). Thus, \(P(2 \leq X \leq 5) = (1 - e^{-5}) - (1 - e^{-2})\). This simplifies to \(e^{-2} - e^{-5}\), which evaluates approximately to \(0.1353 - 0.0067 = 0.1286\).

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Key Concepts

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

Probability Calculations
In probability theory, calculating probabilities involving random variables is a crucial skill. For the exponential distribution, the focus is often on intervals and specific values. Here, the probability of variable outcomes can tell us how likely certain events are to occur. In this exercise, the random variable \(X\) represents the time between arrivals at a bank's drive-thru window.
One of the tasks is to find the probability \(P(X \leq 4)\). This is a cumulative probability, which calculates the likelihood that \(X\) will be less than or equal to 4 minutes. Similarly, calculating \(P(2 \leq X \leq 5)\) gives the probability that \(X\) falls between 2 and 5 minutes.
The process involves the cumulative distribution function (CDF), which helps sum up the probabilities across intervals. Thus, probability calculations using the exponential distribution involve identifying intervals and utilizing the right functions.
Expected Value
Expected value is an essential concept in probability and statistics. It provides the mean or average value of a random variable in the long run. For an exponential distribution, the expected value is simply the reciprocal of the rate parameter \(\lambda\).
So if \(\lambda = 1\), as given in this scenario, the expected value \(E(X)\) is calculated as follows:
  • Formula: \(E(X) = \frac{1}{\lambda}\)
  • Calculation: \(E(X) = \frac{1}{1} = 1\)
This indicates that on average, the time between two successive arrivals is 1 minute. It's important to recognize that while individual intervals may vary, over a very large number of intervals, the average interval will converge to this expected value.
The expected value not only serves as a measure of central tendency but also aids in predicting future outcomes based on past distributions.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of an exponential distribution, it quantifies how much the time between arrivals varies around the mean expected value.
The standard deviation for an exponential distribution is calculated using the formula: \(\text{SD}(X) = \frac{1}{\lambda}\). With a rate \(\lambda = 1\), the calculation becomes straightforward:
  • Formula: \(\text{SD}(X) = \frac{1}{\lambda}\)
  • Calculation: \(\text{SD}(X) = \frac{1}{1} = 1\)
Thus, this indicates that the variation or spread of time between arrivals around the expected 1 minute is also 1 minute. Understanding standard deviation helps in assessing the reliability and consistency of the intervals.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a key tool in probability to determine the probability that a random variable is less than or equal to a specific value. For the exponential distribution, the CDF is given by the formula \(F(x) = 1 - e^{-\lambda x}\).
In this exercise, the CDF helps compute probabilities for time intervals. For example:
  • Calculating \(P(X \leq 4)\): Use \(F(x) = 1 - e^{-4}\) which evaluates to approximately 0.9817.
  • For \(P(2 \leq X \leq 5)\), calculate \(F(5) - F(2)\), which results in \(e^{-2} - e^{-5}\), or roughly 0.1286.
These functions effectively show the likelihood of \(X\) being within certain bounds. The CDF smoothly integrates the probability density function (PDF), making it easier to assess cumulative probabilities over intervals. Understanding and using the CDF effectively can significantly aid in solving real-world probability problems.

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Most popular questions from this chapter

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