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The probability density function of the time customers arrive at a terminal (in minutes after 8.00 A.M.) is \(f(x)=\) \(e^{-x / 10} / 10\) for \(0

Short Answer

Expert verified
(a) \(\approx 0.9975\); (b) \(\approx 0.2231\); (c) Calculate using binomial probability; (d) From CDF, \(\approx 0.2231\).

Step by step solution

01

Understanding the Probability Density Function

The given probability density function is \( f(x) = \frac{e^{-x / 10}}{10} \). This function describes an exponential distribution where \( x \) is the time in minutes after 8:00 A.M. The parameter \( \lambda \) for the exponential distribution is \( \lambda = \frac{1}{10} \).
02

Calculate the Probability (a)

(a) To find the probability that the first customer arrives by 9:00 A.M., we need to calculate \( P(X \leq 60) \) since 9:00 A.M. is 60 minutes after 8:00 A.M. The probability of an event in an exponential distribution is given by the cumulative distribution function (CDF): \[ F(x) = 1 - e^{-x/10} \]Apply this formula: \( F(60) = 1 - e^{-60/10} = 1 - e^{-6} \approx 0.9975 \).
03

Calculate the Probability (b)

(b) For a customer arriving between 8:15 A.M. and 8:30 A.M., find \( P(15 < X < 30) \). This is calculated as:\[ P(15 < X < 30) = P(X < 30) - P(X < 15) = (1 - e^{-30/10}) - (1 - e^{-15/10}) \]\( = e^{-1.5} - e^{-3} \approx 0.2231 \).
04

Calculate the Probability (c)

(c) We need the probability that two or more customers arrive by 8:40 A.M., i.e., \( P(X < 40) \). Since the time is 40 minutes, we calculate the CDF for a single customer as \( F(40) = 1 - e^{-4} \). For five customers, this is a binomial problem:Let \( p = F(40) \). The probability of 2 or more customers arriving is:\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]For binomial \( P(X = k) = \binom{5}{k} p^k (1-p)^{5-k} \), Calculate:\( P(X = 0) = (1-p)^5 \) and \( P(X = 1) = 5p(1-p)^4 \). Subtract from 1 to find the probability of two or more arrivals.
05

Determine the Cumulative Distribution Function (d)

The cumulative distribution function (CDF) for this exponential distribution is:\[ F(x) = 1 - e^{-x/10} \]Now, using the CDF, calculate the probability for the first customer arriving between 8:15 A.M. and 8:30 A.M., which we already calculated in step (b) using:\( P(15 < X < 30) = F(30) - F(15) \approx 0.2231 \).

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Key Concepts

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

Probability Density Function
The concept of a probability density function, often abbreviated as PDF, is essential to understanding continuous probability distributions. For an exponential distribution like in this exercise, the PDF describes the likelihood of a particular event occurring at a specific point in time. Here, the PDF is given by the function \( f(x) = \frac{e^{-x / 10}}{10} \). This indicates that the variable \( x \) represents time in minutes after 8 a.m., and the parameter \( \lambda = \frac{1}{10} \) governs the exponential nature of the distribution.

In exponential distributions, the parameter \( \lambda \) determines the rate at which events occur. A larger \( \lambda \) implies more frequent events. The probability density function itself does not give us probabilities directly, but it helps us to understand how probabilities are determined over continuous intervals of time.

In practice, we use the PDF to derive another important function called the cumulative distribution function (CDF), which is crucial for calculating the probabilities over intervals.
Cumulative Distribution Function
To understand or calculate probabilities for a range of values in a continuous distribution, like the arrival times in this exercise, we use the cumulative distribution function (CDF). The CDF provides the probability that our random variable \( x \) will take a value less than or equal to a certain value. For the exponential distribution defined here, the CDF is:

\[ F(x) = 1 - e^{-x/10} \]

This function emerges by integrating the probability density function over the range of interest. It simplifies the process of finding probabilities for intervals and is particularly useful for computations. For instance, to find out when the first customer arrives before 9:00 A.M., which is 60 minutes after 8:00 A.M., we plug in \( x = 60 \) in the CDF as shown:
  • \( F(60) = 1 - e^{-60/10} = 1 - e^{-6} \approx 0.9975 \)
So, there's a 99.75% chance of a customer's arrival before 9:00 A.M.
Probability Calculation
Probability calculations is using the CDF to calculate the likelihood of certain events. In this scenario, we aim to compute probabilities between time intervals. To find the probability that a customer arrives between 8:15 and 8:30 A.M. (15 to 30 minutes after 8:00 A.M.), we can use:
  • \( P(15 < X < 30) = F(30) - F(15) \)
Where,
  • \( F(30) = 1 - e^{-3} \)
  • \( F(15) = 1 - e^{-1.5} \)
The probability that a customer's arrival falls within this interval is then:
  • \( P(15 < X < 30) = e^{-1.5} - e^{-3} \approx 0.2231 \)
Understanding how to perform probability calculation using combined CDFs aids in determining the chance of independent arrivals, such as calculating the probability that two or more customers arrive before 8:40 A.M. from a group of five. This step often requires an understanding of the binomial distribution, where cases like 2 or more out of 5 are considered by summing respective probabilities that result from CDF estimations for each scenario.

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